Classiﬁcation of some groups of order pqr Adam Burley

Classification of some groups of order pqr

Adam Burley

Contents

Preliminaries 50.1. Results from [1] 50.2. Other Results 5

Chapter 1. Solvability results 71.1. Subgroups of quotient groups 71.2. Facts about derived subgroups 91.3. Abelian series 131.4. An important condition for solvability 141.5. Groups of squarefree order 15

Chapter 2. Hall Theory 172.1. π-numbers 172.2. Hall’s Theorems 192.3. Elementary Abelian p-groups 192.4. Products of subgroups and The Frattini Argument 202.5. An important special case of the Hall Theory 252.6. The Remaining Case 27

Chapter 3. The Extension Classification Schema 333.1. Isomorphism Classes 333.2. Semidirect Products 343.3. The Schema itself 423.4. Proof of Theorem 3.3.3(i) and Theorem 3.3.3(ii) 453.5. Proof of Theorem 3.3.3(iii) 46

Chapter 4. Some groups of order pqr 514.1. Classification of automorphisms of the Cyclic Group 514.2. The case where Y is cyclic of prime order 524.3. Classification of groups of order pq 534.4. The case where X is cyclic of squarefree order 604.5. Classification of groups of order pqr where q - r − 1 734.6. Examples of the classification 774.7. Conclusion 79

Index 81

Bibliography 83

Appendix A. Solvability without the transfer 85

3

4 CONTENTS

A.1. The full Feit-Thompson Theorem 85A.2. Solvability of groups of order pqr, without the transfer 85A.3. Solvability of groups of order p2q 87

Preliminaries

0.1. Results from [1]

The following is the main result proved in [1]. There, it is numbered asCorollary 1.5, and proved in section 7.

Lemma 0.1.1. There are no non-Abelian simple groups of order less than200.

Removing the last 3 sentences of the proof of Theorem 6.2 in [1], we obtainthe following.

Theorem 0.1.2. Let G be a group, G not a p-group, and suppose that Ghas a Sylow p-subgroup P (of order ps, say) with P 6 Z(NG(P )). Then G hasa normal subgroup of index ps.

The following is the core argument of the proof of Corollary 7.1 in [1]. Itcan be extracted from the proof of this, by taking only paragraphs 2 and 3, andthe first 4 sentences of paragraph 5, and making very minor modifications tothe wording.

Lemma 0.1.3. Let G be a group and P a Sylow p-subgroup. If there existsno prime q such that q > p and q | |Aut(P )|, then P 6 Z(NG(P )).

Combining the above two lemmas, we obtain:

Theorem 0.1.4. Let G be a group, G not a p-group, and suppose that Ghas a Sylow p-subgroup P (of order ps, say), such that there exists no prime qwith q > p and q | |Aut(P )|. Then G has a normal subgroup of index ps.

The following is Proposition 3.2 of [1].

Lemma 0.1.5. Let G be a group of order p, where p is a prime. Then

|Aut(G)| = p− 1.

0.2. Other Results

The following is from elementary Group Theory.

Lemma 0.2.1. If X is a G-set, then the map

ρ : G 7→ Sym(X); ρ(g)(x) = gx

is a homomorphism.

5

CHAPTER 1

Solvability results

1.1. Subgroups of quotient groups

The following lemma establishes the form of subgroups of a quotient group.

Lemma 1.1.1. Let G be a finite group and N a normal subgroup. IfK 6 G/N then K = H/N for some H 6 G with N E H.

Proof. Let G be a finite group and N a normal subgroup. Also, letK 6 G/N . Let H = {g ∈ G : gN ∈ K}. Then if x ∈ K then x ∈ G/N , sox = gN for some g ∈ G, and so this g ∈ H. Therefore, x = gN ∈ H/N , sowe have shown K ⊆ H/N . If hN ∈ H/N , then this h ∈ H, so hN ∈ K. SoH/N ⊆ K, and so K = H/N .

Next we show that H 6 G. Clearly we have H ⊆ G. Now take h1, h2 ∈ H.Then h1N,h2N ∈ K, so (h1h2)N = (h1N)(h2N) ∈ K, so h1h2 ∈ H. Also ifh ∈ H then hN ∈ K, so h−1N = (hN)−1 ∈ K, so h−1 ∈ H. Therefore, H 6 G.

Now, if n ∈ N then nN = N = idG/N ∈ K, so n ∈ H (as n ∈ G andnN ∈ K). So N ⊆ H. Now if n ∈ N , h ∈ H then h ∈ G so nhn−1 ∈ N (asN E G). So we have shown that N E H. �

We can establish a stronger result if we have a normal subgroup of thequotient group. The following is the key step.

Lemma 1.1.2. Let G be a finite group with a subgroup H and a normalsubgroup N . Suppose H/N 6 G/N . Then H E G.

Proof. Assume H/N E G/N . So for all gN ∈ G/N (or in other words,for all g ∈ G), we have (gN)(H/N)(gN)−1 = H/N , so

{(gN)(hN)(gN)−1 : h ∈ H} = {hN : h1 ∈ H} for all g ∈ G

so{(ghg−1)N : h ∈ H} = {hN : h1 ∈ H} for all g ∈ G

so for all g ∈ G, h ∈ H, we have that (ghg−1)N = h1N for some h1 ∈ H, sothat (ghg−1) = h1, so ghg−1 ∈ H. Therefore, we have proved that H E G. �

This lemma corresponds to Lemma 1.1.1, in the case where we have a normalsubgroup. The proof is immediate from our previous two lemmas, but we giveit in any case.

Lemma 1.1.3. Let G be a finite group and N a normal subgroup. IfK E G/N then K = H/N for some H E G with N E H.

7

8 1. SOLVABILITY RESULTS

Proof. Let G be a finite group and N a normal subgroup. If K E G/Nthen K = H/N for some H 6 G with N E H, by Lemma 1.1.1. But sinceK = H/N we have H/N E G/N , and we already have that N E G and H 6 G.So we can apply Lemma 1.1.2 to obtain that H E G. �

We may use this powerful lemma to prove an important result about max-imal normal subgroups.

Lemma 1.1.4. Suppose G is a finite group and N is a maximal proper nor-mal subgroup of G. Then G/N is simple.

Proof. Suppose G is a finite group and N is a maximal proper normal sub-group of G. Suppose for a contradiction that G/N were not simple. Then thereexists a proper non-trivial normal subgroup of G/N , say K. By Lemma 1.1.3,K has the form H/N for some H E G, where also N E H. However H is aproper subgroup of G, since if H = G then |H| = |G| so

|H/N | = |H||N |

=|G||N |

= |G/N |

and soH/N = G/N , a contradiction asH/N is proper. SinceN is the maximalproper normal subgroup of G, we must have N = H. But then

|H/N | = |H||N |

= 1

so H/N is a trivial subgroup, a contradiction. �

Finally, we shall prove a lemma concerning quotients of quotient groups.

Lemma 1.1.5. Suppose G is a finite group, and N and H are finite withN E H, H 6 G, N E G, and H/N E G/N (so that H E G by Lemma 1.1.2).Then

(G/N)/(H/N) ∼= G/H.

Proof. Suppose G is a finite group, and N and H are finite with N E H,H 6 G, N E G, and H/N E G/N (so that H E G by Lemma 1.1.2). Take thefunction defined by:

φ : (G/N)/(H/N) 7→ G/H; (gN)(H/N) 7→ gH

We must check that this function is well-defined. There are two thingswhich could change: first of all, we could take a different representative of thecoset gN , or in other words take a g1 ∈ gN and consider φ((g1N)(H/N)). Thiswould equal g1H. But since g1 = gn1 for some n1 ∈ N , and N ⊆ H, we havethat g1 = gn1 for n1 ∈ H. So we have

φ((g1N)(H/N)) = g1H = gH = φ((gN)(H/N))

and the function is well-defined in that sense.The other way is if we took a different representative for the coset

(gN)(H/N), or in other words take (g2N) ∈ (gN)(H/N) and considerφ((g2N)(H/N)). This equals g2H. However g2N = (gN)(hN) = (gh)N for

1.2. FACTS ABOUT DERIVED SUBGROUPS 9

some hN ∈ H/N (so h ∈ H), and so g2 = ghn2 for some n2 ∈ N . But n2 ∈ Has N ⊆ H, so hn2 ∈ H, and so

φ((g2N)(H/N)) = g2H = gH = φ((gN)(H/N))

and the function is well-defined in that sense too.Since

|(G/N)/(H/N)| = |G/N ||H/N |

=|G||N ||H||N |

=|G||H|

= |G/H|

the domain and codomain of φ have the same size, so to show it is a bijectionwe need only check it is onto. But this is clear, as for any gH ∈ G/H, theelement (gN)(H/N) maps onto gH. So φ is a bijection.

We shall check that φ is an homomorphism. Take

(g1N)(H/N), (g2N)(H/N) ∈ (G/N)/(H/N); .

Then:

φ((g1N)(H/N))φ((g2N)(H/N)) = (g1H)(g2H)

= (g1g2)H

= φ(((g1g2)N)(H/N))

= φ(((g1N)(g2N))(H/N))

= φ(((g1N)(H/N))((g2N)(H/N)))

so that φ is a homomorphism. So φ is an isomorphism. �

1.2. Facts about derived subgroups

In this section we shall prove some lemmas about derived subgroups. Recallthat the derived subgroup is the subgroup generated by all elements xyx−1y−1.The derived series is the sequence G,G′, G′′, .... If this series terminates, thegroup is said to be solvable. For a more precise account, see my other docu-ment [1].

These elements xyx−1y−1 are called commutators. Throughout this sectionwe shall denote [x, y] = xyx−1y−1. Note that the inverse of a commutator is acommutator.

Lemma 1.2.1. Let G be a group and x, y ∈ G. Then [x, y]−1 = [y, x].

Proof.

[x, y]−1 = (xyx−1y−1)−1 = yxy−1x−1 = [y, x]

�

First of all, a general picture of what elements of G′ look like is very easyto derive.

Lemma 1.2.2. Let G be a group and x ∈ G′. Then there exist c1, ..., cn withx = c1c2 · · · cn, such that for all 0 ≤ i ≤ n, ci = [ui, vi], where ui, vi ∈ G.


Proof. If x ∈ G′ then x = ar11 · · · arm

m by definition. Let c1, ..., cn be ob-tained by regarding ri = 0 as the element idG, ri < 0 as a product(ai)−1 · · · (ai)−1, and ri > 0 as a product ai · · · ai. Note that the elementsobtained in this method are all commutators, as ai

−1 is a commutator fromLemma 1.2.1, and idG = [idG, idG]. So x = c1 · · · cn where for all 0 ≤ i ≤ n, ciis a commutator, or in other words, ci = [ui, vi] where ui, vi ∈ G. �

We shall now prove, in three main parts, that the derived subgroup is thesmallest normal subgroup N such that G/N is Abelian. The first of these partsis split into two lemmas, as we shall need a stronger result than simply G′ E Glater on.

Lemma 1.2.3. Let G be a finite group and N E G. Then N ′ E G.

Proof. First notice that if u, v ∈ N , and g ∈ G, then

g([u, v])g−1 = g(uvu−1v−1)g−1

= g(ug−1gvg−1gu−1g−1gv−1)g−1

= (gug−1)(gvg−1)(gug−1)−1(gvg−1)−1

= [gug−1, gvg−1]

which is itself a commutator (since N is normal, gug−1 and gvg−1 are in N),so g([u, v])g−1 ∈ N ′. Now, if x ∈ N ′ then x = c1 · · · cn, where the ci are allcommutators, by Lemma 1.2.2. Now:

gxg−1 = g(c1 · · · cn)g−1

= g(c1g−1gc2g−1 · · · gcn−1g

−1gcn)g−1

= (gc1g−1)(gc2g−1) · · · (gcn−1g−1)(gcng−1)

but the last term is a product of commutators, so gxg−1 ∈ N ′, and the resultis proved. �

Lemma 1.2.4. Let G be a group. Then G′ E G.

Proof. We know G E G, so we apply Lemma 1.2.3 with N replaced by G,to obtain the result. �

Lemma 1.2.5. If G is a group, then G/G′ is Abelian.

Proof. Let (xG′) and (yG′) be arbitrary elements of G/G′. Then[x, y] ∈ G′, as x, y ∈ G. So [x, y]G′ = G′, and:

(xG′)(yG′)(xG′)−1(yG′)−1 = (xG′)(yG′)(x−1G′)(y−1G′)

= (xyx−1y−1)G′

= [x, y]G′

= G′

= idG/G′

so that(xG′)(yG′) = (yG′)(xG′)

as required. �

1.2. FACTS ABOUT DERIVED SUBGROUPS 11

Lemma 1.2.6. If N is a normal subgroup of a group G such that G/N isAbelian, then G′ 6 N .

Proof. Let x, y ∈ G, then [x, y] ∈ G′. Now,

[x, y]N = (xyx−1y−1)N

= (xN)(yN)(x−1N)(y−1N)

= (xN)(yN)(xN)−1(yN)−1

= (xN)(xN)−1(yN)(yN)−1

= idG/N

= N

so that [x, y] ∈ N . Since N contains all commutators, it contains all productsof commutators, and therefore G′ 6 N . �

Finally, we present some results on derived subgroups of subgroups andof isomorphic groups. The first of these states that taking derived subgroupspreserves the subgroup relation.

Lemma 1.2.7. If G and H are groups with G 6 H, then G′ 6 H ′.

Proof. Suppose x ∈ G′. Then, by Lemma 1.2.2

x = c1 · · · cnwhere the ci = [ui, vi] for ui, vi ∈ G. So ui, vi ∈ H, and so

ci = uiviu−1i v−1

i ∈ H ′

So we must havex = c1 · · · cn ∈ H ′

which gives the result. �

We may extend this result by induction.

Lemma 1.2.8. If G and H are groups with G 6 H, then G(r) 6 H(r) for allintegers r ≥ 0.

Proof. Let G and H be groups with G 6 H. We shall prove the statement“G(r) 6 H(r)” by induction, for all integers r ≥ 0. First take the case r = 0.Then the statement reads G(0) 6 H(0), i.e. G 6 H, which is one of ourhypotheses. Now suppose the statement is true for r = k. Then G(k) 6 H(k)

by the inductive hypothesis, and we can apply Lemma 1.2.7 to obtain(G(k))′ 6 (H(k))′. But this means G(k+1) 6 H(k+1), so the statement is true forr = k + 1. Therefore, by the principle of induction, the statement holds for allintegers r ≥ 0. �

This enables us to get a result on solvability.

Lemma 1.2.9. If G is a solvable group and H 6 G, then H is solvable.

Proof. Let G be a solvable group and H 6 G. Then there exists Nsuch that G(N) = {idG}. Therefore H(N) 6 G(N) = {idG}, by Lemma 1.2.8.Therefore, H(N) = {idG}, and so H is solvable. �


Another, similar result on solvability may be obtained in a similar fashion.

Lemma 1.2.10. If G and H are groups such that G ∼= H, then G′ ∼= H ′.

Proof. Let G and H be groups such that G ∼= H. Let φ : G 7→ H be anisomorphism. We shall show that φ(G′) = H ′. Therefore, the restriction of φto G′ is an onto map from G′ to H ′. We already know that this restriction is a1-1 homomorphism, so this will give G′ ∼= H ′ as required.

(⊆) First suppose that c ∈ G′ is a commutator. Then c = xyx−1y−1 forsome x, y ∈ G. Now

φ(c) = φ(x)φ(y)φ(x−1)φ(y−1) = φ(x)φ(y)φ(x)−1φ(y)−1 ∈ H ′

as φ(x), φ(y) ∈ H.Now suppose that g ∈ G′ is any element. Then, by Lemma 1.2.2,

g = c1c2 · · · cnwhere c1, ..., cn ∈ G′ are commutators. Now

φ(g) = φ(c1)φ(c2) · · ·φ(cn) ∈ H ′

and so φ(G′) ⊆ H ′.(⊇) First suppose that c ∈ H ′ is a commutator. Then c = xyx−1y−1 for

some x, y ∈ H. Since φ is onto, there exist x1, y1 ∈ G such that φ(x1) = x andφ(y1) = y. Let c1 = x1y1x

−11 y−1

1 . Then c1 is a commutator, c1 ∈ G′ and

φ(c1) = φ(x1)φ(y1)φ(x−11 )φ(y−1

1 )

= φ(x1)φ(y1)φ(x1)−1φ(y1)−1

= xyx−1y−1

= c.

Now suppose that h ∈ H ′ is any element. Then, by Lemma 1.2.2,

h = c1c2 · · · cnwhere c1, c2, ..., cn ∈ H ′ are commutators. Now there exist commutatorsd1, d2, ..., dn ∈ G′ such that φ(di) = ci for each i with 1 ≤ i ≤ n. Letg = d1d2 · · · dn ∈ G′. Then φ(g) ∈ φ(G′), and

φ(g) = φ(d1)φ(d2) · · ·φ(dn) = c1c2 · · · cn = h.

Thus H ′ ⊆ φ(G′), i.e. φ(G′) ⊇ H ′. �

Lemma 1.2.11. If G and H are groups such that G ∼= H, then G(r) ∼= H(r)

for all integers r ≥ 0.

Proof. Let G and H be groups such that G ∼= H. We shall prove thestatement “G(r) ∼= H(r)” by induction, for all integers r ≥ 0. First take thecase r = 0. Then the statement reads G(0) ∼= H(0), i.e. G ∼= H, which isone of our hypotheses. Now suppose the statement is true for r = k. ThenG(k) ∼= H(k) by the inductive hypothesis, and we can apply Lemma 1.2.10 toobtain (G(k))′ ∼= (H(k))′. But this means G(k+1) ∼= H(k+1), so the statement istrue for r = k+1. Therefore, by the principle of induction, the statement holdsfor all integers r ≥ 0. �

1.3. ABELIAN SERIES 13

Lemma 1.2.12. If G is a solvable group and H is a group such that G ∼= H,then H is solvable.

Proof. Let G be a solvable group and H be a group such that G ∼= H.Since G is solvable, there exists N such that G(N) = {idG}. Therefore,|G(N)| = 1. By Lemma 1.2.11, G(N) ∼= H(N), and so |H(N)| = 1. Therefore,H(N) = {idH}, so H is solvable. �

1.3. Abelian series

We met solvability briefly in [1]. However, here it will be much more im-portant. We start by providing a new equivalent definition of solvability. Thisstarts with the definition of a normal series.

Definition 1.3.1. Let G be a group. A normal series of subgroups forG is a sequence

H0,H1, ...,Hn

of subgroups of G such that H0 = {idG}, Hn = G, and Hi E Hi+1 for all0 ≤ i < n. n is the length of the normal series of subgroups.

Definition 1.3.2. Let G be a group. A normal series for G is a sequence

G0, G1, ..., Gn

of groups such that there exists a normal series of subgroups

H0,H1, ...,Hn

for G with Gi∼= Hi for all 0 ≤ i ≤ n. n is the length of the normal series.

Now we define the idea of an Abelian series, which leads directly to our newdefinition of solvability.

Definition 1.3.3. Let G be a group. An Abelian series of subgroupsfor G is a normal series of subgroups

H0,H1, ...,Hn

for G such that Hi+1/Hi is Abelian for all 0 ≤ i < n. n is the length of theAbelian series of subgroups.

Definition 1.3.4. LetG be a group. An Abelian series forG is a sequence

G0, G1, ..., Gn

of groups such that there exists an Abelian series of subgroups

H0,H1, ...,Hn

for G with Gi∼= Hi for all 0 ≤ i ≤ n. n is the length of the Abelian series.

While our definitions of “normal series of subgroups” and “Abelian seriesof subgroups” are non-standard, it is clear that any normal series of subgroupsis a normal series, and any Abelian series of subgroups is an Abelian series.

Our new definition of solvability is “having an Abelian series”.

Proposition 1.3.5. Let G be a group. Then G is solvable iff it has anAbelian series.


Proof. ( =⇒ ) If G is a solvable group then there exists some N withG(N) = {idG}. Let n be the smallest suchN . Then letGn = G andGr = G(n−r)

for all 0 ≤ r < n. By construction we have G0 = G(n) = {idG} and Gn = G.By Lemma 1.2.4, Gi E Gi+1. Also, for all 0 ≤ i < n:

Gi+1/Gi = Gi+1/(Gi+1)′

which is Abelian by Lemma 1.2.5.( ⇐= ) If G has an Abelian series G0, G1, ..., Gn, then there is an Abelian

series of subgroups H0,H1, ...,Hn. Now Hn/Hn−1 is Abelian (by Lemma 1.2.6),so G′ = (Hn)′ 6 Hn−1. Next note that Hn−1/Hn−2 is Abelian, so again byLemma 1.2.6, (Hn−1)′ 6 Hn−2. But by Lemma 1.2.7, we have that

G′′ 6 (Hn−1)′ 6 Hn−2.

Continuing in this way, we obtain that G(r) 6 Hn−r for all 0 < r ≤ n. Inparticular, G(n) 6 H0 = {idG}, so G(n) = {idG}, and so G is solvable. �

1.4. An important condition for solvability

The new definition of solvability we have presented makes the followingresult immediate.

Lemma 1.4.1. If N is a solvable normal subgroup of a group G, and G/Nis Abelian, then G is solvable.

Proof. If N is a solvable normal subgroup of a group G then N has anAbelian series, and therefore an Abelian series of subgroups, say N0, N1, ..., Nr.Let n = r + 1, and let Gi = Ni for all 0 ≤ i < n, and Gn = G. ThenG0 = {idN} = {idG} also. Furthermore, Ni E Ni+1 for all 0 ≤ i < n − 1 = r,and Ni = N E G = Gi+1 for i = n − 1. Finally, Ni+1/Ni is Abelian for all0 ≤ i < n− 1 = r, and Ni+1/Ni = G/N is Abelian for i = n− 1. So G0, ..., Gn

is an Abelian series for G, and hence G is solvable. �

The apparent weakness of this result is deceptive; it may in fact be usedin conjunction with induction to prove a much stronger result. We shall firstrequire an easy lemma.

Lemma 1.4.2. Suppose G is a finite Abelian group and H is a finite group,with G ∼= H. Then H is Abelian.

Proof. Suppose G is a finite Abelian group and H is a finite group, withG ∼= H. Let

φ : G 7→ H

be an isomorphism.Now take any x, y ∈ H. There are x1, y1 ∈ G with φ(x1) = x, φ(y1) = y (as

φ is onto). Now,

xy = φ(x1)φ(y1) = φ(x1y1) = φ(y1x1) = φ(y1)φ(x1) = yx.

So H is Abelian. �

Now the stronger version of Proposition 1.4.1 can be proved.

1.5. GROUPS OF SQUAREFREE ORDER 15

Proposition 1.4.3. If N is a solvable normal subgroup of a finite group G,and G/N is solvable, then G is solvable.

Proof. Let G be a finite group. We shall use induction to prove the state-ment “If N E G, N is solvable and G/N has an Abelian series of length k, thenG is solvable” for all k.

First of all consider the case k = 0. So suppose that N E G, N is solvableand G/N has an Abelian series of length 0. So G/N has an Abelian series ofsubgroups of length 0. This series contains exactly one element. This element isthe first in the series and so must equal {idG/N}, but is also the last in the seriesso must equal G/N . So G/N = {idG/N} and G/N is then Abelian. Therefore,G is solvable by Lemma 1.4.1.

Now assume truth for k = m and suppose that N E G, N is solvable andG/N has an Abelian series of length m + 1. Then G/N has an Abelian seriesof subgroups of length m+ 1, say K0, ...,Km,Km+1. By Lemma 1.1.3, we havethat Km = H/N , where N E H and H E G. Now H/N = Km is clearlysolvable (as it has an Abelian series K0,K1, ...,Km), and N is solvable, so H issolvable (as we have assumed truth for k = m).

We have that (G/N)/(H/N) = Km+1/Km is Abelian. By Lemma 1.1.5,(G/N)/(H/N) ∼= G/H. By Lemma 1.4.2, G/H is Abelian. So we have H E G,with H solvable and G/H is Abelian. Therefore, G is solvable by Lemma 1.4.1,and the statement is true for k = m+ 1.

By the principle of induction, we have truth for all k, namely, the originalstatement of the proposition. �

1.5. Groups of squarefree order

The aim of this section is to show that all groups of squarefree order aresolvable. The following is a cut down version of Corollary 7.1 from [1].

Proposition 1.5.1. Let G be a group of order p1 · · · pr, where p1, ..., pr areprimes. Then G has a normal subgroup of index p1.

Proof. Let G be a group of order p1 · · · pr, where p1, ..., pr are primes.Let P be a Sylow p1-subgroup, which clearly has order p1. We have that|Aut(P )| = p1− 1, by Lemma 0.1.5. Suppose there were a prime q with q > p1

and q | |Aut(P )|. Then q < |Aut(P )| = p1−1 < p1, a contradiction. Therefore,there is no such q, and so, by Theorem 0.1.4 (and G is clearly not a p-group),G has a normal subgroup of index p1. �

We may now prove that all groups of squarefree order are solvable.

Corollary 1.5.2. Let G be a group of order p1 · · · pr, where p1 < ... < pr

are primes. Then G is solvable.

Proof. Suppose for a contradiction that the statement of the result is false,and let r be the smallest positive integer such that there exists a non-solvablegroup of order p1 · · · pr (p1 < ... < pr primes). Let G be such a group. ByProposition 1.5.1, G has a normal subgroup of index p1, say N . Now N hasorder p2 · · · pr and hence is solvable, otherwise r would not be the smallestpositive integer satisfying its defining property, as r − 1 would also satisfy it.


In addition, G/N is Abelian (having prime order p1) and therefore solvable. SoG is solvable by Lemma 1.4.3. This is a contradiction as G was assumed not tobe solvable. Therefore, the statement of the result is true. �

CHAPTER 2

Hall Theory

2.1. π-numbers

The concept of a π-number is the foundation of the Hall Theory.

Definition 2.1.1. Let π be a finite set of primes. A π-number is either 1,or an integer divisible by only those primes in π. A π′-number is either 1, oran integer divisible by none of the primes in π.

This leads to the idea of a π-group, a generalisation of the p-group, and theHall subgroup, a generalisation of the Sylow subgroup.

Definition 2.1.2. A π-group is a group G such that |G| is a π-number. Aπ′-group is a group G such that |G| is a π′-number.

Definition 2.1.3. Let G be a finite group and π a finite set of primes. Aπ-subgroup of G is a subgroup H of G such that H is a π-group. A π′-subgroupof G is a subgroup H of G such that H is a π′-group.

Definition 2.1.4. Let G be a finite group and π a finite set of primes. AHall π-subgroup of G is a π-subgroup H such that [G : H] is a π′-number. AHall π′-subgroup of G is a π′-subgroup H such that [G : H] is a π-number.

It can be seen that this is a generalisation of the Sylow subgroup.

Lemma 2.1.5. Let G be a finite group and π a finite set of primes. Thena Hall π-subgroup of G is a maximal π-subgroup, and a Hall π′-subgroup is amaximal π′-subgroup.

Proof. We shall show the π case; the π′ case is similar. Suppose G is afinite group, π a finite set of primes, and H a Hall π-subgroup of G. Supposefor a contradiction that there existed a π-subgroup K with |K| > |H|. Thereare two cases. Firstly, we could have that there was some prime p dividing|K| such that p does not divide |H|. However then p ∈ π, as p divides |K|,a π-number. Now p divides |G| = |H|[G : H], but p does not divide |H|, sop | [G : H]. However this is a contradiction as p ∈ π, and [G : H] is meant tobe a π′-number.

The second case is ifH andK are divisible by all the same primes. Then, wemust have at least one prime p which divides |K| to a higher power, say s, thanit divides |H| to (say r). We must have p ∈ π, as p divides |K|, a π-number.Now ps−r does not divide p−r|H|, as p divides |H| to the rth power only. Also,ps divides |G| = |H|[G : H], so ps−r divides p−r|G| = (p−r|H|)[G : H]. Thusps−r divides [G : H], and as s > r, we have p | [G : H]. However this is acontradiction as p ∈ π, and [G : H] is meant to be a π′-number. �

17

18 2. HALL THEORY

Then, the following lemma, allowing us to calculate the orders ofHall π-subgroups, is immediate.

Lemma 2.1.6. Let G be a finite group and π a finite set of primes. Supposep1, ..., pr is an enumeration of the primes in π, and q1, ...., qs are the primesdividing |G| which are not in π. Suppose

|G| = pa11 p

a22 · · · par

r qb11 q

b22 · · · qbs

s .

Then a Hall π-subgroup of G has order pa11 p

a22 · · · par

r , and a Hall π′-subgrouphas order qb1

1 qb22 · · · qbs

s .

We present some examples.

Examples 2.1.7.

(1) A Hall {5, 7}-subgroup of a group of order 4725 = 33527 would haveorder 527 = 175.

(2) A Hall {3}-subgroup of a group of order 4725 = 33527 would haveorder 33 = 27. This is the same order as a Sylow 3-subgroup wouldhave. In fact, it is easy to see that a Hall {p}-subgroup is the same asa Sylow p-subgroup.

(3) A Hall {11, 13}-subgroup of a group of order 4725 = 33527 would haveorder 1.

Finally, the intersection of a π-subgroup and a π′-subgroup (whether Hallor not) must be trivial.

Proposition 2.1.8. Let G be a group with subgroups A and B, thenA∩B 6 A and A∩B 6 B. Furthermore, if A E G and B E G then A∩B E G.

Proof. Let G be a group with subgroups A and B. We have that A ∩ Bis non-empty, as both A and B contain the identity of G. It is also clearly asubset of G (as e.g. A ∩B ⊆ A ⊆ G).

Now take x, y ∈ A ∩B. Then x ∈ A and y ∈ A so xy ∈ A. Also x ∈ B andy ∈ B so xy ∈ B. Therefore, xy ∈ A ∩B.

Take any x ∈ A ∩ B. Then x ∈ A so x−1 ∈ A, and x ∈ B so x−1 ∈ B.Therefore, x−1 ∈ A ∩B.

Therefore, A ∩B is a subgroup of G, hence a group. Since A ∩B ⊆ A andA ∩B ⊆ B, we have that A ∩B 6 A and A ∩B 6 B.

Suppose A E G and B E G. Take n ∈ A ∩ B, and g ∈ G. Then n ∈ Aso gng−1 ∈ A, and n ∈ B so gng−1 ∈ B. Therefore, gng−1 ∈ A ∩ B, and soA ∩B E G. �

Corollary 2.1.9. Let G be a group, let A be a π-subgroup and B aπ′-subgroup. Then A ∩B is trivial.

Proof. Let G be a group, let A be a π-subgroup and B a π′-subgroup.Then A ∩ B 6 A and A ∩ B 6 B, by Proposition 2.1.8. Therefore, |A ∩ B|divides both |A| and |B|, hence is both a π-number and a π′-number. Therefore,it must be 1, and so A ∩B is trivial. �

2.3. ELEMENTARY ABELIAN p-GROUPS 19

2.2. Hall’s Theorems

The object of this chapter is to prove the following theorems, known asHall’s First, Second and Third Theorems. The first and second of these areessentially generalisations of Sylow’s First and Second Theorems.

Theorem 2.2.1. Let G be a finite solvable group, and π be a set of primes.(1) G has a Hall π-subgroup.(2) Any two Hall π-subgroups of G are conjugate.(3) Any π-subgroup is contained within a Hall π-subgroup.

In fact, for π-groups or π′-groups, the theorems are easy.

Lemma 2.2.2. Let G be a π-group or a π′-group. Then Theorem 2.2.1 holdsfor G.

Proof. Suppose first that G is a π-group. Then G is a π-subgroup of itself,and therefore any Hall π-subgroup of G must have order |G|, by Lemma 2.1.5.Hence, G is the only Hall π-subgroup of G. Clearly, Hall’s First Theorem holdsthen. Also, G is conjugate to itself by the identity. Therefore, Hall’s SecondTheorem holds. Finally, Hall’s Third Theorem holds (as any π-subgroup of Gis contained in G, a Hall π-subgroup)

Suppose now that G is a π′-group. Then the order of any subgroup ofG must be a π′-number, as it must divide the order of G. Therefore, everysubgroup of G is a π′-subgroup, and so any π-subgroup of G must also be aπ′-subgroup. The order of such a subgroup must be both a π and a π′-number,in other words, it must be 1. Therefore, H = {idG} is the only π-subgroup, andit is a Hall π-subgroup. Thus, Hall’s First Theorem is proved. H is conjugateto itself by the identity and it is the only Hall π-subgroup, giving Hall’s SecondTheorem. Finally, any π-subgroup must equal H, therefore it is contained inH, which gives Hall’s Third Theorem. �

2.3. Elementary Abelian p-groups

The following definition is perhaps at first interesting as it is the exact caseto which we can generalise the arguments in Section 4 of [1] to. However thiswill not be the purpose we will require it for.

Definition 2.3.1. Let p be a prime. An elementary Abelian p-group is anAbelian p-group in which all non-identity elements have order p.

We have a simpler condition for an elementary Abelian p-group.

Lemma 2.3.2. Let p be a prime, and G an Abelian p-group such that for allx ∈ G, we have xp = idG. Then G is an elementary Abelian p-group.

Proof. Let p be a prime, and G an Abelian p-group such that for all x ∈ G,we have xp = idG. Let x be in G. Then the order of x divides the order of G,a power of p, therefore the order of x is 1, p, p2, ... However as xp = idG, thenthe order of x is 1 or p. If x is a non-identity element then its order is not 1, soit must be p. �

20 2. HALL THEORY

The following lemma states that (non-trivial) normal subgroups of smallestorder are elementary Abelian.

Lemma 2.3.3. Let G be a solvable group and A a non-trivial normal subgroupof G of smallest order. Then A is an elementary Abelian p-group, for someprime p.

Proof. Let G be a solvable group and A a non-trivial normal subgroupof G of smallest order. A is solvable by Lemma 1.2.9. Therefore, A′ 6= A(so |A′| < |A|). If A′ is non-trivial, then A′ is a normal subgroup of G (byLemma 1.2.3), non-trivial, and has smaller order than A. This contradicts theminimality of A. Therefore, A′ must be trivial, i.e. A is Abelian.

Choose any prime p such that p | |A|. Let

B = {x ∈ A : xp = idG}Then B is non-trivial, as by Cauchy’s group theorem, A contains an element oforder p; this element is not the identity, and is in B.

We shall show that B E G. So take b ∈ B, g ∈ G. Then b ∈ A, sogbg−1 ∈ A, as A E G. Also

(gbg−1)p = gbpg−1 = gidGg−1 = idG

(noting for the first equality that the inner g−1g terms all cancel) so thatgbg−1 ∈ B. Therefore, B E G.

So B is a non-trivial normal subgroup of G, which is a subgroup of A.Therefore, we must have A = B, or else B would contradict the minimality ofA. Suppose A were not a p-group. Then there exists some other prime q dividingthe order of A. So by Cauchy’s group theorem, there is an element a of A whichhas order q. Since ap = idG, we must have q | p, a contradiction. Therefore,A must be a p-group, and so we have all we need to apply Lemma 2.3.2, andobtain that A is an elementary Abelian p-group. �

2.4. Products of subgroups and The Frattini Argument

We can take the product of two subgroups simply by taking the productsof their elements.

Definition 2.4.1. Let G be a group with subgroups H and K. We define

HK = {hk : h ∈ H, k ∈ K}

While this product will not be a group in general, it will be a group if oneof the subgroups is normal.

Proposition 2.4.2. Let G be a group with subgroups H and K, and eitherH E G or K E G. Then HK is a group.

Proof. Let G be a group with subgroups H and K. The set HK is non-empty as it contains idG = idGidG, as idG ∈ H, idG ∈ K. It is clearly a subsetof G.

Firstly suppose H E G. To show closure under the binary operation, takeh1k1, h2k2 ∈ HK, and let h3 = k1h2k

−11 ∈ H. Then:

h1k1h2k2 = h1k1h2k−11 k1k2 = h1h3k1k2 ∈ HK

2.4. PRODUCTS OF SUBGROUPS AND THE FRATTINI ARGUMENT 21

so the set is closed under the binary operation. To show closure under inverses,take hk ∈ HK and let h1 = k−1h−1k = k−1h−1(k−1)−1 ∈ H. Then:

(hk)−1 = k−1h−1 = k−1h−1kk−1 = h1k−1 ∈ HK

so the set is closed under inverses. Therefore, HK is a group.Secondly suppose K E G. To show closure under the binary operation, take

h1k1, h2k2 ∈ HK, and let k3 = h−12 k1h2 = h−1

2 k1(h−12 )−1 ∈ K. Then:

h1k1h2k2 = h1h2h−12 k1h2k2 = h1h2k3k2 ∈ HK

so the set is closed under the binary operation. To show closure under inverses,take hk ∈ HK and let k1 = hk−1h−1 ∈ K. Then:

(hk)−1 = k−1h−1 = h−1hk−1h−1 = h−1k1 ∈ HK

so the set is closed under inverses. Therefore, HK is a group. �

We can compute the order of this set.

Proposition 2.4.3. Let G be a group with subgroups H and K. Then:

|HK| = |H||K||H ∩K|

Proof. Let G be a group with subgroups H and K. Let S = H × K,T = HK and A = H ∩K. A acts on S by

a ∗ (h, k) = (ha−1, ak)

and if a ∗ (h, k) = (h, k) then a = idA, so the stabilizer always consists onlyof the identity. Therefore, by the orbit-stabilizer theorem, every orbit has size|A|.

Consider the mapφ : S 7→ T ; (h, k) 7→ hk

which is onto, since if t ∈ T , then t = hk for some h ∈ H, k ∈ K, so thatφ(h, k) = t.

We shall show that φ is 1-1 within its orbits: Suppose φ(s1) = φ(s2) wheres1 = h1k1 and s2 = h2k2. Then h1k1 = h2k2 so that h−1

2 h1 = k2k−11 . The

left-hand side is in H and the right is in K, so both are in H ∩K = A (beingequal). Let a = h−1

2 h1 = k2k−11 ∈ A. Then

a ∗ s1 = a(h1, k1) = (h1a−1, ak1) = (h1h

−11 h2, k2k

−11 k1) = (h2, k2) = s2

and so s1 and s2 lie in the same orbit.Converseley, suppose that s1 = (h, k) and s2 lie in the same orbit. Then

s2 = a ∗ s1 for some a ∈ A. Now

φ(s1) = hk = φ(ha, a−1k) = φ(a ∗ s1) = φ(s2)

and taking this together with the previous argument gives that φ(s1) = φ(s2)if and only if s1 and s2 lie in the same orbit.

Enumerate the elements of T = HK by {t1, ..., tn} where n = |T |. LetSi = φ−1(ti), for 1 ≤ i ≤ n. Then the Si are disjoint, and S is the union ofthem (as each s must map to a ti and it cannot map to more than one).

22 2. HALL THEORY

Take any si ∈ Si, and notice that

s ∈ Si ⇐⇒ φ(s) = ti ⇐⇒ φ(s) = φ(si) ⇐⇒ s ∈ Orb(si)

and so each Si = Orb(si), so |Si| = |A| for each i.Now we get

|H| · |K| = |S|= |S1|+ |S2|+ ...+ |Sn|= |A|+ |A|+ ...+ |A|= n|A|= |HK| · |H ∩K|

and so

|HK| = |H| · |K||H ∩K|

as required. �

The following chain of lemmas lead to the computation of the derived sub-group of such a product, under some (rather restrictive) conditions.

Lemma 2.4.4. Let G be a group with subgroups H and K. Then(HK)/K = H/K.

Proof. Let G be a group with subgroups H and K. Take x ∈ H/K. Thenx = hK for some h ∈ H, and so

x = (h idG)K = (h idK)K ∈ (HK)/K

so H/K ⊆ (HK)/K.Now take x ∈ (HK)/K. Then x = (hk)K = h(kK) for some h ∈ H, k ∈ K.

Clearly kK ⊆ K (as any element of kK is a product of two elements of K). Butalso K ⊆ kK, since if y ∈ K then y = k(k−1x) ∈ kK. Therefore, kK = K, andx = h(kK) = hK ∈ H/K. So (HK)/K ⊆ H/K, and the result follows. �

Lemma 2.4.5. If G is a group with subgroups H and K. Then(HK/K)′ = (H ′K)/K.

Proof. Let G be a group with subgroups H and K. By Lemma 2.4.4,(HK/K)′ = (H/K)′ and (H ′K)/K = H ′/K. It remains to show that(H/K)′ = H ′/K

(⊆) Let x ∈ (H/K)′ be a commutator. Then x = [u, v] for some u, v ∈ H/K.We have u = h1K and v = h2K for some h1, h2 ∈ H. Then

x = [h1K,h2K]

= (h1K)(h2K)(h1K)−1(h2K)−1

= (h1h2h−11 h−1

2 )K

= [h1, h2]K ∈ H ′/K.

Now let x ∈ (H/K)′ be any element. Then x = c1c2 · · · cn forc1, c2, ..., cn ∈ H/K all commutators (by Lemma 1.2.2). Each ci ∈ H ′/K, sox ∈ H ′/K. Therefore, (H/K)′ ⊆ H ′/K.

2.4. PRODUCTS OF SUBGROUPS AND THE FRATTINI ARGUMENT 23

(⊇) Let x ∈ H ′/K be such that x = hK where h = [u, v] ∈ H ′ is acommutator (where u, v ∈ H). Then

x = [u, v]K

= (uvu−1v−1)K

= (uK)(vK)(uK)−1(vK)−1

= [uK, vK] ∈ (H/K)′

Now let x ∈ H ′/K be any element. Then x = hK for some h ∈ H ′. Then,by Lemma 1.2.2, h = c1 · · · cn for c1, ..., cn ∈ H all commutators. Then eachciK ∈ (H/K)′, and

x = (c1c2 · · · cn)K = (c1K)(c2K) · · · (cnK) ∈ (H/K)′

so that (H/K)′ ⊇ H ′/K.Therefore, (H/K)′ = H ′/K, as required. �

Lemma 2.4.6. Suppose G is a group. Let H and K be subgroups of G, suchthat K is a normal subgroup of G of smallest order. Suppose further that ifN1, N2 are normal subgroups of HK then |N1| cannot divide |(HK)/N2|. Then(HK)′ = H ′K.

Proof. Suppose G is a group such that if N1, N2 are normal subgroupsof G then |N1| cannot divide |G/N2|. Let H and K be subgroups of G, suchthat K is a normal subgroup of G of smallest order. We wish to show that(HK)′ = H ′K.

HK and H ′K are groups by Proposition 2.4.2. If k ∈ K and g ∈ HK theng = hk1 for some h ∈ H and k1 ∈ K, and so

(hk1)k(hk1)−1 = h(k1kk−11 )h−1 ∈ K

as K E G. Thus K E HK. Therefore, K E H ′K, as H ′K ⊆ HK (sinceH ′ ⊆ H).

If n ∈ H ′K and g ∈ HK then n = h1k1 and g = hk for some h1 ∈ H ′,h ∈ H, k1, k ∈ K. Let

h2 = hh1h−1 ∈ H ′

k2 = hkh−1 ∈ Kk3 = hk1k

−1h−1 ∈ Kk4 = h−1

2 k2h2 ∈ K

24 2. HALL THEORY

because H ′ E H and K E HK (note H ⊆ HK so e.g. h ∈ HK for the secondequality here). We have:

gng−1 = hkh1k1k−1h−1

= hkh−1hh1k1k−1h−1

= k2hh1k1k−1h−1

= k2hh1h−1hk1k

−1h−1

= k2h2k3

= h2h−12 k2h2k3

= h2k4k3 ∈ H ′K

and so H ′K E HK.(⊆) We have that

(HK)/(H ′K) ∼= ((HK)/K)/((H ′K)/K) = ((HK)/K)/(((HK)/K)′)

the isomorphism being by Lemma 1.1.5, and the equality by Lemma 2.4.5. Thelast term is Abelian by Lemma 1.2.5, and hence the first term is Abelian byLemma 1.4.2. Therefore, by Lemma 1.2.6, (HK)′ ⊆ H ′K.

(⊇) Suppose K 6⊆ (HK)′. Then K ∩ (HK)′ 6= K, and clearlyK ∩ (HK)′ ⊆ K, so |K ∩ (HK)′| < |K|. But K E (HK) and (HK)′ E (HK),so K ∩ (HK)′ E (HK), by Proposition 2.1.8. Therefore, K ∩ (HK)′ must betrivial, or it would contradict the minimality of K.

Define the map

φ : (HK)′ 7→ (HK)/K;x 7→ xK

which is a homomorphism since if x, y ∈ (HK)′ then

φ(x)φ(y) = (xK)(yK) = (xy)K = φ(xy).

Also we have

ker(φ) = {x ∈ (HK)′ : xK = K}= {x ∈ (HK)′ : x ∈ K}= (HK)′ ∩K= {idG}.

Therefore, by the First Isomorphism Theorem,

(HK)′ ∼= (HK)′/ker(φ) ∼= im(φ)

and so|(HK)′| = | im(φ)| | |(HK)/K|.

However (HK)′ E HK and K E HK, so this is a contradiction. Therefore wemust have K ⊆ (HK)′.

As H 6 (HK)′ then H ′ 6 (HK)′. If x ∈ H ′K then x = hk for someh ∈ H ′, k ∈ K. Thus h ∈ (HK)′, k ∈ (HK)′, so x = hk ∈ (HK)′, and so(HK)′ ⊇ H ′K. �

2.5. AN IMPORTANT SPECIAL CASE OF THE HALL THEORY 25

The following is known as The Frattini Argument. Since we will use aslightly modified version of it later, we shall “wrap up” the core of the argumentin an initial lemma, with less restrictive conditions.

Lemma 2.4.7. If G is a finite group, K is a normal subgroup of G and P isa subgroup of K, such that for all g ∈ G, gPg−1 is conjugate to P in K, thenG = K NG(P ).

Proof. Let G be a finite group, K be a normal subgroup of G and Pbe a subgroup of K, such that for all g ∈ G, gPg−1 is conjugate to P in K.Let g ∈ G. So for some k ∈ K we have kPk−1 = gPg−1. Let x = k−1g,then xPx−1 = (k−1g)P (g−1k) = k−1(kPk−1)k = P , and so x ∈ NG P . Butg = kx, so g ∈ K NG(P ). Therefore G ⊆ K NG(P ). However it is clear thatK NG(P ) ⊆ G, so the result follows. �

Proposition 2.4.8 (The Frattini Argument). If G is a finite group, K is anormal subgroup of G and P is a Sylow p-subgroup of K, then G = K NG(P ).

Proof. Let G be a finite group, K be a normal subgroup of G and P be aSylow p-subgroup of K. Take any g ∈ G, then P and gPg−1 are conjugate inK, by Sylow’s Second Theorem. The result follows by Lemma 2.4.7. �

We state a useful lemma which sometimes works in unison with the FrattiniArgument.

Lemma 2.4.9. If G is a finite group and H a subgroup, thenNG(H) = H NG(H).

Proof. Let G be a finite group and H a subgroup. Clearly

NG(H) ⊆ H NG(H).

Now if h1 ∈ H,h2 ∈ NG(H), then

(h1h2)H(h1h2)−1 = h1h2Hh−12 h−1

1 = h1Hh−11

since h2 is in the normalizer. But this last term is just h1Hh−11 = H, as h1 ∈ H.

Therefore, h1h2 ∈ NG(H), and so NG(H) ⊇ H NG(H). �

2.5. An important special case of the Hall Theory

The following case will be dealt with separately.

Lemma 2.5.1. Let G be a finite solvable group and π a set of primes. Sup-pose that G has a non-trivial normal subgroup N such that G/N is not a π-group. Then Theorem 2.2.1 holds for G.

Proof. Suppose the statement of the result were false, and let G be acounterexample of minimum order. Let N be a non-trivial normal subgroupof G such that G/N is not a π-group. Since N is non-trivial, |G/N | < |G|,so Theorem 2.2.1 must hold for G/N , otherwise G/N would contradict theminimality of G.

(i) As Hall’s First Theorem holds for G/N , G/N has a Hall π-subgroup, sayK/N (here we are invoking Lemma 1.1.1, and we have thatK 6 G andN E K).Since K/N is a π-group and G/N is not, then K/N 6= G/N , so K 6= G. Let

26 2. HALL THEORY

H be a Hall π-subgroup of K. Then |H| is a π′-number, and [K : H] is aπ′-number. Note also that [G : K] = [G/N : K/N ] is a π′-number. Therefore,[G : H] = [G : K][K : H] is a π′-number, and so H is a Hall π-subgroup of G.Therefore, Hall’s First Theorem holds for G.

(ii) Let H1,H2 be two Hall π-subgroups of G. By Proposition 2.4.2, NH1

is a group, and since N E G, N E NH1. Therefore, (NH1)/N is a group,and furthermore it has order equal to that of H1, therefore it is a π-group.Also [G/N : (NH1)/N ] = [G : NH1] and [G : NH1][NH1 : H1] = [G : H1]so [G/N : (NH1)/N ] | [G : H1], a π′-number. Therefore [G/N : (NH1)/N ] isitself a π′-number, and so (NH1)/N is a Hall π-subgroup of G/N . Similarly,(NH2)/N is a Hall π-subgroup of G/N .

As Hall’s Second Theorem holds for G/N , (NH1)/N and (NH2)/N areconjugate in G/N . Therefore there exists gN ∈ G/N with

(gN)((NH1)/N)(gN)−1 = (NH2)/N

. We have:

(NH2)/N = (gN)((NH1)/N)(gN)−1

= {(gN)(hN)(gN)−1 : hN ∈ (NH1)/N}= {ghg−1N : hN ∈ (NH1)/N}= {ghg−1N : h ∈ NH1}= {xN : x ∈ g(NH1)g−1}= (g(NH1)g−1)/N

so that NH2 = g(NH1)g−1. Then:

NH2 = g(NH1)g−1

= {gnh1g−1 : n ∈ N,h1 ∈ H1}

= {gng−1gh1g−1 : n ∈ N,h1 ∈ H1}

= {n′gh1g−1 : n′ ∈ N,h1 ∈ H1}

= N(gH1g−1)

the penultimate inequality being because gNg−1 = N , as N is normal. Thegroup gH1g

−1 is a subgroup of N(gH1g−1) and hence of NH2. Furthermore, it

is a π-group (so is H2). Also, [G : (gH1g−1)] = [G : H2], a π′-number, and so

[NH2 : (gH1g−1)] = [NH2 : H2] is a π′-number, as it divides

[G : H2] (the quotient is [G : NH2], which is an integer as NH2 is a group byProposition 2.4.2). Therefore, H2 and (gH1g

−1) are Hall π-subgroups of NH2.Suppose G = NH2. Then, by Proposition 2.4.3,

|G| = |N ||H2||N ∩H2|

so

|G/N | = |G||N |

=|H2|

|N ∩H2|.

2.6. THE REMAINING CASE 27

Now if p | |G/N | then

p | |H2||N ∩H2|

so p divides |H2|, so p ∈ π. Therefore, G/N is a π-group, a contradiction.Therefore, NH2 6= G. So |NH2| < |G|, and so Theorem 2.2.1 holds for

NH2. Therefore, H2 and (gH1g−1) are conjugate in NH2, i.e. there exists

k ∈ H2 with k(gH1g−1)k−1 = H2, and so H1 and H2 are conjugate in G (by

kg). Therefore, Hall’s Second Theorem holds for G.(iii) Let J be a π-subgroup of G. Then NJ is a group, and since N E G,

then N E NJ . So (NJ)/N is a subgroup of G/N , and since Theorem 2.2.1holds for the latter, then (NJ/N) is contained in a Hall π-subgroup of G/N .By Lemma 1.1.1, this π-subgroup can be written as L/N , where N E L andL 6 G. Then J ⊆ NJ ⊆ L and so J 6 L. L is not equal to G, as otherwiseG/N = L/N would be a π-group. Therefore, |L| < |G|, and Theorem 2.2.1holds for L. So there is a Hall π-subgroup of L, say H, which contains J . Wehave that

[G : H] = [G : L][L : H] = [G/N : L/N ][L : H]

a product of two π′-numbers, hence a π′-number itself. Now H is aHall π-subgroup of G and J 6 H. Therefore, Hall’s Third Theorem holds forG. �

2.6. The Remaining Case

Before tackling the remaining cases, we shall give a preliminary lemma.

Lemma 2.6.1. If G is a group, A,B 6 G and ([G : A], [G : B]) = 1 then[G : A ∩B] = [G : A][G : B].

Proof. Let G be a group, A,B 6 G with ([G : A], [G : B]) = 1. Then:

[G : A ∩B] = [G : A][G : A ∩B]

= [G : A]|G|

|A ∩B|

= [G : A]|AB||B|

≤ [G : A]|G||B|

= [G : A][G : B]

the third equality being from Proposition 2.4.3.Also, since [G : A], [G : B] | [G : A ∩ B] (the quotients being [A : A ∩ B]

and [B : A ∩ B], respectively). Therefore, [G : A][G : B] | [G : A ∩ B], and so[G : A][G : B] ≤ [G : A ∩B].

Taking these two inequalities together gives [G : A∩B] = [G : A][G : B]. �

The following argument will be used more than once in the proof, so it isconvenient to give it here.

28 2. HALL THEORY

Lemma 2.6.2. Let G be a group, π a set of primes, H a Hall π-subgroup ofG, and K a Hall π′-subgroup of G. Then G = KH = HK.

Proof. Let G be a group, π a set of primes, H a Hall π-subgroup of G,and K a Hall π′-subgroup of G.

Suppose that H is trivial. Then none of the primes in π divide the orderof G, and so G is a π′-group. Then K, being a maximal Hall π′-subgroup ofG (by Lemma 2.1.5), has to equal G. We have that G = {idG}G = HK andG = G{idG} = KH, and so we are done. For the rest of the proof therefore,we can assume that H is non-trivial.

First, as [G : K] is a π-number, and divides [G : K]|K| = |G|, then wemust have [G : K] ≤ |H|, as |H| is the maximal π-number dividing |G|, byLemma 2.1.5.

Secondly, as H is a subgroup of G, |H| divides |G| = [G : K]|K|, and so|H| divides either |K| or [G : K]. However |H| is a π-number (and does notequal 1 as H is non-trivial), and |K| is a π′-number, so we cannot have that|H| divides |K|. Therefore, |H| divides [G : K], so |H| ≤ [G : K].

Taking these two inequalities together gives [G : K] = |H|, that is,

|G| = |K|[G : K] = |K| · |H|.

However, as |K| is a π′-number and |H| is a π-number, then by Corollary 2.1.9,|K ∩H| = 1. Therefore, by Proposition 2.4.3,

|G| = |K| · |H| = |KH| · |K ∩H| = |KH|.

Similarly,

|G| = |H| · |K| = |HK| · |H ∩K| = |HK| · |K ∩H| = |HK|.

Clearly KH ⊆ G and HK ⊆ G, so the fact that they have the same ordergives that G = KH = HK, as required. �

We can now deal with the remaining case.

Proof of Theorem 2.2.1. Let G be a finite solvable group, and π bea set of primes. If G is a π-group or a π′-group then the theorem holds byLemma 2.2.2, and if G has a non-trivial normal subgroup N with G/N not aπ-group, then the theorem holds by Lemma 2.5.1. Therefore, we may assumethat neither of these hold. That is, G is not a π-group or a π′-group, and everynon-trivial normal subgroup N of G is a π-group.

We have that for any p ∈ π, and a non-trivial Sylow p-subgroup P of G,that Theorem 2.2.1 holds for NG(P ). To see this, suppose for a contradictionthat G = NG(P ). Then P must be a normal subgroup, as any element ofG normalizes P . Also, P is non-trivial. Therefore G/P is a π-group, and so|G| = |G/P | · |P | is a π-number (as also p ∈ π). Therefore, G is a π-group,contradiction. So we must have that NG(P ) 6= G, and so |NG(P )| < |G|, soTheorem 2.2.1 holds for NG(P ).

Let A be a non-trivial normal subgroup of G of smallest order. Then G/Ais a π-group. Also, A is an elementary Abelian p-group for some prime p, byLemma 2.3.3. If p ∈ π, then |G| = |G/A||A| would be a π-number, and so G


would be a π-group, which we have assumed it to not be. Therefore, p 6∈ π, andso A is a π′-group.

(i) Take a normal subgroup of G/A of smallest order, which by Lemma 1.1.3,we can write as B/A, where A E B and B E G. Then, B/A is a q-group, byLemma 2.3.3. Since q | |G/A|, we must have q ∈ π, and so q 6= p.

Let Q be a Sylow q-subgroup of B. Then Q is non-trivial, as

q | |B/A| | |B/A||A| = |B|.

We have that B = AQ, by Lemma 2.6.2. Therefore, by the Frattini Argumentand then Lemma 2.4.9, we have:

G = BNG(Q) = AQNG(Q) = ANG(Q).

Notice that

[G : NG(Q)] =|G|

|NG(Q)|=|ANG(Q)||NG(Q)|

=|A|

|A ∩NG(Q)|the last equality being because of Proposition 2.4.3. However, considering thelast term on the right, the numerator is a power of p and the denominator isalso a power of p (as it must divide |A| by Proposition 2.1.8 and Lagrange’sTheorem). Therefore, [G : NG(Q)] is a power of p. By an argument at the startof this proof, Theorem 2.2.1 holds for NG(Q).

By Hall’s First Theorem in NG(Q), there exists a Hall π-subgroup H ofNG(Q). H is a π-group, and

[G : H] = [G : NG(Q)][NG(Q) : H].

The first term in the product is π′ as it is a power of p and p /∈ π, and thesecond term is π′ as H is a Hall π-subgroup of NG(Q). Therefore, [G : H] is π′,and H is a Hall π-subgroup of G. Therefore, Hall’s First Theorem holds for G.

(ii) We shall split this into two cases, depending on whether G/A is Abelian.(ii.1) First consider the case where G/A is Abelian. Take any

Hall π-subgroup of G, say H. Consider the map

φ : H 7→ G/A;x 7→ xA.

This map is an homomorphism: take x, y ∈ H and then

φ(xy) = (xy)A = (xA)(yA) = φ(x)φ(y).

It is onto: take any B ∈ G/A, and some x ∈ B, then φ(x) = xA = B, the lastequality being as x ∈ xA. G/A has the same order as a Hall π-subgroup, as|G/A| is a π-number, and

|G||G/A|

= |A|

which is a power of p, hence a π′-number (since p /∈ π). Therefore, φ is order-preserving, hence 1-1, hence an isomorphism. We shall use this to show that His Abelian: take any x, y ∈ H. Then φ(xy) = φ(x)φ(y) = φ(y)φ(x) = φ(yx), sotaking inverses (which we can do since φ is a bijection) gives xy = yx.

Let q be any prime dividing the order of G/A. Let Q be a Sylow q-subgroupof G (then Q is non-trivial since q divides |G/A|, which divides |G/A||A| = |G|),

30 2. HALL THEORY

and Q be a Sylow q-subgroup of H. By an argument at the start of this proof,Theorem 2.2.1 holds for NG(Q).

The group Q is a Sylow q-subgroup of G, since q ∈ π and so cannot divide[G : H], a π′-number, hence q cannot divide [G : H][H : Q] (as Q is a Sylowq-subgroup of H, so q does not divide [H : Q]). Therefore, by Sylow’s SecondTheorem, Q and Q are conjugate in G.

Take any h ∈ H. We wish to show that h ∈ NG(Q). So take any q ∈ Q.Then q ∈ H, and as H is Abelian, qh = hq so hqh−1 = q. Therefore, as q wasarbitrary, hQh−1 = Q, and so h ∈ NG(Q). Therefore, H 6 NG(Q).

Now let g ∈ G be such that Q and Q are conjugate by g. Then Q = gQg−1,and so

gNG(Q)g−1 = {gng−1 : n ∈ NG(Q)}= {gng−1 : n ∈ G, (∀h ∈ Q)(hnh−1 = n)}= {m : g−1mg ∈ G, (∀h ∈ Q)(hg−1mgh−1 = g−1mg)}= {m : m ∈ G, (∀h ∈ Q)(hg−1mgh−1 = g−1mg)}= {m : m ∈ G, (∀h ∈ Q)(ghg−1mgh−1g−1 = m)}= {m : m ∈ G, (∀q ∈ Q)(qmq−1 = m)}= NG(Q)

therefore H is conjugate to a subgroup of NG(Q), say K. K has the same orderas H, and hence is a π-group; also [NG(Q) : K] divides [G : K] = [G : H] (thequotient is [G : NG(K)]) and therefore is a π′-number. Hence H is conjugateto a Hall π-subgroup of NG(Q).

Now take any two Hall π-subgroups of G, say H1 and H2. H1 is conjugateto some Hall π-subgroup of NG(Q), say H3, and H2 is conjugate to some otherHall π-subgroup of NG(Q), say H4. By Hall’s Second Theorem in NG(Q), H3

and H4 are conjugate. Therefore, H1 is conjugate to H3, which is conjugate toH4, which is conjugate to H2. So H1 and H2 are conjugate, and Hall’s SecondTheorem is true for G.

(ii.2) Suppose G/A is non-Abelian and let H be a Hall π-subgroup of G.Then G = HA, by Lemma 2.6.2. We have that A E G. IfN1, N2 E HA = G then |G/N2| is a π-number. If |N1| divides |(HA)/N2| =|G/N2|, then |N1| would also be a π-number. However, |G/N1| is a π-number,and so |G| = |G/N1||N1| is a π-number, a contradiction. Therefore, we canapply Lemma 2.4.6 to obtain G′ = H ′A.

Now, G′ E G, H ′ 6 G′ by Lemmas 1.2.4 and 1.2.7. We have G′ 6= G, sinceotherwise G would not be solvable. So |G′| < |G|, and Theorem 2.2.1 holds forG′. Now H ′ is a π-group, as if p divides |H ′| then p divides |H| (as H ′ 6 Hand we can use Lagrange) and so p ∈ π. Furthermore ([G′ : H ′] is π′, as if itwasn’t, there would exist a prime in π dividing [G′ : H ′], but then this primewould also divide [G′ : H ′][H ′ : H] = [G : H], a contradiction as H is a Hallπ-subgroup of G). Therefore H ′ is a Hall π-subgroup of G′.

Since H ′ ⊆ G′ and G′ E G, then gH ′g−1 6 G′ for any g ∈ G. ThereforegH ′g−1, having the same order as H ′, is also a Hall π-subgroup of G′, and Hall’s


Second Theorem in G′ gives that H is conjugate to gHg−1 in G′. Then thecore Frattini Argument, Lemma 2.4.7, gives that G = G′ NG(H ′). This givesG = AH ′ NG(H ′) = ANG(H ′), the second equality being from Lemma 2.4.9.

Now take any two Hall π-subgroups of G, say H1 and H2. From the abovework, we have that H ′

1 and H ′2 are Hall π-subgroups of G′ and Theorem 2.2.1

holds for G′, so that H ′1 and H ′

2 are conjugate in G′. Let h ∈ G′ be an elementthey are conjugate by, so that hH ′

1h−1 = H ′

2.Note that if u, v ∈ H1 then

h[u, v]h−1 = huvu−1v−1h−1

= huh−1hvh−1hu−1h−1hv−1h−1

= (huh−1)(hvh−1)(huh−1)−1(hvh−1)−1

= [huh−1, hvh−1]

so that

(hH1h−1)′

= {[hu1h−1, hv1h

−1][hu2h−1, hv2h

−1] · · · [hunh−1, hvnh

−1] : ui, vi ∈ H1}= {(h[u1, v1]h−1)(h[u, v]h−1) · · · (h[un, vn]h−1) : ui, vi ∈ H1}= {h[u1, v1][u2, v2] · · · [un, vn]h−1 : ui, vi ∈ H1}= hH ′

1h−1

where the first and last equalities come from Lemma 1.2.2.Now H2 is a π-group and a subgroup of NG(H ′

2) (as H ′2 E H2), and

[NG(H ′2) : H2] divides [G : H2] (the quotient is [G : NG(H ′

2)]), a π′-number.Therefore, H2 is a Hall π-subgroup of NG(H ′

2). Since |hH1h−1| = |H1| = |H2|,

and hH1h−1 is also a subgroup of NG((hH1h

−1)′) = NG(hH ′1h

−1) = NG(H ′2) (as

(hH1h−1)′ E hH1h

−1), we have that hH1h−1 is a Hall π-subgroup of NG(H2)

as well.Suppose NG(H2) = G. Then H2 is a normal subgroup of G. Furthermore,

it is a π-group. We have that [G : H2] is a π-number, and so |G| = [G : H2]|H2|is also a π-number, contradiction. Thus NG(H2) 6= G, and since NG(H2) ⊆ G,we have that |NG(H2)| < |G|. Therefore, Hall’s Second Theorem holds forNG(H2).

Thus hH1h−1 and H2 are conjugate in NG(H2), say by an element n, so

that n(hH1h−1)n−1 = H2. Then

(nh)H1(nh)−1 = nhH1h−1n−1 = H2

so that H1 and H2 are conjugate in G (namely by nh). Therefore, Hall’s SecondTheorem holds for G.

(iii) Let J be a π-subgroup of G. Suppose G = AJ . Then

|G| = |A| · |J ||A ∩ J |

= |A| · |J |

by Corollary 2.1.9. Therefore,

[G : J ] =|G||J |

=|A| · |J ||J |

= |A|

32 2. HALL THEORY

which is a π′-number. So J is itself a Hall π-subgroup of G, and certainly J ⊆ J ,so Hall’s Third Theorem is true for G in this case.

Now suppose G 6= AJ . We have that |AJ | = |A| · |J |, by Proposition 2.4.3(as |A ∩ J | = 1 by Corollary 2.1.9). So

[G : AJ ] =|G||AJ |

=|G|

|A| · |J |=

[G : A]|J |

the quotient of two π-numbers, hence a π-number.Let H be any Hall π-subgroup of G. Then [G : H] is a π′-number. By

Lemma 2.6.1, we have that [G : H ∩AJ ] = [G : H][G : AJ ], and so[G : H ∩ AJ ] = |A|[G : AJ ] = [G : J ] (as [G : A], being a π number with |G|

[G:A]

a π′-number, is the order of a Hall π-subgroup, and hence [G : A] = |H|, so[G : H] = |A|). So |H ∩AJ | = |J |.

We know A is normal in G so AJ is a group, and H ∩ J 6 J 6 AJ . Alsonote that |H ∩ AJ | = |J | is a π-number and [AJ : H ∩ J ] = [AJ : J ] = |A|is a π′-number, so J and H ∩ AJ are Hall π-subgroups of AJ . As G 6= AJ ,AJ ⊆ G, we have |AJ | < |G|. Therefore, Theorem 2.2.1 holds for AJ , and so Jand H ∩AJ are conjugate in AJ , hence in G. Let g be an element which theyare conjugate by. Then

J = g(H ∩AJ)g−1 ⊆ gHg−1

and the latter term has the same order as H, so is a Hall π-subgroup of G.Therefore, Hall’s Third Theorem holds for G. �

CHAPTER 3

The Extension Classification Schema

3.1. Isomorphism Classes

Recall that two groups are isomorphic if there exists a bijective homomor-phism between them. We say that G ∼= H if the groups are isomorphic, and wesay that the bijective homomorphism is an isomorphism.

Lemma 3.1.1. ∼= is an equivalence relation.

Sketch of Proof. For groups G, H, K, and isomorphisms φ1 : G 7→ H,φ2 : H 7→ K, the identity map G 7→ G is an isomorphism, the inverse map(φ1)−1 : H 7→ G is an isomorphism and the composition (φ2 ◦ φ1) : G 7→ K isan isomorphism. �

Because of this, ∼= splits the class of all groups into “equivalence classes”.However, these are not true equivalence classes, as the collection of all groupscannot ever form a set in the traditional sense. We can consider them to beequivalence classes, however, and certainly we can take a complete set of repre-sentatives from the equivalence classes. This set will in fact only be countablyinfinite, in light of the following result:

Lemma 3.1.2. If n is a positive integer, there exists a finite set πn of groupssuch that if G is any group of order n, then G is isomorphic to one of the groupsin πn.

Proof. Define a magma as an arbitrary set X equipped with a binaryoperation X × X 7→ X. The idea of a group isomorphism is identical formagmas, and any group must itself be a magma. Now, any magma is completelydescribed (up to isomorphism) by a Cayley table showing the result of applyingthe operation to any pair of elements, and each Cayley table corresponds to amagma (e.g by taking the set {1, ..., n}). There are n2 elements in the tableif |X| = n, and each element in the table can be chosen in n ways, so thenumber of possible tables is n multiplied by itself n2 times, or nn2

possibletables. Therefore, the set of all magmas of order n (say, Tn) has size nn2

, andhence is finite. Now let πn be the set of magmas in Tn which are also groups.This set is clearly finite. �

So therefore we have the existence of a finite complete set of representativesfor the “isomorphism classes”, because it is just equal to

⋃n πn above.

We call each set πn a classification of the groups of order n, or an enumer-ation of them a list of groups of order n up to isomorphism:

33

34 3. THE EXTENSION CLASSIFICATION SCHEMA

Definition 3.1.3. Let G1, ..., Gn be a sequence of groups. We say thatG1, ..., Gn is a list of groups with property P (for some property P ) up toisomorphism, if:

(1) For each i with 1 ≤ i ≤ n, Gi has property P .(2) Every group which has property P is isomorphic to Gi for some i with

1 ≤ i ≤ n.(3) For each i, j with 1 ≤ i ≤ n, 1 ≤ j ≤ n and i 6= j, Gi and Gj are not

isomorphic.

Definition 3.1.4. Let S be a finite set of groups. We say that S is aclassification of groups with property P (for some property P ), if there is anenumeration G1, ..., Gn of the groups in S which is a list of groups with propertyP up to isomorphism.

We shall give some examples of this:

Examples 3.1.5.(i) The sequence C6, S3 is a list of groups of order 6 up to isomorphism.(ii) The set {C4, C2 × C2} is a classification of groups of order 4.(iii) The sequence C10, C5×C2, D5 is not a list of groups of order 10 up to

isomorphism, as C10∼= C5 × C2.

(iv) The empty set is a classification of non-Abelian prime-order groups.

Our aim is to classify some groups of order pqr. In this chapter, we shallpresent The Extension Classification Schema, which is a much more powerfulresult than we need for the order pqr case. However, it shall give us all we needin order to present the classification of groups of order pqr.

3.2. Semidirect Products

We know the direct product well, both in its internal and external forms. Ifa group is the internal direct product of two of its subgroups, then both have tobe normal. If we relax the condition so that only one of them has to be normal,we get something called the semidirect product, and this has a fairly simpleexternal analog. (In fact, if we require neither to be normal, we get somethingcalled the Zappa-Szep Product (discovered independently by Zappa and Szepin the 1940s), but the external analog of this is much more complicated).

First we consider the internal semidirect product.

Definition 3.2.1. Let G be a group with subgroups N and H. Then G issaid to be the internal semidirect product of N and H, if N E G, G = NH,and N ∩H is trivial.

The analog to this is the external semidirect product.

Definition 3.2.2. Let H,K be groups, and σ ∈ Hom(K,Aut(H)). Thenthe external semidirect product ofH andK with respect to σ, is the magmawhose underlying set is H ×K (where × is the cartesian product of sets) andwhose operation is defined by:

(h1, k1)(h2, k2) = (h1σ(k1)(h2), k1k2).

3.2. SEMIDIRECT PRODUCTS 35

We denote the external semidirect product of H and K with respect to σby H oσ K.

Proposition 3.2.3. Let H,K be groups, and σ ∈ Hom(K,Aut(H)). ThenH oσ K is a group.

Proof. Let H,K be groups, and σ ∈ Hom(K,Aut(H)). Let G = H oσ K.Since the underlying set of G is the set cartesian product of two non-emptysets, G is non-empty.

We shall check associativity for the group operation. Suppose x1, x2, x3 ∈ G.Then x1 = (h1, k1), x2 = (h2, k2)andx3 = (h3, k3) for some h1, h2, h3 ∈ H andk1, k2, k3 ∈ K. Now:

(x1x2)x3 = ((h1, k1)(h2, k2))(h3, k3)

= (h1σ(k1)(h2), k1k2)(h3, k3)

= (h1σ(k1)(h2)σ(k1k2)(h3), k1k2k3)

= (h1σ(k1)(h2)(σ(k1) ◦ σ(k2))(h3), k1k2k3)

= (h1σ(k1)(h2)σ(k1)(σ(k2)(h3)), k1k2k3)

= (h1σ(k1)(h2σ(k2)(h3)), k1k2k3)

= (h1, k1)(h2σ(k2)(h3), k2k3)

= (h1, k1)((h2, k2)(h3, k3))

and so the group operation is associative.Let x ∈ G, then x = (h, k) for some h ∈ H, k ∈ K. Now:

x(idH , idK) = (h, k)(idH , idK)

= (hσ(k)(idH), k idK)

= (h idH , k)

= (h, k)= x

and

(idH , idK)x = (idH , idK)(h, k)

= (idH σ(idK)(h), idK k)

= (idH h, k) = (h, k)= x

so the element (idH , idK) acts like an identity.


Finally, again let x ∈ G so that x = (h, k) for some h ∈ H, k ∈ K. Lety = ((σ(k−1))(h−1), k−1). Then:

yx = ((σ(k−1))(h−1), k−1)(h, k)

= ((σ(k−1))(h−1)σ(k−1)(h), k−1k)

= ((σ(k−1))(h−1)σ(k−1)(h), idK)

= (σ(k−1)(h−1h), idK)

= (σ(k−1)(idH), idK)

= (idH , idK)

and

xy = (h, k)((σ(k−1))(h−1), k−1)

= (hσ(k)((σ(k−1))(h−1)), kk−1)

= (h(σ(k) ◦ σ(k−1))(h−1), idK)

= (hσ(kk−1)(h−1), idK)

= (hσ(idK)(h−1), idK)

= (hh−1, idK)

= (idH , idK)

so the element y acts like an inverse for x.Therefore, all the group axioms are satisfied, and so G is a group. �

We can see that the external semidirect product is indeed an analog of theinternal one by the following proposition and Proposition 3.2.6.

Proposition 3.2.4. Let G be a group with subgroups N and H. Sup-pose G is the internal semidirect product of N and H. Then there existsσ ∈ Hom(H,Aut(N)) with G ∼= N oσ H.

Proof. Let G be a group with subgroups N and H. Suppose G is theinternal semidirect product of N and H. For each h ∈ H let λh be defined by

λh : N 7→ N ;n 7→ hnh−1.

Clearly λ ∈ Aut(N). Now define a map σ by:

σ : H 7→ Aut(N);h 7→ λh.

Clearly σ ∈ Hom(H,Aut(N)). Define a map φ by:

φ : N oσ H 7→ G; (n, h) 7→ nh

Our goal is to show that φ is an isomorphism.First of all φ is onto. Take any g ∈ G. Then as G is the internal semidirect

product of N and H then G = NH so g = nh for some n ∈ N , h ∈ H. Now(n, h) ∈ N oσ H, and φ(n, h) = nh.

Also |N oσ H| = |N | · |H| = |N |·|H||N∩H| = |NH| = |G| since G = NH and

|N ∩H| = 1. Therefore, taken with the fact that φ is onto, φ is a bijection andhence 1-1.


Finally we must show that φ is a homomorphism. Take x1, x2 ∈ N oσ H.Then x1 = (n1, h1) and x2 = (n2, h2) for n1, n2 ∈ N and h1, h2 ∈ H. Now

φ(x1x2) = φ((n1, h1)(n2, h2))

= φ(n1σ(h1)(n2), h1h2)

= φ(n1(h1n2h−11 ), h1h2)

= n1h1n2h−11 h1h2

= n1h1n2h2

= φ(n1, h1)φ(n2, h2)

= φ(x1)φ(x2)

so that φ is a homomorphism. Therefore, φ is an isomorphism, and hence weare done. �

We cannot have that an external semidirect product is an internal semidirectproduct of the two original groups, since these are not even subsets (note thatan external semidirect product is a group of ordered pairs). However, it is aninternal semidirect product of groups isomorphic to the original groups, givenby the following definition.

Definition 3.2.5. Suppose H and K are groups. Then we define thecanonical semidirect product embeddings:

(H, idK) = {(h, idK) : h ∈ H}and

(idH ,K) = {(idH , k) : k ∈ K}.

Proposition 3.2.6. Let H and K be groups and σ ∈ Hom(K,Aut(H)).Let

H1 = (H, idK)and

K1 = (idH ,K).Then H1 and K1 are subgroups of H oσ K, H1

∼= H and K1∼= K, and H oσ K

is the internal semidirect product of H1 and K1.

Proof. Let H and K be groups and σ ∈ Hom(K,Aut(H)). Let

H1 = {(h, idK) : h ∈ H}and

K1 = {(idH , k) : k ∈ K}.First note that H1 and K1 are non-empty, as they both contain the element(idH , idK).

Let x1, x2 ∈ H1. Then x1 = (h1, idK) and x2 = (h2, idK). σ(idK) is anautomorphism, and so σ(idK)(h2) ∈ H, and so

x1x2 = (h1, idK)(h2, idK)

= (h1σ(idK)(h2), idK idK)

= (h1σ(idK)(h2), idK) ∈ H1.


Similarly, if y1, y2 ∈ H2, then y1 = (idH , k1) and y2 = (idH , k2). σ(k2) is anautomorphism, and so σ(k2)(idH) = idH , and so

y1y2 = (idH , k1)(idH , k2)

= (idH σ(k2)(idH), k1k2)

= (idH idH , k1k2)

= (idH , k1k2) ∈ K1.

Now let x ∈ H1. Then x = (h, idK). Let a = (h−1, idK). Since σ is ahomomorphism, σ(idK) is the identity automorphism, and

ax = (h−1, idK)(h, idK)

= (h−1σ(idK)(h), idK idK)

= (h−1h, idK)

= (idH , idK)

and also

xa = (h, idK)(h−1, idK)

= (hσ(idK)(h−1), idK idK)

= (hh−1, idK)

= (idH , idK)

so that a is the inverse of x. However clearly a ∈ H1. This taken with previousarguments gives that H1 is a subgroup of H oσ K.

Let y ∈ K1. Then y = (idH , k). Let b = (idH , k−1). Since σ(k) and σ(k−1)

are a automorphisms, σ(k)(idH) = idH and σ(k−1)(idH) = idH , and

by = (idH , k−1)(idH , k)

= (idH σ(k)(idH), k−1k)

= (idH idH , idK)

= (idH , idK)

and also

yb = (idH , k)(idH , k−1)

= (idH σ(k−1)(idH), kk−1)

= (idH idH , idK)

= (idH , idK)

so that b is the inverse of y. However clearly b ∈ K1. This taken with previousarguments gives that K1 is a subgroup of H oσ K.

Construct maps from H1 to H and K1 to K as follows:

φ : H1 7→ H; (h, idK) 7→ h

ψ : K1 7→ K; (idH , k) 7→ k

Then φ and ψ are onto by definition. Suppose φ(x) = φ(y). Then x = (h1, idK)and y = (h2, idK) for some h1, h2 ∈ H, so we get h1 = φ(x) = φ(y) = h2, so


x = y and φ is 1-1, hence a bijection. Suppose ψ(x) = ψ(y). Then x = (idH , k1)and y = (idH , k2) for some k1, k2 ∈ K, so we get k1 = ψ(x) = ψ(y) = k2, sox = y and ψ is 1-1, hence a bijection. From this we get that |H1| = |H| and|K1| = |K|.

In fact we have that φ and ψ are homomorphisms. First take x, y ∈ H1.Then x = (h1, idK) and y = (h2, idK) for some h1, h2 ∈ H. Then

φ(xy) = φ((h1, idK)(h2, idK))

= φ(h1σ(idK)(h2), idK idK)

= φ(h1h2, idK)= h1h2

= φ(h1, idK)φ(h2, idK)

= φ(x)φ(y)

as σ(idK) is the identity automorphism. Therefore φ is a homomorphism. Sec-ondly, take x, y ∈ K1. Then x = (idH , k1) and y = (idH , k2) for some k1, k2 ∈ K.Then

ψ(xy) = ψ((idH , k1)(idH , k2))

= ψ(idH σ(k1)(idH), k1k2)

= ψ(idH , k1k2)= k1k2

= ψ(idH , k1)ψ(idH , k2)

= ψ(x)ψ(y)

as any automorphism must map the identity to the identity. Therefore ψ is ahomomorphism. Since φ and ψ are also bijections, they are isomorphisms, andthis gives H1

∼= H and K1∼= K.

It is clear that if g ∈ H1∩K1 then g = (h, idK) = (idH , k), so g = (idH , idK).Therefore, H1 ∩K1 is trivial. By Proposition 2.4.3, we have that|H1K1| = |H1| · |K1| = |H| · |K| = |H oσ K|, and so since H1K1 ⊆ H oσ K, wemust have H1K1 = H oσ K.

Finally, we show that H1 E H oσ K. We do this by showing that every leftcoset is equal to every right coset. Let (s, t) ∈ H oσ K. Then

H1(s, t) = {(h, idH)(s, t) : h ∈ H}= {(hσ(idH)(s), idH t) : h ∈ H}= {(hs, t) : h ∈ H}= {(x, t) : x ∈ H}


since every element of H can be written as x = (xs−1)s = hs for h = xs−1.Also

(s, t)H1 = {(s, t)(h, idH) : h ∈ H}= {(sσ(t)(h), t idH) : h ∈ H}= {(sσ(t)(h), t) : h ∈ H}= {(x, t) : x ∈ H}

since every element of H can be written as

x = ss−1x = s(σ(t)((σ(t))−1(s−1x))) = sσ(t)(h)

for h = (σ(t))−1(s−1x). This is what we wanted.Therefore, H1 E H oσ K is the internal semidirect product of H1 and

K1. �

We would be questioning our definition of the semidirect product if semidi-rect products of isomorphic groups were not isomorphic. We wrap up the coreof the proof in a preliminary lemma, which is slightly stronger than we requirehere.

Lemma 3.2.7. Suppose H1,K1,H2,K2 are groups, such that H1∼= H2 and

K1∼= K2. Let α : H1 7→ H2 and β : K1 7→ K2 be isomorphisms. Define

λα : Aut(H1) 7→ Aut(H2); f 7→ α ◦ f ◦ α−1.

Take σ1 ∈ Hom(K1,Aut(H1)), σ2 ∈ Hom(K2,Aut(H2)). Define

φ : H1 oσ1 K1 7→ H2 oσ2 K2; (h1, k1) 7→ (α(h1), β(k1)).

Then φ is a homomorphism if and only if σ2 = λα ◦ σ1 ◦ β−1.

Proof. Suppose H1,K1,H2,K2 are groups, such that H1∼= H2 and


λα : Aut(H1) 7→ Aut(H2); f 7→ α ◦ f ◦ α−1.

Take σ1 ∈ Hom(K1,Aut(H1)), σ2 ∈ Hom(K2,Aut(H2)). Define

φ : H1 oσ1 K1 7→ H2 oσ2 K2; (h1, k1) 7→ (α(h1), β(k1)).

Let x1, x2 ∈ H1 be arbitrary. Then x1 = (h1, k1) and x2 = (h2, k2) for someh1, h2 ∈ H1 and k1, k2 ∈ K1. Note that:

φ(x1x2) = φ((h1, k1)(h2, k2))

= φ(h1σ1(k1)(h2), k1k2)

= (α(h1σ1(k1)(h2)), β(k1k2))

= (α(h1)α(σ1(k1)(h2)), β(k1)β(k2))

= (α(h1)α(σ1(k1)(α−1(α(h2)))), β(k1)β(k2))

= (α(h1)λα(σ1(k1))(α(h2)), β(k1)β(k2))

= (α(h1)λα(σ1(β−1(β(k1))))(α(h2)), β(k1)β(k2))


and

φ(x1)φ(x2) = φ(h1, k1)φ(h2, k2)

= (α(h1), β(k1))(α(h2), β(k2))

= (α(h1)σ2(β(k1))(α(h2)), β(k1)β(k2))

( ⇐= ) Suppose σ2 = λα ◦ σ1 ◦ β−1. Then

φ(x1)φ(x2) = (α(h1)σ2(β(k1))(α(h2)), β(k1)β(k2))

= (α(h1)λα(σ1(β−1(β(k1))))(α(h2)), β(k1)β(k2))

= φ(x1x2)

and so φ is a homomorphism.( =⇒ ) Suppose φ is a homomorphism. Then

(α(h1)λα(σ1(β−1(β(k1))))(α(h2)), β(k1)β(k2))

= φ(x1x2)

= φ(x1)φ(x2)

= (α(h1)σ2(β(k1))(α(h2)), β(k1)β(k2))

Now we have

α(h1)λα(σ1(β−1(β(k1))))(α(h2)) = α(h1)σ2(β(k1))(α(h2))

and soλα(σ1(β−1(β(k1))))(α(h2)) = σ2(β(k1))(α(h2)).

This formula holds true for all h2 ∈ H1 and k1 ∈ K1. Let h ∈ H2, k ∈ K2.Then since α and β are onto, there exist some h2 ∈ H1 and k1 ∈ K1 such thatα(h2) = h and α(k1) = k. Now the above formula becomes:

λα(σ1(β−1(k)))(h) = σ2(k)(h)

for all h ∈ H2, k ∈ K2. We therefore have

(λα ◦ σ1 ◦ β−1)(k) = σ2(k)

and soσ2 = λα ◦ σ1 ◦ β−1

as required. �

Proposition 3.2.8. Suppose H1,K1,H2,K2 are groups, such that H1∼= H2

and K1∼= K2. Let α : H1 7→ H2 and β : K1 7→ K2 be isomorphisms. Define

λα : Aut(H1) 7→ Aut(H2); f 7→ α ◦ f ◦ α−1.

Take any σ1 ∈ Hom(K1,Aut(H1)) and define

σ2 : K2 7→ Aut(H2)

by σ2 = λα ◦ σ1 ◦ β−1. Then H1 oσ1 K1∼= H2 oσ2 K2.


Proof. Suppose H1,K1,H2,K2 are groups, such that H1∼= H2 and


λα : Aut(H1) 7→ Aut(H2); f 7→ α ◦ f ◦ α−1.

Take any σ1 ∈ Hom(K1,Aut(H1)) and define

σ2 : K2 7→ Aut(H2)

by σ2 = λα ◦ σ1 ◦ β−1. Define

φ : H1 oσ1 K1 7→ H2 oσ2 K2; (h, k) 7→ (α(h), β(k)).

First let us show that φ is onto. Take any x ∈ H2 oσ2K2. Then x = (h2, k2)for some h2 ∈ H2, k2 ∈ K2. Since α and β are onto, there exist h1 ∈ H1, k1 ∈ K1

such that α(h1) = h2 and α(k1) = k2. Now

φ(h1, k1) = (α(h1), β(k1)) = (h2, k2) = x

and so φ is onto.Since α and β are isomorphisms, they are bijections and so |H1| = |H2| and

|K1| = |K2|. Therefore

|H1 oσ1 K1| = |H1| · |H2| = |K1| · |K2| = |H2 oσ2 K2|and taking this with the fact that φ is onto gives that it is a bijection, hence1-1.

Finally, φ is a homomorphism by Lemma 3.2.7. Therefore, φ is an isomor-phism, and so H1 oσ1 K1

∼= H2 oσ2 K2. �

3.3. The Schema itself

The schema works by classifying groups of a particular type into semidi-rect products: however such groups are sometimes isomorphic to each other.Therefore we group the ones which are isomorphic into orbits under a particu-lar action specific to this Schema. We shall therefore call this action the ECSAction:

Proposition 3.3.1. Let X and Y be groups, then the map

∗ : (Aut(X)×Aut(Y ))×Hom(Y,Aut(X)) 7→ Hom(Y,Aut(X));

((α, φ), ρ) 7→ λα ◦ ρ ◦ φ−1

whereλα : Aut(X) 7→ Aut(X); f 7→ αfα−1

is an action.

Proof. Let X and Y be groups, and define the map ∗ as in the statementof the proposition. First notice that for any x ∈ Hom(Y,Aut(X)):

∗(idAut(X)×Aut(Y ), x) = ∗((idAut(X), idAut(Y )), x)

= λidAut(X)◦ x ◦ (idAut(Y ))

−1

= λidAut(X)◦ x

= x

3.3. THE SCHEMA ITSELF 43

sinceλidAut(X)

(f) = idAut(X) ◦f ◦ (idAut(X))−1 = f

Now take any g, h ∈ Aut(X) × Aut(Y ). Let g = (g1, g2) and h = (h1, h2).Then gh = (g1◦h1, g2◦h2). Again, take x ∈ Hom(Y,Aut(X)), and f ∈ Aut(X).We have:

λg1◦h1(f) = (g1 ◦ h1) ◦ f ◦ (g1 ◦ h1)−1

= (g1 ◦ h1) ◦ f ◦ ((h1)−1 ◦ (g1)−1)

= g1 ◦ (h1 ◦ f ◦ (h1)−1) ◦ (g1)−1

= g1 ◦ λh1(f) ◦ (g1)−1

= (λg1 ◦ λh1)(f)

so that λg1h1 = λg1λh1 , and so

∗(gh, x) = ∗((g1 ◦ h1, g2 ◦ h2), x)

= λg1◦h1 ◦ x ◦ (g2 ◦ h2)−1

= λ(g1 ◦ h1) ◦ x ◦ h−12 ◦ g−1

2

= λ(g1) ◦ λh1 ◦ x ◦ h−12 ◦ g−1

2

= λ(g1) ◦ (λh1 ◦ x ◦ h−12 ) ◦ g−1

2

= λ(g1) ◦ ∗((h1, h2), x) ◦ g−12

= λ(g1) ◦ ∗(h, x) ◦ g−12

= ∗((g1, g2), ∗(h, x))= ∗(g, ∗(h, x)).

Therefore, both the conditions for an axioms are satisfied, and so ∗ is anaction. �

Definition 3.3.2. The action of the above proposition is called the ECSAction. We denote as usual ∗((α, φ), ρ) by (α, φ) ∗ ρ.

Remarks.

(i) Note that it is perhaps easier to see what is happening with the actionby looking at the result of the action applied to an element of Y . Thenwe get:

((α, φ) ∗ ρ)(y) = λα(ρ(φ−1(y))) = α ◦ ρ(φ−1(y)) ◦ α−1.

(ii) Note also how similar this action is to the function σ2 defined in Propo-sition 3.2.8. This is no coincidence, as the proof of part (iii) of thebelow shows.

We are now in a position to state the actual schema.

Theorem 3.3.3 (The Extension Classification Schema). Let X and Y besolvable groups of coprime orders. Then:

(i) For any ρ ∈ Hom(Y,Aut(X)), the semidirect product G = X oρ Y hasa normal subgroup N isomorphic to X with G/N isomorphic to Y .


(ii) For any group G with a normal subgroup N isomorphic to X and G/Nisomorphic to Y , there exists some ρ ∈ Hom(Y,Aut(X)) such that Gis isomorphic to the semidirect product X oρ Y .

(iii) Let σ, ρ ∈ Hom(Y,Aut(X)). Then the semidirect products Xoσ Y andX oρ Y are isomorphic if and only if σ and ρ lie in the same orbit ofthe ECS Action for X and Y .

Once we have proved this, we have classified all groups with a certain prop-erty.

Corollary 3.3.4. Let X and Y be solvable groups of coprime orders. LetT be a complete set of representatives for the orbits of the ECS Action for Xand Y . Then the set

{X oρ Y : ρ ∈ T}

is a classification of the groups G with a normal subgroup N isomorphic to X,and G/N isomorphic to Y .

Proof. Let X and Y be solvable groups of coprime orders. Let T be acomplete set of representatives for the orbits of the ECS Action for X and Y .Let G1, ..., Gn be an enumeration of the elements of the set

{X oρ Y : ρ ∈ T}

where n is the size of the above set. Take any Gi, then Gi = X oρ Y forsome ρ ∈ T ⊆ Hom(Y,Aut(X)), and so, by Theorem 3.3.3(i), Gi has a normalsubgroup N isomorphic to X, and G/N isomorphic to Y .

Take any group G with a normal subgroup N isomorphic to X and G/Nisomorphic to Y . By Theorem 3.3.3(ii), there is some ρ ∈ Hom(Y,Aut(X))with G ∼= X oρ Y . Since T is a complete set of representatives for the orbitsof the ECS Action for X and Y , there is some σ ∈ T such that σ ∈ Orb(ρ)under the ECS Action for X and Y . Therefore, σ and ρ lie in the same orbitof the ECS Action for X and Y , and so by the “if” part of Theorem 3.3.3(iii),X oσ Y ∼= X oρ Y ∼= G. Also, since σ ∈ T , there is some i with 1 ≤ i ≤ n andGi = X oσ Y . For this i, we therefore have Gi = X oσ Y ∼= G.

Finally, take any Gi, Gj , with 1 ≤ i ≤ n, 1 ≤ j ≤ n and i 6= j. The groupGi = X oσ Y for some σ ∈ T and the group Gj = X oρ Y for some ρ ∈ T . Infact, σ 6= ρ, otherwise Gi = Gj , so i = j. Therefore, σ and ρ lie in differentorbits of the ECS Action. By the “only if” part of Theorem 3.3.3(iii), we cannothave that X oσ Y ∼= X oρ Y , that is, Gi is not isomorphic to Gj . �

Remark. As we shall see later, the above result allows us to classify allgroups of squarefree order, in a very theoretical sense. This is because any suchgroup (of order say p1 · · · pn) has a normal subgroup of index p1 (by Proposi-tion 1.5.1), and we can take this for X, and Cp1 for Y (as there is only onegroup of order p1 up to isomorphism). Proceeding inductively, we can go froma classification of groups of order p2 · · · pn to a classification of groups of orderp1 · · · pn (taking X to be each possibility for a group of order p2 · · · pn in turn),and in this way classify all groups of squarefree order.

3.4. PROOF OF THEOREM 3.3.3(I) AND THEOREM 3.3.3(II) 45

3.4. Proof of Theorem 3.3.3(i) and Theorem 3.3.3(ii)

Before proving parts (i) and (ii) of Theorem 3.3.3, we must first prove someinitial lemmas.

Lemma 3.4.1. Suppose G is a group with subgroups N and H, such that Gis the internal semidirect product of N and H. Then NH = HN .

Proof. Suppose G is a group with subgroups N and H, such that G is theinternal semidirect product of N and H. Then G = NH, so that |G| = |NH|.Furthermore, |N∩H| = 1, so that |H∩N | = 1. Therefore, by Proposition 2.4.3,

|HN | = |H| · |N ||H ∩N |

=|N | · |H||N ∩H|

= |NH| = |G|.

But also HN ⊆ G. Therefore, we must have HN = G = NH. �

Lemma 3.4.2. Suppose G is a group with subgroups N and H, such that Gis the internal semidirect product of N and H. Then H ∼= G/N .

Proof. Suppose G is a group with subgroups N and H, such that G is theinternal semidirect product of N and H. Since H ⊆ G, if h ∈ H then h ∈ Gand so hN ∈ G/N . Define the following map:

φ : H 7→ G/N ;h 7→ hN.

By the above, this is well-defined.We shall first show that φ is onto. Take any x ∈ G/N , then x = gN for

some g ∈ G. Since G = NH, then by Lemma 3.4.1, G = HN . Therefore,g = hn for some h ∈ H, n ∈ N . Now gN = (hn)N = h(nN) = hN and soφ(h) = hN = gN = x. Therefore, φ is onto.

Since G = NH, then by Proposition 2.4.3, |G| = |NH| = |N |·|H||N∩H| = |N | · |H|,

as |N∩H| = 1. Therefore, |H| = |G||N | , and so since φ is also onto, it is a bijection,

hence 1-1.Finally let h1, h2 ∈ H. Then

φ(h1h2) = (h1h2)N = (h1N)(h2N) = φ(h1)φ(h2)

so that φ is a homomorphism.Therefore, φ is an isomorphism, and we are done. �

We are now in a position to prove Theorem 3.3.3, parts (i) and (ii).

Proof of Theorem 3.3.3(i). Let X and Y be solvable groups of coprimeorders, and let ρ ∈ Hom(Y,Aut(X)). Let G = X oρ Y . Let X1 = (X, idY )and Y1 = (idX , Y ). Then, by Proposition 3.2.6, X1 and Y1 are subgroups of G,X1

∼= X and Y1∼= Y , and G is the internal semidirect product of X1 and Y1.

Let N = X1. Then N = X1 E G, as G is the internal semidirect productof X1 and Y1. Also N = X1

∼= X, and Y ∼= Y1∼= G/X1 = G/N , the second

isomorphism being by Lemma 3.4.2. Therefore, we are done. �

Proof of Theorem 3.3.3(ii). LetX and Y be solvable groups of coprimeorders, and let G be a group with a normal subgroup N isomorphic to X and


G/N isomorphic to Y . N and G/N are solvable by Lemma 1.2.12, and henceG is solvable by Proposition 1.4.3.

Let π be the set of primes dividing both |G| and |Y |. Note |X| = |N | asN is isomorphic to X. Therefore, if p is a prime dividing the order of N thenp divides the order of X. However since X and Y have coprime orders, p doesnot divide the order of Y , so p is not in π. The upshot is that N is actually aπ′-group.

By Proposition 1.2.12, N and G/N are solvable. Therefore, by Proposi-tion 1.4.3, G is solvable. By Theorem 2.2.1, G has a Hall π-subgroup, sayH. By Lemma 2.6.2, G = NH. Since also N E G and |N ∩ H| = 1 (byCorollary 2.1.9), we have that G is the internal semidirect product of N andH.

By Proposition 3.2.4, there is some σ ∈ Hom(H,Aut(N)) such thatG ∼= N oσ H. By Lemma 3.4.2, H ∼= G/N , and so Y ∼= G/N ∼= H. Therefore,by Proposition 3.2.8, G ∼= X oρ Y , where ρ = λα ◦ σ ◦ β−1. Here β : G/N 7→ Yis an isomorphism, and

λα : Aut(N) 7→ Aut(X); f 7→ α ◦ f ◦ α−1

where α : N 7→ X is an isomorphism. �

3.5. Proof of Theorem 3.3.3(iii)

The core proof of this part of the theorem has already been done, inLemma 3.2.7 and Lemma 3.2.8. We proved a stronger version of this thanwe required, and we may now reap the benefits of that.

Proof of Theorem 3.3.3(iii). Let X and Y be solvable groups of co-prime orders. Let σ, ρ ∈ Hom(Y,Aut(X)). Define Gσ = Xoσ Y , Gρ = Xoρ Y .Define Nσ = Nρ = (X, idY ) as sets, and Hσ = Hρ = (idX , Y ) as sets. ByProposition 3.2.6, Nσ and Hσ are subgroups of Gσ, Nσ

∼= X and Hσ∼= Y ,

and Gσ is the internal semidirect product of Nσ and Hσ. By Proposition 3.2.6again, Nρ and Hρ are subgroups of Gρ, Nρ

∼= X and Hρ∼= Y , and Gρ is the

internal semidirect product of Nρ and Hρ.( ⇐= ) Suppose σ and ρ lie in the same orbit of the ECS Action for X and

Y . Then in particular σ ∈ Orb(ρ), and so we have

σ = (α, β) ∗ ρ = λα ◦ ρ ◦ β−1

(where λα is defined as in the ECS Action, and ∗ denote the ECS Action itself),for some (α, β) ∈ Aut(X) × Aut(Y ). Therefore α ∈ Aut(X) and β ∈ Aut(Y ),and so by Proposition 3.2.8, we have that Gσ

∼= Gρ.( =⇒ ) Suppose Gσ

∼= Gρ. Then there exists some isomorphismψ : Gσ 7→ Gρ. We know that ψ could be split into component functionsψ1 and ψ2, however in general these functions will depend on both inputs, i.e.ψ1 = ψ1(x, y). We want to find an isomorphism such that that these componentfunctions in fact only depend on one of the inputs, so they become like α andβ above.

First of all, since Gρ is the internal semidirect product of Nρ and Hρ,then Nρ E Gρ. Also, Nρ is solvable, since Nρ

∼= X, and we can apply

3.5. PROOF OF THEOREM 3.3.3(III) 47

Lemma 1.2.12. The groupGρ/Nρ is solvable, because it is isomorphic toHρ∼= Y

(by Lemma 3.4.2), and we can again apply Lemma 1.2.12. Therefore, Gρ is solv-able by Proposition 1.4.3.

Now let π be the set of primes dividing Hσ. Then since |ψ(Hσ)| = |Hσ|,ψ(Hσ) is also a π-group. Also,

[Gρ : ψ(Hσ)] =|Gρ|

|ψ(Hσ)|=|X| · |Y ||Hσ|

=|X| · |Y ||Y |

= |X|.

Suppose p | [Gρ : ψ(Hσ)]. Then p divides the order of X, so it cannot divide theorder of Y , since X and Y have coprime orders. Therefore, it does not divide|Hσ| = |Y |, and so is not in π. So ψ(Hσ) is a Hall π-subgroup of Gρ. Since|ψ(Hσ)| = |Hσ| = |Hρ|, Hρ is also a Hall π-subgroup of Gρ. By Theorem 2.2.1,Hρ and ψ(Hσ) are conjugate by some element of Gρ, say a.

Define a map

φ : Gσ 7→ Gρ; t 7→ aψ(t)a−1.

φ is a bijection as its inverse is

f : Gρ 7→ Gσ; t 7→ a−1ψ−1(t)a

and is a homomorphism: take t1, t2 ∈ Gσ and then

φ(t1t2) = aψ(t1t2)a−1 = aψ(t1)ψ(t2)a−1 = aψ(t1)a−1aψ(t2)a−1 = φ(t1)φ(t2).

Therefore, it is an isomorphism. We also have that φ(Hσ) = aψ(Hσ)a−1 = Hρ.Now let π be redefined as the set of primes dividing Nσ. Then since

|φ(Nσ)| = |Nσ|, φ(Nσ) is also a π-group. Also,

[Gρ : φ(Nσ)] =|Gρ|

|φ(Nσ)|=|X| · |Y ||Nσ|

=|X| · |Y ||X|

= |Y |.

Suppose p | [Gρ : φ(Nσ)]. Then p divides the order of Y , so it cannot divide theorder of X, since X and Y have coprime orders. Therefore, it does not divide|Nσ| = |X|, and so is not in π. So φ(Nσ) is a Hall π-subgroup of Gρ. Since|φ(Nσ)| = |Nσ| = |Nρ|, Nρ is also a Hall π-subgroup of Gρ. By Theorem 2.2.1,φ(Nσ) and Nρ are conjugate by some element of Gρ, say a. But as Nρ is normalin Gρ, Nρ = aNρa

−1 = φ(Nσ).We have that, for all x ∈ X, y ∈ Y , φ(x, y) = (φ1(x, y), φ2(x, y)) for some

φ1 : X oσ Y 7→ X and φ2 : X oσ Y 7→ Y . Let

α : X 7→ X;x 7→ φ1(x, idY )

and

β : Y 7→ Y ; y 7→ φ2(idX , y).

Since φ(Nσ) = Nρ and φ(Hσ) = Hρ, we have that for all x ∈ X, y ∈ Y ,φ2(x, idY ) = idY and φ1(idX , y) = idX (the reader is advised to look at the


definitions of Nσ, Hσ, etc if still unsure). Now we have that, for all (x, y) ∈ Gσ,

φ(x, y) = φ((xσ(idY )(idX), idY y))

= φ((x, idY )(idX , y))

= φ(x, idY )φ(idX , y)

= (φ1(x, idY ), idY )(idX , φ2(idX , y))

= (α(x), idY )(idX , β(y))

= (α(x)σ(idY )(idX), idY β(y))

= (α(x) idX , β(y))

= (α(x), β(y))

We would like to show that α and β are automorphisms. First, they are 1-1since if α(x) = α(y) for some x, y ∈ X, then

φ(x, idY ) = (φ1(x, idY ), φ2(x, idY ))

= (φ1(y, idY ), idY )

= (φ1(y, idY ), φ2(y, idY ))

= φ(y, idY )

and so (x, idY ) = (y, idY ) (since φ is 1-1), and so x = y. If β(x) = β(y) forsome x, y ∈ Y , then

φ(idX , x) = (φ1(idX , x), φ2(idX , x))

= (idX , φ2(idX , y))

= (φ1(idX , y), φ2(idX , y))

= φ(idX , y)

and so (idX , x) = (idX , y) (since φ is 1-1), and so x = y. Hence α and β are1-1. Since their domain and codomains are equal, they have the same size, andso α and β are bijections, hence onto.

We shall now show that α and β are homomorphisms. Take x, y ∈ X. Then:

(α(xy), idY ) = (φ1(xy, idY ), φ2(xy, idY ))

= φ(xy, idY )

= φ(xσ(idY )(y), idY idY )

= φ((x, idY )(y, idY ))

= φ(x, idY )φ(y, idY )

= (φ1(x, idY ), φ2(x, idY ))(φ1(y, idY ), φ2(y, idY ))

= (α(x), idY )(α(y), idY )

= (α(x)ρ(idY )(α(y)), idY idY )

= (α(x)α(y), idY )

3.5. PROOF OF THEOREM 3.3.3(III) 49

and so α(xy) = α(x)α(y), thus α is a homomorphism, hence an isomorphism.Now take any x, y ∈ Y .

(idX , α(xy)) = (φ1(idX , xy), φ2(idX , xy))

= φ(idX , xy)

= φ(idX σ(x)(idX), xy)

= φ((idX , x)(idX , y))

= φ(idX , x)φ(idX , y)

= (φ1(idX , x), φ2(idX , x))(φ1(idX , y), φ2(idX , y))

= (idX , β(x))(idX , β(y))

= (idX ρ(β(x))(idX), β(x)β(y))

= (idX idX , β(x)β(y))

= (idX , β(x)β(y))

and so β(xy) = β(x)β(y), thus β is a homomorphism, hence an isomorphism.Since φ is an isomorphism, it is a homomorphism, and so by Lemma 3.2.7,

ρ = λα ◦ σ ◦ β−1, where λα is defined as in the ECS Action. Therefore,ρ = (α, β) ∗ σ, where ∗ denotes the ECS Action. Thus ρ ∈ Orb(σ), and so ρand σ lie in the same orbit under the ECS Action. �

CHAPTER 4

Some groups of order pqr

4.1. Classification of automorphisms of the Cyclic Group

We wish to classify the automorphisms of the cyclic group of order n, Cn.The following result is a generalisation of our results in [1], Section 3.

Proposition 4.1.1. Suppose G is a cyclic group of order n, say G = 〈c〉where c has order n. Then

Aut(G) = {(β : G 7→ G) : (∃i : β(cr) = cri, 1 ≤ i < n, (n, i) = 1)}

Proof. Suppose G is a cyclic group of order n, say G = 〈c〉 where c hasorder n. Let

X = {(β : G 7→ G) : (∃i : β(cr) = cri, 1 ≤ i < n, (n, i) = 1)}

for brevity.(⊇) Let α ∈ X. Then ∃i : α(cr) = cri, (n, i) = 1. α is a homomorphism,

sinceα(crcs) = cricsi = cri+si = c(r+s)i = α(cr+s)

Now suppose α(cr) = α(cs), then cri = csi, so ri = si+kn for some k, so i | kn,so i | k, as (n, i) = 1. Therefore,

cr = cs+kin = cs

(noting here that i 6= 0) and so α is injective. Finally, if y ∈ G then byLemma 0.2 of [1], we have that (ci)k = y, so that if we set x = ck thenα(x) = α(ck) = (ck)i = (ci)k = y. So α is an automorphism of G and soα ∈ Aut(G). As α was an arbitrary element of X, we have that X ⊆ Aut(G).

(⊆) Let α ∈ Aut(G). Then α(c) ∈ G, so α(c) = ci for some 0 ≤ i < n, as cgenerates the group. This i 6= 0, otherwise we would have

α(c) = c0 = idG = α(idG)

and so α would not be 1-1. So 1 ≤ i < n. We have that α(cr) = (α(c))r =(ci)r = cri. It remains to show that (n, i) = 1. Suppose not, and let a | n, a | i,where a > 1. Now i = am, say, and n = ar. Note r | n so r < n, and r 6= 1otherwise n = a so n | i, but this is a contradiction as i < n. So

α(cr) = cri = cram = cnm = (cn)m = (idG)m = idG = α(idG)

but since 1 < r < n, we have cr 6= idG, so α is not injective and so cannot bean automorphism. �

51

52 4. SOME GROUPS OF ORDER pqr

4.2. The case where Y is cyclic of prime order

Computation of the set Hom(Y,Aut(X)) is crucial for any application ofTheorem 3.3.3. However, in general this set is quite hard to compute. Forour purposes, the case where Y is cyclic of prime order shall be quite impor-tant, as the remark immediately after Corollary 3.3.4 shows. In this case, thecomputation of the set is simplified a little.

Proposition 4.2.1. Suppose X is a solvable group of order n, and Y is acyclic (and therefore Abelian, therefore solvable) group of prime order p, suchthat p does not divide n (so that X and Y have coprime orders). Let c be agenerator of Y . Then the map

φ : Hom(Y,Aut(X)) 7→ {f ∈ Aut(X) : fp = idAut(X)}; g 7→ g(c)

where fp denotes f composed with itself p times, is a bijection.

Proof. Let X be a solvable group of order n, and Y a cyclic group of primeorder p, such that p does not divide n. Let c be a generator of Y , and define

S = {f ∈ Aut(X) : fp = idAut(X)}

and

φ : Hom(Y,Aut(X)) 7→ S; g 7→ g(c)

where fp denotes f composed with itself p times.This map is well-defined, since for any g ∈ Hom(Y,Aut(X)), g(c) ∈ Aut(X),

and also

(g(c))p = g(cp) = g(idY ) = idAut(X)

as a homomorphism must map the identity to the identity. Therefore,φ(g) = g(c) is in S.

We shall show that φ is 1-1. Let g1, g2 ∈ Hom(Y,Aut(X)) be such thatφ(g1) = φ(g2). Then g1(c) = g2(c). Take any y ∈ Y , then y = ci for someinteger i. Now

g1(y) = g1(ci) = (g1(c))i = (g2(c))i = g2(ci) = g2(y)

so that g1 = g2 (since y was arbitrary). Therefore, φ is 1-1.We show that φ is onto. Let f ∈ S. Define the map

g : Y 7→ Aut(X); ci 7→ f i

where ci denotes the unique such representation with i an integer and 0 ≤i < p. Clearly φ(g) = g(c) = g(c1) = f1 = f . We need to show that g is ahomomorphism. Take y1, y2 ∈ Y , then y1 = ci and y2 = cj for some uniqueintegers i, j with 0 ≤ i < p, 0 ≤ j < p. Now if i+ j < p then

g(y1y2) = g(cicj) = g(ci+j) = f i+j = f if j = g(y1)g(y2).

4.3. CLASSIFICATION OF GROUPS OF ORDER pq 53

But if i + j ≥ p then certainly i + j < 2p, so i + j = p + r for some r with0 ≤ r < p. In this case,

g(y1y2) = g(cicj)

= g(ci+j)

= g(cp+r)

= g(cpcr)

= g(cr)= f r

= fpf r

= fp+r

= f i+j

= f if j

= g(y1)g(y2).

In either case, g is a homomorphism, and so φ is onto.Therefore, φ is a bijection. �

Example 4.2.2 (Groups of order 2qr). Let G be a group of order 2qr. LetX = Cqr and Y = C2. Y is Abelian therefore solvable, and X is solvable byProposition 1.5.2. Furthermore, (2, qr) = 1. Applying the Extension Classifi-cation Schema, G corresponds to a homomorphism from Y to Aut(X), and byProposition 4.2.1, the set Hom(Y,Aut(X)) is in 1-1 correspondence with theset of automorphisms σ of X which have the property σ2 = idAut(X). Thusfor any G we get an automorphism of X, and G is uniquely determined bythis automorphism. However the automorphism does not uniquely determineG - if any two σ1 and σ2 produce isomorphic groups, then the correspondinghomomorphisms ρ1 and ρ2 must be in the same orbit of the ECS Action. Ifthis happens, then we must have ρ2 = λα ◦ ρ1 ◦ β−1 for some automorphismsα : X 7→ X and β : Y 7→ Y . However any automorphism of Y = C2 must bethe identity, so we get that ρ2 = λα ◦ ρ1, or in other words, ρ2 = α ◦ ρ1 ◦ α−1.Since the ρi are homomorphisms, this property must translate to the automor-phisms σi, in other words, σ1 and σ2 give rise to isomorphic groups iff they areconjugate by some element of Aut(X), i.e. if they are conjugate in Aut(X).

4.3. Classification of groups of order pq

We state an initial lemma concerning composition of automorphisms ofcyclic groups.

Lemma 4.3.1. Suppose that G is a cyclic group with order n and generatorg, that f ∈ Aut(G), that i is an integer such that 0 ≤ i < n and f(g) = gi, andthat m is a non-negative integer. Then

fm(x) = xim

for all x ∈ G.


Proof. Suppose that G is a cyclic group with order n and generator g,that f ∈ Aut(G), and that i is an integer such that 0 ≤ i < n and f(g) = gi.

We shall prove the statement “fm(x) = xim for all x ∈ G” by induction, forall integers m ≥ 0. The basis case is easy, since f0(x) = x = x1 = xi0 .

Suppose the statement is true for m = k. Take any x ∈ G. Then x = gr forsome r with 0 ≤ r < n and

fk+1(x) = f(fk(x))

= f(xik)

= f((gr)ik)

= f(grik)

= f(g)rik

= (gi)rik

= grik+1

= (gr)ik+1

= xik+1

so that the statement is true for m = k + 1.By the principle of induction, the statement holds for all integers m ≥ 0,

i.e. the lemma is true. �

When we have a group of order pq, X is also cyclic, and so we have a muchsimpler situation again. However we shall prove the following results in thecase where X is just cyclic, not necessarily of prime order. As we have donebefore, we shall split the proof into two results, so that we can use the core ofthe argument again later under different circumstances.

Lemma 4.3.2. Suppose that X is a cyclic group (therefore Abelian, thereforesolvable) of order n and with generator d. For any x ∈ X, let I(x) denote theunique i with 0 ≤ i < n such that x = di. Define

S = {f ∈ Aut(X) : fp = idAut(X)}T = {i ∈ Z : 1 ≤ i < n, (n, i) = 1, ip ≡ 1 (modn)}

where fp denotes f composed with itself p times. Then the map

φ : S 7→ T ; f 7→ I(f(d))

is a bijection.

Proof. Suppose that X is a cyclic group of order n and with generator d.For any x ∈ X, let I(x) denote the unique i with 0 ≤ i < n such that x = di.Define



where fp denotes f composed with itself p times, and

φ : S 7→ T ; f 7→ I(f(d)).

First of all φ is well-defined. Take any f ∈ S and let i = I(f(d)), clearly aninteger with 0 ≤ i < n. By Proposition 4.1.1, we have (n, i) = 1 and 1 ≤ i < nfrom the fact that f ∈ Aut(X). By Lemma 4.3.1, we have fp(d) = dip , but asf ∈ S we also have fp(d) = d. Therefore, dip−1 is equal to the identity, so ip−1must be a multiple of n, the order of d. Therefore, i ≡ 1 (modn), and so i ∈ T .

Next φ is 1-1. Suppose f, g ∈ S and φ(f) = φ(g). Then f(d) = gi forsome i with 0 ≤ i < n and g(d) = gj for some j with 0 ≤ j < n. But i = j,so f(d) = g(d). Now let x ∈ X be arbitrary. Then x = dr for some r with0 ≤ r < n. Now

f(x) = f(dr) = f(d)r = g(d)r = g(dr) = g(x)

and so f = g. Hence φ is 1-1.Finally, we show φ is onto. Suppose i ∈ T . Define the map

f : X 7→ X; dr 7→ dir.

As i ∈ T , Proposition 4.1.1 gives that f is an automorphism. Also, if x is anyelement of X then x = dr for some r with 0 ≤ r < n. Then, by Lemma 4.3.1,fp(x) = xip = x, since ip ≡ 1 (modn). Therefore, f ∈ S, and so φ is onto.

Therefore, φ is a bijection. �

Proposition 4.3.3. Suppose that X is a cyclic group (therefore Abelian,therefore solvable) of order n and with generator d, and Y is a cyclic group(therefore Abelian, therefore solvable) of prime order p and with generator c,such that p does not divide n (so that X and Y have coprime orders). For anyx ∈ X, let I(x) denote the unique i with 0 ≤ i < n such that x = di. Then themap

φ : Hom(Y,Aut(X)) 7→ {i ∈ Z : 1 ≤ i < n, (n, i) = 1, ip ≡ 1 (modn)};g 7→ I(g(c)(d))

is a bijection.

Proof. Suppose that X is a cyclic group of order n and with generator d,and Y is a cyclic group of prime order p and with generator c, such that p doesnot divide n. For any x ∈ X, let I(x) denote the unique i with 0 ≤ i < n suchthat x = di. Define


where fp denotes f composed with itself p times, and

φ : Hom(Y,Aut(X)) 7→ T ; g 7→ I(g(c)(d)).

Now letφ1 : Hom(Y,Aut(X)) 7→ S; g 7→ g(c)

andφ2 : S 7→ T ; f 7→ I(f(d))


so that φ = φ2 ◦ φ1. Since X is a cyclic group, it is Abelian and thereforesolvable, and so Proposition 4.2.1 gives that φ1 is a bijection. By Lemma 4.3.2,φ2 is a bijection. Therefore, φ = φ2 ◦ φ1 is a bijection. �

Remark. This proposition shows that any element σ ∈ Hom(Y,Aut(X))is uniquely determined by its value σ(c)(d). For given some x ∈ X, we canconstruct an element of Hom(Y,Aut(X)) which has σ(c)(d) is equal to x, as φis onto. Furthermore, if σ1 and σ2 are two elements of Hom(Y,Aut(X)) withthe same value σ(c)(d), then they are in fact equal, as φ is 1-1.

The set constructed above is in fact rather special: it forms a group.

Lemma 4.3.4. Let p, q be primes, p < q. Then the set

G = {i ∈ Z : 1 ≤ i < q, ip ≡ 1 (mod q)}

forms a group with respect to multiplication modulo q. Furthermore, this groupis cyclic, and is either trivial or has order p.

Proof. Let p, q be primes, p < q, and

G = {i ∈ Z : 1 ≤ i < q, ip ≡ 1 (mod q)}

under the operation of multiplication modulo q.First, G is closed, as if i, j ∈ G then ip ≡ 1 (mod q) and jp ≡ 1 (mod q), so

(ij)p = ipjp ≡ 1 (mod q).Secondly, the operation is associative on G, as modular multiplication is

associative.Thirdly, G contains an identity, namely 1, and 1i = i1 = i for i ∈ G under

normal multiplication, therefore this remains true under modular multiplica-tion.

Finally, if i ∈ G then iq−1i = iiq−1 = 1 under normal multiplication, there-fore also under modular multiplication, and so i has an inverse (namely iq−1).

G is cyclic since it is a subgroup of Z∗q , a cyclic group.

If G is non-trivial then there is some non-identity i ∈ G and ip ≡ 1 (mod q),that is, ip = 1 in the group. The order of i must therefore divide p, so must be1 or p. Since i is not the identity, it does not have order 1, so it has order p.The order of i divides the order of the group, so the order of the group is atleast p.

Since i was arbitrary, we have shown that every non-identity element hasorder p. Since G is cyclic, it must contain an element with order equal to theorder of the group. The maximum order of elements is p, so G must have orderat most p. Therefore, |G| = p. �

We shall also need a result on cyclic groups of squarefree order.

Lemma 4.3.5. Let p1, ..., pr be primes, then

Cp1 × Cp2 × · · · × Cpr∼= Cp1p2···pr .

Proof. Let p1, ..., pr be primes, and let

G = Cp1 × Cp2 × · · · × Cpr


Further, letn = |G| = p1p2 · · · pr

We need only show that the group G is cyclic. Then it must be isomorphic tothe unique (up to isomorphism) cyclic group of order n.

Suppose ai ∈ Cpi are generators. Let

a = (a1, a2, ..., ar) ∈ G.Now we have that

an = (an1 , a

n2 , ..., a

nr )

= ((ap11 )

np1 , (ap2

1 )np2 , ..., (apr

r )npr )

= (idnp1Cp1

, idnp2Cp2

, ..., idnprCpr

)

= (idCp1, idCp2

, ..., idCpr)

= idG

and so the order of a must divide n. However divisors of n have the formm = q1q2 · · · qs where each qi = pj for some j, and no two qi are equal. Wecertainly have m ≤ n. If m < n then there must exist some k such that pk doesnot divide m. Now if am = idG then

idG = (idCp1, idCp2

, ..., idCpr) = am = (am

1 , am2 , ..., a

mr )

and so amk = idCpk

. However as ak generates Cpk, it has order pk, so pk must

dividem, a contradiction. Therefore n = |G| is the order of a, and so a generatesG, hence G is cyclic. �

We are now in a position to apply the Extension Classification Schema toclassify all groups of order pq.

Proposition 4.3.6. Let p and q be primes with p < q.(i) If there does not exist an integer i with 1 < i < q and ip ≡ 1 (mod q)

then any group of order pq is isomorphic to the cyclic group Cpq.(ii) If there exists an integer i with 1 < i < q and ip ≡ 1 (mod q) then

define a map

σ : Cp 7→ Aut(Cq); ar 7→ fir

where a is a generator of Cp, and

fj : Cq 7→ Cq; bs 7→ bjs

where b is a generator of Cq. Then “Cpq, Cq oσ Cp” is a list of groupsof order pq up to isomorphism.

Proof. Let p and q be primes with p < q. Let G be any group of order pq.Then, by Proposition 1.5.1, G has a normal subgroup of index p, say N . LetX = Cq (with generator b) and Y = Cp (with generator a). X and Y are bothcyclic, therefore Abelian, therefore solvable. They also have coprime orders,since p and q are primes with p 6= q. We have that

|N | = |G||G||N |

=|G|

[G : N ]=pq

p= q


and so N ∼= X. Furthermore, |G/N | = [G : N ] = p, and so G/N ∼= Y .Let

S = {i ∈ Z : 1 ≤ i < q, ip ≡ 1 (mod q)}.By Proposition 4.3.3, Hom(Y,Aut(X)) has the same size as S (we have (i, q) = 1for all i ∈ S, because q is prime and i < q).

(i) Suppose that there does not exist an integer i with 1 < i < q andip ≡ 1 (mod q). Then there cannot exist any elements in S which are not equalto 1. But in fact 1 ∈ S, because 1 ≤ 1 < q, (1, q) = 1 and 1p = 1 ≡ 1 (mod q).Therefore, S contains only 1 element, namely 1. So

|Hom(Y,Aut(X))| = |S| = 1.

By Theorem 3.3.3(ii), G ∼= XoσY , for some σ ∈ Hom(Y,Aut(X)). Howeveralso Cpq

∼= X oρ Y for some ρ ∈ Hom(Y,Aut(X)). Since Hom(Y,Aut(X)) hasonly one element, we must have σ = ρ, and so G ∼= X oσ Y = X oρ Y ∼= Cpq.

(ii) Suppose that there exists an integer i with 1 < i < q and ip ≡ 1 (mod q).Define

fj : X 7→ X;x 7→ xj

for any j, as in the statement of the result, and let

σi : Y 7→ Aut(X); ak 7→ fik

for any integer i.Now take any r, s such that 1 < r < q, 1 < s < q and rp ≡ 1 (mod q),

sp ≡ 1 (mod q). By Theorem 3.3.3(iii), σr and σs give rise to isomorphic groupsX oσr Y and X oσs Y if and only if

σs = λα ◦ σr ◦ β−1 (†)for some α ∈ Aut(X), β ∈ Aut(Y ). By Proposition 4.3.3, σr, σs are com-pletely determined by their values σ(a)(b). By Proposition 4.1.1, α and β arecompletely determined by their values at b and a respectively. Therefore, letβ(a) = ai and α(b) = bj (by Proposition 4.1.1 we also have that 1 ≤ i < p,1 ≤ j < q). The formula (†) is then equivalent to:

bs = σs(a)(b)

= (λα(σr(β−1(a))))(b)

= (λα(σr(ap−i)))(b)

= (λα(frp−i))(b)

= α(frp−i(α−1(b)))

= α(frp−i(bq−j))

= α(brp−i(q−j))

= ((brp−i

)q−j)j

= brp−i

which is in turn equivalent to s = rp−i (mod q). So σr and σs give rise toisomorphic groups if and only if there exists some i with 1 ≤ i < p such thats = rp−i (mod q), in other words, if there exists some i with 0 < i ≤ p − 1


such that s = ri (mod q). However, by Lemma 4.3.4, this is true if r, s 6= 1,the identity of the group, as then r must generate the group. This is true byhypothesis, and so σr and σs give rise to isomorphic groups. Therefore, there isat most two groups of order pq in this case. However in fact, there are exactlytwo, as we cannot have i such that s = 1i for some s 6= 1. Therefore, the r, sabove give rise to a different group from 1. Let σ = σr as in the statementof the result, and θ = σ1, the trivial map corresponding to 1. By applyingTheorem 3.3.3(i) and Theorem 3.3.3(ii), we obtain that “X oθ Y , X oσ Y ” is acomplete list of groups of order pq up to isomorphism (because any such groupmust be isomorphic to some X oρ Y for ρ ∈ Hom(Y,Aut(X)), and this groupis isomorphic to either X oθ Y , if ρ = θ, or X oσ Y , otherwise).

It remains to show that X oθ Y is isomorphic to the cyclic group of orderpq. First note that this group has underlying set X ×Y , and if we take (x1, y1)and (x2, y2) in the group, then the operation is

(x1, y1)(x2, y2) = (x1θ(y1)(x2), y1y2) = (x1f1(x2), y1y2) = (x1x2, y1y2)

in other words, the group is isomorphic to the direct product X×Y = Cq×Cp.By Lemma 4.3.5, this group is cyclic of order qp = pq. �

We may indeed get a much easier condition for the previous result to hold.For the proof of the following I was helped by Bart Goddard ([3]).

Lemma 4.3.7. Suppose p and q are primes. Then there exists i with1 < i < q and ip ≡ 1 (mod q) iff p | q − 1.

Proof. Suppose p and q are primes. Let G = {1, 2, ..., q − 1} under multi-plication modulo q.

( =⇒ ) Suppose there exists i with 1 < i < q and ip ≡ 1 (mod q). Theni ∈ G, and ip ≡ 1 within the group. Therefore, the order of i is at most p.Suppose the order is m. Then m | p, but m 6= 1 otherwise i = 1. Therefore,m = p. Since the order of i divides the order of the group, we have that p | q−1.

( ⇐= ) Suppose p | q − 1. Then G contains an element of order p byCauchy’s Group Theorem. This element, say i, has ip ≡ 1 (mod q). Also i 6= 1,since otherwise i would have order 1. �

The proof of the following involves an easy combination of the above tworesults, and so is omitted.

Proposition 4.3.8. Let p and q be primes with p < q.(i) If p - q−1 then any group of order pq is isomorphic to the cyclic group

Cpq.(ii) If p | q−1 then by Lemma 4.3.7, there exists an integer i with 1 < i < q

and ip ≡ 1 (mod q). Define a map

σ : Cp 7→ Aut(Cq); ar 7→ fir

where a is a generator of Cp, and

fj : Cq 7→ Cq; bs 7→ bjs

where b is a generator of Cq. Then “Cpq, Cq oσ Cp” is a list of groupsof order pq up to isomorphism.


4.4. The case where X is cyclic of squarefree order

The initial lemmas of this section concern the automorphism group of aproduct. The first of these is simply an intermediate step of the second, which isconvenient to give outside of the main body of the proof for notational purposes.

Lemma 4.4.1. Let G1, G2, ..., Gr be groups of coprime orders n1, n2, ..., nr,and G = G1 ×G2 × · · · ×Gr. For each i, let

Hi = {(idG1 , idG2 , ..., idGi−1 , x, idGi+1 , ..., idGr) : x ∈ Gi}

andJi = {x ∈ G : xni = idG}.

Then Hi = Ji.

Proof. Let G1, G2, ..., Gr be groups of coprime orders n1, n2, ..., nr, andG = G1 ×G2 × · · · ×Gr. For each i, let


andJi = {x ∈ G : xni = idG}.

(⊆) Let i be an integer and h ∈ Hi. Then hnii = idGi , since hi ∈ Gi and

|Gi| = ni. Therefore,

hni = (idG1 , idG2 , ..., idGi−1 , x, idGi+1 , ..., idGr)ni

= (idniG1, idni

G2, ..., idni

Gi−1, x, idni

Gi+1, ..., idni

Gr)

= (idG1 , idG2 , ..., idGi−1 , idGi , idGi+1 , ..., idGr)= idG

so that h ∈ Ji.(⊇). Let i be an integer and j ∈ Ji. Then jni = idG, so

(jni1 , j

ni2 , ..., j

nir ) = (j1, j2, ..., jr)ni

= jni

= idG

= (idG1 , idG2 , ..., idGr)

and so jnik = idGk

for each integer k.Take an integer k 6= i. Now jni

k = idGk, so the order of jk must divide ni.

However certainly the order of jk divides |Gk| = nk. But nk and ni are coprime(as k 6= i), therefore the order of jk must be 1. Thus jk = idGk

for each k 6= i,that is, j has the form

j = (idG1 , idG2 , ..., idGi−1 , ji, idGi+1 , ..., idGr)

or in other words, j ∈ Hi. �

Lemma 4.4.2. Let G1, G2, ..., Gr be groups of coprime orders. Then

Aut(G1 ×G2 × · · · ×Gr) = Aut(G1)×Aut(G2)× · · · ×Aut(Gr).

4.4. THE CASE WHERE X IS CYCLIC OF SQUAREFREE ORDER 61

Proof. Let G1, G2, ..., Gr be groups of coprime orders n1, n2, ..., nr, andG = G1 ×G2 × · · · ×Gr.

(⊇) Let f ∈ Aut(G1) × Aut(G2) × · · · × Aut(Gr). Then f = (f1, f2, ..., fr)where each fi ∈ Aut(Gi). We regard f as a function from G to G, by

f(g1, g2, ..., gr) = (f1(g1), f2(g2), ..., fr(gr)).

We must check that f is an automorphism of G. Clearly its domain andcodomain have the same order, so we need only check that it is onto to seeit is a bijection. Take some y ∈ G, then

y = (y1, y2, ..., yr)

for yi ∈ Gi. Now each yi = fi(xi) for some xi ∈ Gi, since fi is onto. Let

x = (x1, x2, ..., xr)

then

f(x) = f(x1, x2, ..., xr) = (f1(x1), f2(x2), ..., fr(xr)) = (y1, y2, ..., yr) = y

and so f is onto, hence a bijection.Finally we show that f is a homomorphism. Take x, y ∈ G, then

x = (x1, x2, ..., xr)

andy = (y1, y2, ..., yr).

Now

f(xy) = f((x1, x2, ..., xr)(y1, y2, ..., yr))

= f(x1y1, x2y2, ..., xryr)

= (f1(x1y1), f2(x2y2), ..., fr(xryr))

= (f1(x1)f1(y1), f2(x2)f2(y2), ..., fr(xr)fr(yr))

= (f1(x1), f2(x2), ..., fr(xr))(f1(y1), f2(y2), ..., fr(yr))

= f(x1, x2, ..., xr)f(y1, y2, ..., yr)

= f(x)f(y)

so that f is a homomorphism.Altogether, f is an automorphism of G.(⊆) Let f ∈ Aut(G). For each i, let


andJi = {x ∈ G : xni = idG}

Then Hi = Ji, by Lemma 4.4.1.We may split f into components fi : G 7→ Gi, so that

f(x1, x2, ..., xr) = (f1(x1, x2, ..., xr), f2(x1, x2, ..., xr), ..., fr(x1, x2, ..., xr)).

Take any x ∈ G. We have that

x = y1y2 · · · yr


where

yi = (idG1 , idG2 , ..., idGi−1 , xi, idGi+1 , ..., idGr).

We can deduce yi ∈ Hi, so yi ∈ Ji, so ynii = idG. We have that

f(yi)ni = f(ynii ) = f(idG) = idG

so f(yi) ∈ Ji, so f(yi) ∈ Hi. Therefore, f(yi) has the form

(f1(yi), f2(yi), ..., fr(yi)) = f(yi) = (idG1 , idG2 , ..., idGi−1 , zi, idGi+1 , ..., idGr)

for some zi ∈ Gi, so that fk(yi) = idGkfor k 6= i. Now,

(f1(x1, x2, ..., xr), f2(x1, x2, ..., xr), ..., fr(x1, x2, ..., xr))

= f(x1, x2, ..., xr)

= f(x)

= f(y1y2 · · · yr)

= f(y1)f(y2) · · · f(yr)

= (f1(y1), f2(y1), ..., fr(y1))

(f2(y1), f2(y2), ..., fr(y2))· · ·(f1(yr), f2(yr), ..., fr(yr))

= (f1(y1), idG2 , ..., idGr)

(idG1 , f2(y2), idG3 , ..., idGr)· · ·(idG1 , idG2 , ..., idGr−1 , fr(yr))

= (f1(y1), f2(y2), ..., fr(yr))

= (f1(x1, idG2 , ..., idGr),

f2(idG1 , x2, idG3 , ..., idGr),...,

fr(idG1 , idG2 , ..., idGr−1 , fr(yr)))

so that for each i,

fi(x1, x2, ..., xr) = fi(idG1 , idG2 , ..., idGi−1 , xi, idGi+1 , ..., idGr)

no matter what xj equals for i 6= j. In other words, fi does not depend on xj

for j 6= i, and we can simply write

fi(x1, x2, ..., xr) = fi(xi)

and regard fi as a function fi : Gi 7→ Gi.


Now take xi, yi ∈ Gi for each i and let x = (x1, x2, ..., xr) andy = (y1, y2, ..., yr). Then

(f1(x1y1), f2(x2y2), ..., fr(xryr)) = (f1((xy)1), f2((xy)2), ..., fr((xy)r))

= f(xy)

= f(x)f(y)

= (f1(x1), f2(x2), ..., fr(xr))

(f1(y1), f2(y2), ..., fr(yr))

= (f1(x1)f1(y1), f2(x2)f2(y2), ..., fr(xr)fr(yr))

and so for each i,fi(xiyi) = fi(xi)fi(yi)

therefore, the fi are homomorphisms.We have that

(f1(idG1), f2(idG2), ..., fr(idGr)) = f(idG1 , idG2 , ..., idGr)

= f(idG)= idG

= (idG1 , idG2 , ..., idGr)

so that fi(idGi) = idGi for each i.Suppose that fi(x) = fi(y) for some integer i and x, y ∈ Gi. Let

g = (idG1 , idG2 , ..., idGi−1 , x, idGi+1 , ..., idGr)

andh = (idG1 , idG2 , ..., idGi−1 , y, idGi+1 , ..., idGr).

Now it is certainly true that

(idG1 , idG2 , ..., idGi−1 , fi(x), idGi+1 , ..., idGr)

= (idG1 , idG2 , ..., idGi−1 , fi(y), idGi+1 , ..., idGr)

and so f(g) = f(h). Therefore g = h, that is, x = y. Thus fi is 1-1 for each i.Let y ∈ Gi for some integer i. Define

h = (idG1 , idG2 , ..., idGi−1 , y, idGi+1 , ..., idGr).

There is some g ∈ G with f(g) = h. Now let x = gi, so that

(f1(g1), f2(g2), ..., fi−1(gi−1), fi(x), fi+1(gi+1), ..., fr(gr)) = f(g) = h

that is, fi(x) = y. Therefore fi is onto for each i.Altogether we have shown that for each i, fi is an automorphism of Gi. But

by definition f = (f1, f2, ..., fr), so

f ∈ Aut(G1)×Aut(G2)× · · · ×Aut(Gr)

as required. �

The following is a generalisation of Proposition 4.3.3, Lemma 4.3.4 and ourresults on two homomorphisms being in the same orbit, taken from the proofof Proposition 4.3.6. Because the pqr case is more notationally complicated,we split this off from the main theorem (which was not so necessary with the


pq case). However both directions of implication for part (iii) of the belowproposition rely on the same algebraic lemma, which we shall state and provebefore the main proposition.

Lemma 4.4.3. Let X be a cyclic group of squarefree order n = p1p2 · · · pr

with generator d, and let Y be a cyclic group of prime order p with generatorc. Let

H = Cp1 × Cp2 × · · · × Cpr .

Let hi be a generator of Cpi for each i, chosen such that

d = ψ−1(h1, h2, ..., hr).

Define for each i, j the function

fi,j : Cpi 7→ Cpi ;x 7→ xj

Let α : X 7→ X and β : Y 7→ Y be automorphisms, and let ψ : X 7→ H be anisomorphism. Let σ1, σ2 ∈ Hom(Y,Aut(X)), and define r-tuples (s1, s2, ..., sr)and (t1, t2, ..., tr) such that

σ1(c) = ψ ◦ (f1,s1 , f2,s2 , ..., fr,sr) ◦ ψ−1

andσ2(c) = ψ ◦ (f1,t1 , f2,t2 , ..., fr,tr) ◦ ψ−1

which we can do by Proposition 4.1.1, Lemma 4.4.2 and Lemma 4.3.5. Also let

λα : Aut(X) 7→ Aut(X); f 7→ α ◦ f ◦ α−1.

Let i be the unique integer such that 1 ≤ i < p and β−1(c) = ci. Then(i)

(ht11 , h

t22 , ..., h

trr ) = ψ(σ2(c)(d))

(ii)

(hsi1

1 , hsi2

2 , ..., hsir

r ) = ψ((λα(σ1(β−1(c))))(d)).

Proof. Let X be a cyclic group of squarefree order n = p1p2 · · · pr withgenerator d, and let Y be a cyclic group of prime order p with generator c. Let

H = Cp1 × Cp2 × · · · × Cpr .

Let hi be a generator of Cpi for each i, chosen such that

d = ψ−1(h1, h2, ..., hr)

which we can do (for the details, see the proof of Lemma 4.3.5). Define for eachi, j the function


Let α : X 7→ X and β : Y 7→ Y be automorphisms, and let ψ : X 7→ H be anisomorphism. Let σ1, σ2 ∈ Hom(Y,Aut(X)), and define r-tuples (s1, s2, ..., sr)and (t1, t2, ..., tr) such that

σ1(c) = ψ ◦ (f1,s1 , f2,s2 , ..., fr,sr) ◦ ψ−1



which we can do by Proposition 4.1.1, Lemma 4.4.2 and Lemma 4.3.5. Also let

λα : Aut(X) 7→ Aut(X); f 7→ α ◦ f ◦ α−1.

Finally, let i be the unique integer such that 1 ≤ i < p and β−1(c) = ci.(i)

(ht11 , h

t22 , ..., h

trr ) = (f1,t1(h1), f2,t2(h2), ..., fr,tr(hr))

= (f1,t1 , f2,t2 , ..., fr,tr)(h1, h2, ..., hr)

= (ψ ◦ σ2(c) ◦ ψ−1)(h1, h2, ..., hr)

= ψ(σ2(c)(ψ−1(h1, h2, ..., hr)))

= ψ(σ2(c)(d))

(ii) Let j be the unique integer such that α(d) = dj and k be the uniqueinteger such that α−1(d) = dk. Now djk = α(α−1(d)) = d. This means that forany x ∈ X, which must be a power of d, we have that xjk = x. By applyingψ to both sides, the same is true in H: raising elements of H to the power jkleaves them unchanged.

(hsi1

1 , hsi2

2 , ..., hsir

r ) = (hsi1

1 , hsi2

2 , ..., hsir

r )jk

= (hksi1

1 , hksi

22 , ..., hksi

rr )j

= ψ(ψ−1((hksi1

1 , hksi

22 , ..., hksi

rr )j))

= ψ((ψ−1(hksi1

1 , hksi

22 , ..., hksi

rr ))j)

= ψ(α(ψ−1(hksi1

1 , hksi

22 , ..., hksi

rr )))

= ψ((α ◦ ψ−1)(hksi1

1 , hksi

22 , ..., hksi

rr ))

= ψ((α ◦ ψ−1)(f i1,s1

(hk1), f

i2,s2

(hk2), ..., f

ir,sr

(hkr )))

= ψ((α ◦ ψ−1)(f i1,s1

, f i2,s2

, ..., f ir,sr

)(hk1, h

k2, ..., h

kr ))

= ψ((α ◦ ψ−1 ◦ (f1,s1 , f2,s2 , ..., fr,sr)i)(h1, h2, ..., hr)k)

= ψ((α ◦ ψ−1 ◦ (f1,s1 , f2,s2 , ..., fr,sr)i)(ψ(d))k)

= ψ((α ◦ ψ−1 ◦ (f1,s1 , f2,s2 , ..., fr,sr)i ◦ ψ)(dk))

= ψ((α ◦ (ψ−1 ◦ (f1,s1 , f2,s2 , ..., fr,sr) ◦ ψ)i)(dk))

= ψ((α ◦ (ψ−1 ◦ (f1,s1 , f2,s2 , ..., fr,sr) ◦ ψ)i)(α−1(d)))

= ψ((α ◦ (ψ−1 ◦ (f1,s1 , f2,s2 , ..., fr,sr) ◦ ψ)i ◦ α−1)(d))

= ψ((λα((ψ−1 ◦ (f1,s1 , f2,s2 , ..., fr,sr) ◦ ψ)i))(d))

= ψ((λα(σ1(c))i)(d))

= ψ((λα(σ1(ci)))(d))

= ψ((λα(σ1(β−1(c))))(d))

�


Proposition 4.4.4. Suppose that X is a cyclic group of squarefree ordern = p1p2 · · · pr with generator d, and that Y is a cyclic group of prime order pwith generator c, such that p 6= pi for any i. Define

S = S1 × S2 × · · · × Sr

whereSi = {j ∈ Z : 1 ≤ j < pi, i

p ≡ 1 (mod p)}.Let

ψ : X 7→ Cp1 × Cp2 × · · ·Cpr

be an isomorphism. For each x ∈ Cp1 ×Cp2 ×· · ·Cpr , denote by I(x) the uniquer-tuple of integers

(i1, i2, ..., ir)

such that 0 ≤ ij < pj and hijj = xj for each j, where xj is the jth component

of x, and hj is a generator for Cpj , chosen such that

d = ψ−1(h1, h2, ..., hr).

Define alsoφ : Hom(Y,Aut(X)) 7→ S; g 7→ I(ψ(g(c)(d))).

(i) The function φ is a bijection.(ii) The set S forms an elementary Abelian p-group under the operation

(x1, x2, ..., xr)(y1, y2, ..., yr) = (x1y1 (mod p1), x2y2 (mod p2), ..., xryr (mod pr))

(iii) Two homomorphisms σ1 and σ2 lie in the same orbit of the ECS Actionfor X and Y iff there exists i with 1 ≤ i < p and

φ(σ2) = (φ(σ1))i

considering φ(σ1) as a group element under the operation defined in(ii).

Proof. Suppose that X is a cyclic group of squarefree order n = p1p2 · · · pr

with generator d, and that Y is a cyclic group of prime order p with generatorc, such that p 6= pi for any i. Define

S = S1 × S2 × · · · × Sr

whereSi = {j ∈ Z : 1 ≤ j < pi, i

p ≡ 1 (mod p)}.(i) X is cyclic, therefore Abelian, therefore solvable. Also, p = |Y | does not

divide the order of X, since otherwise p would have to equal pi for some i. Let

T1 = {f ∈ Aut(X) : fp = idAut(X)}.By Proposition 4.2.1, there is a bijection

φ1 : Hom(Y,Aut(X)) 7→ T1; g 7→ g(c).

Now, by Lemma 4.3.5, X is isomorphic to the group

H = Cp1 × Cp2 × · · · × Cpr .

Letψ : X 7→ H


be such an isomorphism, and let

T2 = {f ∈ Aut(H) : fp = idAut(H)}.For each f ∈ T1, define

φ2 : T1 7→ T2; f 7→ ψ ◦ f ◦ ψ−1

noting that ψ−1 exists as ψ is an isomorphism, hence a bijection.First φ2 is well-defined. Given f in T1, ψ◦f◦ψ−1 is certainly an isomorphism

as it is a composition of isomorphisms, and it is also an automorphism as clearlyψ ◦ f ◦ ψ−1 : X 7→ X. Now

(ψ ◦ f ◦ ψ−1)p = ψ ◦ (f ◦ ψ−1 ◦ ψ)p−1 ◦ f ◦ ψ−1

= ψ ◦ fp−1 ◦ f ◦ ψ−1

= ψ ◦ fp ◦ ψ−1

= ψ ◦ idAut(X) ◦ψ−1

= ψ ◦ ψ−1

= idAut(H)

so that ψ ◦ f ◦ ψ−1 ∈ T2.We shall show that φ2 is 1-1: take any f, g ∈ T1 such that φ2(f) = φ2(g).

Let x ∈ X be arbitrary, then

ψ(f(x)) = (ψ ◦ f ◦ ψ−1 ◦ ψ)(x)

= (φ2(f))(ψ(x))

= (φ2(g))(ψ(x))

= (ψ ◦ g ◦ ψ−1 ◦ ψ)(x)

= ψ(g(x))

so that f(x) = g(x), since ψ is 1-1. Since x was arbitrary, we have f = g, andso φ2 is 1-1.

We shall show that φ2 is onto: take any g ∈ T2. Let f = ψ−1 ◦ gψ. Now fis a composition of isomorphisms, hence an isomorphism. Clearly f : X 7→ X,so f is an automorphism of X. Next note that

fp = (ψ−1 ◦ g ◦ ψ)p

= ψ−1 ◦ (g ◦ ψ ◦ ψ−1)p−1 ◦ g ◦ ψ= ψ−1 ◦ gp−1 ◦ g ◦ ψ= ψ−1 ◦ gp ◦ ψ= ψ−1 ◦ idAut(H) ◦ψ= ψ−1 ◦ ψ= idAut(X)

so f ∈ T1. We haveφ2(f) = ψ ◦ ψ−1 ◦ g = g.

Thus φ2 is onto, hence a bijection.


By Lemma 4.4.2,

Aut(H) = Aut(Cp1)×Aut(Cp2)× · · ·Aut(Cpr).

Therefore,

T2

= {(f1, f2, ..., fr) ∈ Aut(Cp1)×Aut(Cp2)× · · ·Aut(Cpr)

: (f1, f2, ..., fr)p = (idAut(C1), idAut(C2), ..., idAut(Cr))}= {(f1, f2, ..., fr) ∈ Aut(Cp1)×Aut(Cp2)× · · ·Aut(Cpr)

: (fp1 , f

p2 , ..., f

pr ) = (idAut(C1), idAut(C2), ..., idAut(Cr))}

= U1 × U2 × · · ·Ur

whereUi = {f ∈ Aut(Cpi) : fp = idAut(Ci)}.

If j ∈ Si, then j < p so (j, p) = 1. Let hi be a generator of Cpi for each i,chosen such that

d = ψ−1(h1, h2, ..., hr).By Lemma 4.3.2, the maps

µi : Ui 7→ Si; f 7→ Ii(f(hi))

where Ii denotes the ith component of I, are bijections (note that Ii is just whatwe usually call the I function, but as a map from the cyclic group of order pi

to Si). Let

φ3 : T2 7→ S; (f1, f2, ..., fr) 7→ (µ1(f1), µ2(f2), ..., µr(fr)).

We show that φ3 is onto. Let (i1, i2, ..., ir) ∈ S. Then as the µj are bijec-tions, therefore onto, and so there is always some fj ∈ Uj with µj(fj) = ij .Now

φ3(f1, f2, ..., fr) = (µ1(f1), µ2(f2), ..., µr(fr)) = (i1, i2, ..., ir)and so φ3 is onto.

We shall now also show that φ3 is 1-1. So let

(f1, f2, ..., fr), (g1, g2, ..., gr) ∈ T2

and suppose thatφ3(f1, f2, ..., fr) = φ3(g1, g2, ..., gr).

Nowφ3(µ1(f1), µ2(f2), ..., µr(fr)) = (µ1(g1), µ2(g2), ..., µr(gr))

so that µj(fj) = µj(gj) for each j. Now we have fj = gj since µj is 1-1, andhence

(f1, f2, ..., fr) = (g1, g2, ..., gr)so that φ3 is 1-1. Therefore, φ3 is a bijection.

Let (f1, f2, ..., fr) ∈ T2 be arbitrary. Now

φ3(f1, f2, ..., fr) = (µ1(f1), µ2(f2), ..., µr(fr))

= (I1(f1(h1)), I2(f2(h2)), ..., Ir(fr(hr)))

= I(f1(h1), f2(h2), ..., fr(hr))


since the Ii are the components of I. Now, by Lemma 4.4.2,

(f1, ..., fr) ∈ Aut(Cp1)×Aut(Cp2)× · · ·Aut(Cpr)

= Aut(Cp1 × Cp2 × · · · × Cpr)

= Aut(H)

so(f1, ..., fr) = f

for some f ∈ Aut(H), and the fi are the components of f . This means that

f(h1, h2, ..., hr) = (f1(h1), f2(h2), ..., fr(hr))

so thatφ3(f) = I(f(h))

whereh = (h1, h2, ..., hr).

Now h generates H (for more details, see the proof of Lemma 4.3.5). Wehave that d = ψ−1(h), thus d has order equal to that of h (since ψ−1 is anisomorphism as ψ is), namely |H| (as h generates H), which is equal to |X|.Now d = ψ(h), and so

φ3(f) = I(f(h)) = I(f(ψ(d)))

Now we may define φ = φ3◦φ2◦φ1. This is a bijection, as it is a compositionof bijections. Clearly φ : Hom(Y,Aut(X)) 7→ S. Also

φ(σ) = I(ψ(σ(c)(ψ−1(ψ(d))))) = I(ψ(σ(c)(d)))

so that φ is indeed the same function given in the statement of the result.(ii) Each Si is a group with respect to multiplication modulo pi, by

Lemma 4.3.4. The set S therefore forms a group under the operation

(x1, x2, ..., xr)(y1, y2, ..., yr) = (x1y1 (mod p1), x2y2 (mod p2), ..., xryr (mod pr))

as it is then the direct product of the Si.Each of the groups Si are cyclic by Lemma 4.3.4, therefore they are all

Abelian. Any elements x and y in S have the form x = (x1, x2, ..., xr) andy = (y1, y2, ..., yr). Now within the group S,

xy = (x1, x2, ..., xr)(y1, y2, ..., yr)

= (x1y1, x2y2, ..., xryr)

= (y1x1, y2x2, ..., yrxr)

= (y1, y2, ..., yr)(x1, x2, ..., xr)= yx

so that S is an Abelian group.Again by Lemma 4.3.4, the groups Si all have order 1 or p. Hence for any

x ∈ Si, x the order of x must divide the order of Si, thus x has order 1 or p.If x has order 1, then it is the identity and idp

Si= idSi . If x has order p then

xp = idSi by definition. Now take any x ∈ S. We know x = (x1, x2, ..., xr), andso

xp = (x1, x2, ..., xr)p = (xp1, x

p2, ..., x

pr) = (idS1 , idS2 , ..., idSr) = idS


and so the order of x must divide p, hence be 1 or p. If x is a non-identityelement, its order cannot be 1, so must be p. Therefore, every non-identityelement of S has order p.

Suppose S were not a p-group. Then there must be some prime q dividingthe order of S, with q 6= p. Now by Cayley’s Group Theorem, there is anelement of S of order q. This element is not the identity as if so it wouldhave order 1 and q 6= 1 as it is a prime. However we have just seen that everynon-identity element of S has order p. This is a contradiction, so S is a p-group.

Therefore, S is an elementary Abelian p-group.(iii) Suppose that σ1 and σ2 are elements of Hom(Y,Aut(X)). Define

r-tuples (s1, s2, ..., sr) and (t1, t2, ..., tr) such that

σ1(c) = ψ ◦ (f1,s1 , f2,s2 , ..., fr,sr) ◦ ψ−1


which we can do by Proposition 4.1.1, Lemma 4.4.2 and Lemma 4.3.5.Define also for each i, j the function


( =⇒ ) Suppose that σ1 and σ2 lie in the same orbit of the ECS Action.Then there exists some α ∈ Aut(X) and β ∈ Aut(Y ) such that

σ2 = (α, β) ∗ σ1

or in other wordsσ2 = λα ◦ σ1 ◦ β−1.

Let i be the unique integer such that 1 ≤ i < p and β−1(c) = ci. Then, byLemma 4.4.3, we have

(ht11 , h

t22 , ..., h

trr ) = ψ(σ2(c)(d)) = ψ((λα(σ1(β−1(c))))(d)) = (hsi

11 , h

si2

2 , ..., hsir

r ).

Therefore, for each j, we have htjj = h

sij

j . Since hj has order pj , this impliestj ≡ si

j (mod pj), which gives

(t1, t2, ..., tr) = (si1, s

i2, ..., s

ir) = (s1, s2, ..., sr)i

regarded as elements of S. However it is clear that φ(σ1) = (s1, s2, ..., sr) andφ(σ2) = (t1, t2, ..., tr), so we have

φ(σ2) = (φ(σ1))i

and 1 ≤ i < p, as required.( ⇐= ) Suppose there exists an integer i with

φ(σ2) = (φ(σ1))i

and 1 ≤ i < p. Let α be any automorphism of X, and define

γ : Y 7→ Y ; y 7→ yi

and β = γ−1. Now β−1(c) = γ(c) = ci. By Lemma 4.4.3, we have

ψ(σ2(c)(d)) = (ht11 , h

t22 , ..., h

trr ) = (hsi

11 , h

si2

2 , ..., hsir

r ) = ψ((λα(σ1(β−1(c))))(d)).


Therefore, as ψ is 1-1,

σ2(c)(d) = (λα(σ1(β−1(c))))(d).

Let x be any element of X. Then x = dk for some integer k, and

σ2(c)(x) = σ2(c)(dk)

= (σ2(c)(d))k

= ((λα(σ1(β−1(c))))(d))k

= (λα(σ1(β−1(c))))(dk)

= (λα(σ1(β−1(c))))(x)

= ((λα ◦ σ1 ◦ β−1)(c))(x)

and so σ2(c) = (λα ◦ σ1 ◦ β−1)(c). Now let y be any element of Y . Then y = ck

for some integer k, and

σ2(y) = σ2(ck)

= (σ2(c))k

= ((λα ◦ σ1 ◦ β−1)(c))k

= (λα ◦ σ1 ◦ β−1)(ck)

= (λα ◦ σ1 ◦ β−1)(y)

and so σ2 = λα ◦ σ1 ◦ β−1. Therefore, σ2 = (α, β) ∗ σ1, and so σ1 and σ2 lie inthe same orbit of the ECS Action. �

Finally, we shall give an algorithm to generate a particular set from anelementary Abelian group; this will turn out to be in correspondence with theset of homomorphisms for one of the cases of the main theorem.

Algorithm 4.4.5. Suppose that G is an elementary Abelian p-group. Pro-duce a finite sequence of ai as follows:

(1) Let i = 0.(2) Let a0 = idG.(3) (LABEL 1)(4) If

i⋃r=0

〈ar〉 = G

then END.(5) Let i = i+ 1.(6) Choose

ai ∈ Gi⋃

r=0

〈ar〉.

(7) GOTO LABEL 1.


Lemma 4.4.6. Let G be an elementary Abelian p-group of order pn. Thenumber of elements constructed by Algorithm 4.4.5 is equal to

pn−1 + pn−2 + ...+ p2 + p+ 2.

Proof. Let G be an elementary Abelian p-group of order pn. SupposeAlgorithm 4.4.5 constructs the m + 1 elements a0, a1, a2, ..., am. After each“GOTO LABEL 1” of the algorithm, we have removed p− 1 elements from thechoices for the next a (these are the p− 1 non-identity elements of 〈ai〉, whichhas order p since we are working in an elementary Abelian p-group. None ofthese have already been removed, as if one had, say b, then b is in 〈aj〉 for somej < i, so ai is a power of this aj , hence in 〈aj〉, a contradiction. We do notremove the identity, as this was never a choice, being in 〈a0〉). After all the ai

have been constructed, we have removed a total of m(p−1)+1 elements (wherethe 1 comes from the identity, which we dealt with separately), and this mustequal |G| = pn, since the condition to end the algorithm was that there are noelements left to choose from. We have

pn − 1 = m(p− 1)

and this gives that

m = pn−1 + pn−2 + ...+ p2 + p+ 1.

Therefore, the number of elements constructed is

m+ 1 = pn−1 + pn−2 + ...+ p2 + p+ 2.

�

4.5. CLASSIFICATION OF GROUPS OF ORDER pqr WHERE q - r − 1 73

4.5. Classification of groups of order pqr where q - r − 1

Theorem 4.5.1 (Classification of groups of order pqr where q - r − 1). Letp, q, r be primes with p < q < r and q - r − 1. Let Y = Cp with generator c,A = Cq with generator a and B = Cr with generator b. Further, define maps

fi : A 7→ A;x 7→ xi

andgi : B 7→ B;x 7→ xi.

Also defineH = A×B

and letψ : X 7→ H

be an isomorphism.(i) Suppose p - q − 1 and p - r − 1. Then any group of order pqr is

isomorphic to the cyclic group Cpqr.(ii) Suppose p - q−1 and p | r−1. There exists an integer i with 1 < i < r

and ip ≡ 1 (mod r). Define a map

σ : Y 7→ Aut(X); cs 7→ ψ−1 ◦ (idAut(A), gjs).

Then “Cpqr, Cp oσ (Cq × Cr)” is a list of groups of order pqr up toisomorphism.

(iii) Suppose p | q−1 and p - r−1. There exists an integer i with 1 < i < qand ip ≡ 1 (mod q). Define a map

σ : Y 7→ Aut(X); cs 7→ ψ−1 ◦ (fis , idAut(B)).

Then “Cpqr, Cp oσ (Cq × Cr)” is a list of groups of order pqr up toisomorphism.

(iv) Suppose p | q − 1 and p | r − 1. Let

S = {(i, j) : i, j ∈ Z, j ∈ Z, 1 ≤ i < q, 1 ≤ j < r, ip ≡ 1 (mod q), jp ≡ 1 (mod r).

Then we can apply Algorithm 4.4.5 to obtain a sequence of p + 2 ele-ments

(i0, j0), (i1, j1)..., (ip+1, jp+1).Define maps

σm : Y 7→ Aut(X); cs 7→ ψ−1 ◦ (fism , gjsm

).

Then “Cpqr, Cpoσ1 (Cq×Cr), Cpoσ2 (Cq×Cr), ..., Cpoσp+1 (Cq×Cr)”is a list of groups of order pqr up to isomorphism.

Proof. Let p, q, r be primes with p < q < r and q - r−1. Let Y = Cp withgenerator c, A = Cq with generator a and B = Cr with generator b. Further,define maps

fi : A 7→ A;x 7→ xi

andgi : B 7→ B;x 7→ xi.

Also defineH = A×B


and letψ : X 7→ H

be an isomorphism. Note that Aut(H) = Aut(A)×Aut(B), by Lemma 4.4.2.Let

S = {(i, j) : i, j ∈ Z, j ∈ Z, 1 ≤ i < q, 1 ≤ j < r, ip ≡ 1 (mod q), jp ≡ 1 (mod r).

It is clear that S contains the element (1, 1), since 1p = 1 ≡ 1 under anymodulus bigger than 1. Let

S1 = {i : i ∈ Z, 1 ≤ i < q, ip ≡ 1 (mod q)}

andS2 = {i : i ∈ Z, 1 ≤ i < r, ip ≡ 1 (mod r)}

so that clearlyS = S1 × S2.

Let X = Cqr with generator d. Let G be a group of order pqr. Then G hasa subgroup of index p, by Proposition 1.5.1. For any such subgroup N , we havethat

|N | = |G||G||N |

=|G|

[G : N ]=pqr

p= qr.

Therefore N is a group of order qr, and hence is isomorphic to the cyclic groupof order qr, by Proposition 4.3.8. Thus N ∼= X. We have that G/N hasorder [G : N ] = p, and so G/N is isomorphic to the cyclic group of orderp, i.e. G/N ∼= Y . This group has order p, and hence (|X|, |Y |) = (qr, p) = 1.Furthermore, X and Y are solvable by Proposition 1.5.2. Since G was arbitrary,we have shown that any group of order pqr with a normal cyclic subgroup oforder qr, has a normal subgroup N isomorphic to X and G/N isomorphic toY .

Defineφ : Hom(Y,Aut(X)) 7→ S; g 7→ I(ψ(g(c)(d)))

which is a bijection by Proposition 4.4.4(i). Let the homomorphism θ be definedby

θ : Y 7→ Aut(X); y 7→ ψ−1 ◦ (idAut(A), idAut(B)).Note that this group has underlying set X × Y and if x1, x2 ∈ X, y1, y2 ∈ Ythen

(x1, y1)(x2, y2) = (x1θ(y1)(x2), y1y2)

= (x1ψ−1(idAut(A), idAut(B))(x2), y1y2)

= (x1ψ−1(idAut(H))(x2), y1y2)

= (x1 idAut(X)(x2), y1y2)

= (x1x2, y1y2)

so X oθ Y is in fact the direct product of X and Y , or Cqr ×Cp. However, Cqr

is isomorphic to Cq × Cr by Lemma 4.3.5, and hence

X oθ Y = X × Y ∼= (Cq × Cr)× Cp∼= Cq × Cr × Cp

∼= Cqrp = Cpqr

4.5. CLASSIFICATION OF GROUPS OF ORDER pqr WHERE q - r − 1 75

the last isomorphism being again by Lemma 4.3.5. Note that φ(θ) = (1, 1)(since θ(c)(d) = (idA, idB)). We also have that

|X oθ Y | = |Cpqr| = pqr.

(i) Suppose p - q − 1 and p - r − 1. By Lemma 4.3.7, there do not exist i,j with 1 < i < q, 1 < j < r and ip ≡ 1 (mod q), jp ≡ 1 (mod r). Therefore,S can only contain the element (1, 1). Thus |S| = 1. By Proposition 4.4.4(i),|Hom(Y,Aut(X))| = |S| = 1, and so Hom(Y,Aut(X)) contains only one ele-ment, say σ.

Now the group Cpqr is of order pqr, hence has a normal subgroup N iso-morphic to X with G/N isomorphic to Y . By Theorem 3.3.3(ii), we have thatthe cyclic group Cpqr is isomorphic to X oσ Y .

Take any group G of order pqr. Then G has a normal subgroup N isomor-phic to X with G/N isomorphic to Y . By Theorem 3.3.3(ii), we have that Gis isomorphic to X oσ Y , and hence to Cpqr.

(ii) Suppose p - q − 1 and p | r − 1. By Lemma 4.3.7, there does not existi with 1 < i < q and ip ≡ 1 (mod q) but there does exist j with 1 < j < rand jp ≡ 1 (mod r). The set S1 has only one element, and the set S2 hasmore than one element and hence has p elements by Lemma 4.3.4. Therefore|S| = |S1||S2| = p. Also, S is a group under the operation

(1, y1)(1, y2) = (1, y1y2 (mod r))

by Proposition 4.4.4(ii).Let the homomorphism σ be defined by

σ : Y 7→ Aut(X); cs 7→ ψ−1 ◦ (idAut(A), gjs).

First of all note that φ(σ) 6= (1, 1) (since σ(c)(d) = (idA, bj)). Since

(1, 1)k = (1, 1) for any k, we have that

X oθ Y 6∼= X oσ Y

by Proposition 4.4.4(iii).We have that

|X oσ Y | = |X| · |Y | = (|Cq| · |Cr|) · |Cp| = pqr.

Now take any group G of order pqr. Then G has a normal subgroup N iso-morphic to X with G/N isomorphic to Y . Therefore, by Theorem 3.3.3(ii), Gis isomorphic to X oρ Y for some ρ ∈ Hom(Y,Aut(X)). Now if ρ = θ thenG ∼= X oθ Y . If ρ 6= θ then φ(ρ) = φ(σ)k for some integer k (since φ(σ) is anon-identity element and hence generates the group). Now as φ(ρ) is not theidentity (as otherwise we could deduce ρ = θ), this k is not 0, and so 1 ≤ k < p(S has order p). By Proposition 4.4.4(iii), ρ and σ lie in the same orbit of theECS Action, and then by Theorem 3.3.3(iii)

G ∼= X oρ Y ∼= X oσ Y.

Therefore X oσ Y and X oθ Y are non-isomorphic groups, and every groupof order pqr is isomorphic to one of them, so “X oθ Y , X oσ Y ” is a list ofgroups of order pqr up to isomorphism. However X oθ Y ∼= Cpqr, so “Cpqr,X oσ Y ” is a list of groups of order pqr up to isomorphism.


(iii) Suppose p | q − 1 and p - r − 1. By Lemma 4.3.7, there does exist iwith 1 < i < q and ip ≡ 1 (mod q) but there does not exist j with 1 < j < rand jp ≡ 1 (mod r). The set S1 has more than one element and hence hasp elements by Lemma 4.3.4, and the set S2 has only one element. Therefore|S| = |S1||S2| = p. Also, S is a group under the operation

(x1, 1)(x2, 1) = (x1x2 (mod q), 1)

by Proposition 4.4.4(ii).Let the homomorphism σ be defined by

σ : Y 7→ Aut(X); cs 7→ ψ−1 ◦ (fis , idAut(B)).

First of all note that φ(σ) 6= (1, 1) (since σ(c)(d) = (ai, idB)). Since(1, 1)k = (1, 1) for any k, we have that

X oθ Y 6∼= X oσ Y

by Proposition 4.4.4(iii).We have that

|X oσ Y | = |X| · |Y | = (|Cq| · |Cr|) · |Cp| = pqr.

Now take any group G of order pqr. Then G has a normal subgroup N iso-morphic to X with G/N isomorphic to Y . Therefore, by Theorem 3.3.3(ii), Gis isomorphic to X oρ Y for some ρ ∈ Hom(Y,Aut(X)). Now if ρ = θ thenG ∼= X oθ Y . If ρ 6= θ then φ(ρ) = φ(σ)k for some integer k (since φ(σ) is anon-identity element and hence generates the group). Now as φ(ρ) is not theidentity (as otherwise we could deduce ρ = θ), this k is not 0, and so 1 ≤ k < p(S has order p). By Proposition 4.4.4(iii), ρ and σ lie in the same orbit of theECS Action, and then by Theorem 3.3.3(iii)

G ∼= X oρ Y ∼= X oσ Y.

Therefore X oσ Y and X oθ Y are non-isomorphic groups, and every groupof order pqr is isomorphic to one of them, so “X oθ Y , X oσ Y ” is a list ofgroups of order pqr up to isomorphism. However X oθ Y ∼= Cpqr, so “Cpqr,X oσ Y ” is a list of groups of order pqr up to isomorphism.

(iv) Suppose p | q − 1 and p | r − 1. By Lemma 4.3.7, there exists i with1 < i < q and ip ≡ 1 (mod q) and j with 1 < j < r and jp ≡ 1 (mod r). The setS1 has more than one element and hence has p elements by Lemma 4.3.4, andthe set S2 has more than one element and hence has p elements by Lemma 4.3.4.Therefore |S| = |S1||S2| = p2. Also, S is an elementary Abelian p-group underthe operation

(x1, y1)(x2, y2) = (x1x2 (mod q), y1y2 (mod r))

by Proposition 4.4.4(ii).Define a sequence a0, a1, ..., ap+1 by Algorithm 4.4.5; this sequence is full and

complete by Lemma 4.4.6 with n = 2. For each integer m with 0 ≤ m ≤ p+ 1,let am = (im, jm) and define homomorphisms

σm : Y 7→ Aut(X); cs 7→ ψ−1 ◦ (fism , gjsm

).

4.6. EXAMPLES OF THE CLASSIFICATION 77

We have that

|X oσm Y | = |X| · |Y | = (|Cq| · |Cr|) · |Cp| = pqr.

Note that as a0 = idS = (1, 1), then if y = cs,

σ0(y) = σ0(cs)

= ψ−1 ◦ (fis0, gjs

0)

= ψ−1 ◦ (f1s , g1s)

= ψ−1 ◦ (f1, g1)

= ψ−1 ◦ (idAut(A), idAut(B))

= θ(y)

and so σ0 = θ, so X oσ0 Y∼= Cpqr.

Let r < s and suppose φ(σs) = φ(σ(r))k for some k with 1 ≤ k < p. Thenas = ak

r . Therefore, as ∈ 〈ar〉, but this is impossible in the algorithm. This is acontradiction, so φ(σs) 6= φ(σ(r))k for any k, and so by Proposition 4.4.4(iii),

X oσr Y 6∼= X oσs Y.

Let G be any group of order pqr. Then G has a normal subgroup N isomor-phic to X with G/N isomorphic to Y . So G is isomorphic to X oρ Y for someρ ∈ Hom(Y,Aut(X)), by Theorem 3.3.3(ii). Suppose ρ = θ, then G ∼= Cpqr. Ifρ 6= θ then φ(ρ) 6= φ(theta) = (1, 1) = idS . Let s = φ(ρ) ∈ S. Then since

S =p+1⋃k=0

〈ak〉

then we must have s ∈ 〈ak〉 for some integer k. Now s = atk for some integer t

with t 6= 0 (since s 6= (1, 1)) and 0 ≤ t < p, or in other words, φ(ρ) = φ(σk)t forsome integer t with 1 ≤ t < p. Thus, by Proposition 4.4.4(iii),

X oρ Y ∼= X oσkY.

Now we have shown that “X oσ0 Y , X oσ1 Y , X oσ2 Y , ..., X oσr Y ”, andhence “Cpqr, X oσ1 Y , X oσ2 Y , ..., X oσr Y ”, is a list of groups of order pqrup to isomorphism. �

4.6. Examples of the classification

This section serves to apply Theorem 4.5.1, in some special cases.

Remark. We know how to classify groups of order pqr (q - r − 1) for ageneral prime p, from Theorem 4.5.1. Therefore, we can classify groups oforder 2qr, and 3qr, specifically. In the case of groups of order 2qr, things areslightly simplified as for any such order, we must have p | q − 1 and p | r − 1,since q and r, being primes greater than 2, must be odd.

Example 4.6.1 (Groups of order 105). Let G be a group of order 105. Wehave 105 = 3 · 5 · 7 so let p = 3, q = 5 and r = 7. We have that q - r − 1,p - q − 1 and p | r − 1. Therefore, applying Theorem 4.5.1(ii), we obtain thatthere are two groups of order 105 up to isomorphism. One of these is the cyclic


group of order 105. To get the other, note that there must be some integer iwith 1 < i < 7 and i3 ≡ 1 (mod 7). For example, we can take i = 2. Now define

σ : Y 7→ Aut(X); yr 7→ f2r

where y is a generator of Y and

fj : X 7→ X;x 7→ xj .

We may calculate the values of this homomorphism. For example,

σ(idY ) = f20 = f1 = idAut(X)

andσ(y) = f21 = f2

which is the map which squares every element of X.We know that any non-cyclic group of order 105 must be isomorphic to

X oσ Y , by Theorem 4.5.1.

Example 4.6.2 (Groups of order 273). Let G be a group of order 273.

(1) We have 273 = 3 · 7 · 13, so let p = 3, q = 7 and r = 13. We have thatq - r − 1, p | q − 1 and q | r − 1.

(2) We may compute the S1 and S2 of Proposition 4.4.4. Each will havesize p, by Lemma 4.3.7. We have that

S1 = {1, 2, 4}

andS2 = {1, 3, 9}.

Now the S from Theorem 4.5.1, namely

S = {(i, j) : i, j ∈ Z, j ∈ Z, 1 ≤ i < q, 1 ≤ j < r, ip ≡ 1 (mod q), jp ≡ 1 (mod r)

has

S = S1 × S2 = {(1, 1), (1, 3), (1, 9), (2, 1), (2, 3), (2, 9), (4, 1), (4, 3), (4, 9)}.

We may apply Algorithm 4.4.5 to S, which is an elementary Abelianp-group by Proposition 4.4.4. We have a0 = (1, 1). Choose a1 not in〈a0〉 = {(1, 1)}, so choose a1 = (2, 1). Choose a2 not in

〈a0〉 ∪ 〈a1〉= {(1, 1)} ∪ {(1, 1), (2, 1), (4, 1)}= {(1, 1), (2, 1), (4, 1)}

so choose a2 = (1, 3). Choose a3 not in

〈a0〉 ∪ 〈a1〉 ∪ 〈a2〉= {(1, 1)} ∪ {(1, 1), (2, 1), (4, 1)} ∪ {(1, 1), (1, 3), (1, 9)}= {(1, 1), (2, 1), (4, 1), (1, 3), (1, 9)}

4.7. CONCLUSION 79

so choose a3 = (2, 3). Choose a4 not in

〈a0〉 ∪ 〈a1〉 ∪ 〈a2〉 ∪ 〈a3〉= {(1, 1)} ∪ {(1, 1), (2, 1), (4, 1)} ∪ {(1, 1), (1, 3), (1, 9)} ∪ {(1, 1), (2, 3), (4, 9)}= {(1, 1), (2, 1), (4, 1), (1, 3), (1, 9), (2, 3), (4, 9)}

so choose a4 = (2, 9). But now 〈a4〉 = {(1, 1), (2, 9), (4, 3)} and so

〈a0〉 ∪ 〈a1〉 ∪ 〈a2〉 ∪ 〈a3〉 ∪ 〈a4〉= {(1, 1), (2, 1), (4, 1), (1, 3), (1, 9), (2, 3), (4, 9), (2, 9), (4, 3)}= S

so the algorithm is over.(3) As predicted, the algorithm has produced p+2 = 5 elements, and each

corresponds to a group of order 273, by Theorem 4.5.1. There are 5such, and each has the form Cp oσk

(Cq × Cr), where

σk : Cp 7→ Aut(Cq)×Aut(Cr); cs 7→ (fisk, gjs

k)

where c is a generator of Cp, we define (ik, jk) = ak, and

fi : Cq 7→ Cq;x 7→ xi

gi : Cr 7→ Cr;x 7→ xi.

The group Cp oσ (Cq × Cr) corresponds to the cyclic group of order273.

(4) Any group of order 273 is isomorphic to one of those constructed above,and no two of the above are isomorphic. Thus they form a list of groupsof order 273 up to isomorphism, and there are 5 such.

4.7. Conclusion

We have presented a general method which can be applied to classificationof groups of any squarefree order. Whereever possible, we have tried to presentour results in the case where G has any squarefree order, not necessarily theproduct of three primes. Theoretically, we could continue to classify groups oforder pqr in the case where q | r−1, by constructing the automorphism group ofthe non-cyclic group of order qr, and then putting this back into the ExtensionClassification Schema, along with Proposition 4.2.1 and other tools. We couldalso try to classify groups of order pqrs, as was done in 1939 byD T Sigley ([2]). However the theorem we have proved (Theorem 4.5.1) is verygeneral and can be used to classify groups of many orders. For example, thereare 9 groups of order 232, 401, 421.

Index

π-group, 17π-subgroup, 17Examples, 18Hall π-subgroup, 17

π-number, 17

Abelian Series, 13Of Subgroups, 13

Classification of Groups, 34Of order 2qr, 53Of order pq, 57Of order pqr, 73

Example of order 105, 77Example of order 273, 78

Elementary Abelian p-group, 19Extension Classification Schema, 43

ECS Action, 43

Feit-Thompson Theorem, 85Frattini Argument, 25

Hall’s Theorems, 19

List of Groups up to Isomorphism, 33Examples, 34

Normal Series, 13Of Subgroups, 13

Semidirect ProductCanonical Embeddings, 37External, 34Internal, 34

Subgroup Product, 20Internal Semidirect Product, 34

81

Bibliography

[1] Burley, A. Application of the transfer homomorphism to prove a famous corollary of theFeit-Thompson theorem for groups of order less than or equal to 200. 2006, Unpublished.

[2] Sigley, D T. An Enumeration of the Groups of Order pqrs. American Journal of Mathe-matics, Vol. 61, No. 1. (Jan., 1939), pp. 102-106.

[3] Goddard, Bart. Re: Number-theoretic question (congruency). sci.math posting, acces-sible at http://mathforum.org/kb/message.jspa?messageID=5474319 (retrieved 24th Jan2007).

83

APPENDIX A

Solvability without the transfer

A.1. The full Feit-Thompson Theorem

In the previous document, [1], we saw that any non-Abelian group of oddorder less than 200 is not simple. We now can extend this result to prove thefull Feit-Thompson theorem for groups of order less than 200; namely, that anygroup of odd order less than 200 is solvable.

Theorem A.1.1 (The Feit-Thompson Theorem). Suppose G is a group withodd order less than 200. Then G is solvable.

Proof. Suppose for a contradiction that there exists a group with oddorder less than 200 which is not solvable. Let n < 200 be the smallest integersuch that there exists a non-solvable group with odd order n. Let G be a groupwith order n. Then G 6= {idG}, since the trivial group is solvable (for such agroup G we have G(1) = G′ = 1). So G has a proper normal subgroup, namely{idG}. Let N be a maximal proper normal subgroup of G. Now, G/N hassmaller order than G and therefore its order is also less than 200. However, asN is maximal normal in G, we must have that G/N is simple by Lemma 1.1.4.So G/N must be Abelian (as otherwise it could not be simple, by Lemma 0.1.1).Also, N is solvable as it has odd order less than n (its order is odd since it dividesthe order of the group G, which is odd), and n is the smallest integer such thatthere exists a non-solvable group with odd order n. So we have found N E G,G/N Abelian, N solvable, and so G is solvable by Lemma 1.4.1. �

A.2. Solvability of groups of order pqr, without the transfer

Since we already know quite a lot about groups of odd order, the proof inthis case will be much easier. We are most interested, therefore, in the caseof even orders. In this case, suppose we have a group of order pqr. We mayassume without loss of generality that p < q < r. So if pqr is even then p = 2,since 2 must divide the group order. Furthermore, q, r 6= 2 as the primes mustbe distinct. So 2 divides the group order and 22 = 4 does not. Therefore thetheorems in this section can be proved in the more general case where the sizeof G is simply twice an odd number. The disadvantage of not employing thetransfer results, is that we are now restricted to groups of order less than 200.

Proposition A.2.1. Let G be a finite group and suppose |G| = 2m wherem is an odd integer. Then G contains a normal subgroup of index 2.

Proof. Let G be a finite group with |G| = 2m where m is an odd integer.Since 2 divides |G|, then G contains an element of order 2 (as 2 is a prime, and

85

86 A. SOLVABILITY WITHOUT THE TRANSFER

by Cauchy’s Group Theorem). Let h be an element of order 2 in G. Choose anyx1 ∈ G and let x2 = hx1. Now repeat this until the elements of G are exhausted,at each stage picking x2i−1 ∈ G {x1, ..., x2i−2} and letting x2i = hx2i−1. Sincethe order of G is even, there will be an exact number of repetitions of this stepuntil the process ends.

Now let X = G with the action of left multiplication. Define

ρ : G 7→ Sym(X); ρ(g)(x) = gx

This is a homomorphism by Lemma 0.2.1. Then

ρ(h)(x2i−1) = hx2i−1 = x2i

andρ(h)(x2i) = hx2i = h2x2i−1 = x2i−1

as h has order 2. Now, ρ(h) swaps consecutive (when indexed by xi) elementsof G. Or another way of saying this is:

ρ(h) = (x1 x2)(x3 x4) · · · (x2m−3 x2m−2)(x2m−1 x2m)

ρ(h) is a product of m transpositions, where m is odd, hence ρ(h) is an oddpermutation. Because of this, it has sign -1. Or more technically, let ε denotethe sign homomorphism, which maps even permutations to +1 and odd per-mutations to -1. Then ε(ρ(h)) = −1, so (ε ◦ ρ)(h) = −1. Also (ε ◦ ρ)(idG) = 1,since a homomorphism must map the identity of one group to the identity ofthe other group (and ε ◦ ρ is the composition of two homomorphisms so mustbe a homomorphism itself). So ε ◦ ρ : G 7→ {+1,−1} is onto.

Let N = ker(ε ◦ ρ). Then N is a normal subgroup of G. Also, by the FirstIsomorphism Theorem,

[G : N ] = |G/N | = |G/ ker(ε ◦ ρ)| = | im(ε ◦ ρ)| = 2

as ε ◦ ρ is onto with a codomain of size 2. Therefore, G has a normal subgroupof index 2 (namely, N). �

The previous result is much more general than our Proposition 1.5.1. Itleads to an alternative proof that all groups of order pqr are solvable.

Corollary A.2.2. Suppose G is a group of order pqr < 200, where p, q, rare distinct primes. Then G is solvable.

Proof. Suppose G is a group of order pqr < 200, where p, q, r are distinctprimes. We may assume without loss of generality that p < q < r. Now if Ghas odd order, we are done by Theorem A.1.1. So suppose G has even order.Therefore, we must have p = 2, since if q or r was 2, then as 2 divides the grouporder, p would have to be a prime less than 2, a contradiction. Now q and rare primes which are not 2, so they are both odd. So the group order is 2mwhere m = qr is odd. By Proposition A.2.1, G contains a normal subgroup ofindex 2, say N . N has order m < |G| < 200, which is odd, and so N is solvableby Theorem A.1.1. G/N has order 2 also, and so is Abelian (as it has primeorder). So we have N E G, N solvable, G/N Abelian, and so G is solvable byTheorem 1.4.1. �

A.3. SOLVABILITY OF GROUPS OF ORDER p2q 87

A.3. Solvability of groups of order p2q

In this section, we shall prove that all groups of order p2q are solvable,using earlier results in the document. There is only one case which has notbeen already proved, and we shall prove this here, after the following initiallemma.

Lemma A.3.1. Suppose G is a group of order 6. Then G is solvable.

Proof. Let G be a group of order 6. By Proposition A.2.1, G has a normalsubgroup of index 2, say N . Then G/N has prime order 2 and so is Abelian,and N has prime order 3 and so is Abelian, so is solvable. So G is solvable byLemma 1.4.1. �

Now we may deal with the remaining case.

Proposition A.3.2. Suppose G is a group of order 4q where q is an oddprime. Then G is solvable.

Proof. Suppose G is a group of order 4q where q is an odd prime. Thennq, the number of Sylow q-subgroups, divides 4, so is 1, 2 or 4. If nq = 2, then asnq ≡ 1 (mod q), we must have 2 ≡ 1 (mod q), so 1 ≡ 0 (mod q). However as q > 1(as it is prime), 1 6≡ 0 (mod q). This is a contradiction, so nq 6= 2. If nq = 1then let Q be a Sylow q-subgroup. Then Q is normal, also |G/Q| = 4 = 22, soG/Q is Abelian (as its order is the square of a prime). Finally, Q is Abelianas it has order q, a prime, therefore it is solvable (as every commutator is theidentity). Hence by Lemma 1.4.1, G is solvable. For the rest of this proof weshall assume the last case, namely that nq = 4. Then, as 4 = nq ≡ 1 (mod q),we must have that q = 3 (as certainly q 6= p = 2, and if q > 3 then q ≥ 5 (as qis prime) but then we do not have 4 ≡ 1 (mod q)). So |G| = 4q = 12.

Then n2, the number of Sylow 2-subgroups, divides 3, so is either 1 or 3. Ifit’s 1, then we get a normal Sylow 2-subgroup P . Also G/P is Abelian as itsorder is q, a prime. P is solvable as it has order 22 = 4 and therefore is Abelian,therefore solvable. So G is solvable by Lemma 1.4.1.

Now we shall assume that n2 = 3. Let the Sylow 2-subgroups be P1, P2

and P3, and let X = {P1, P2, P3}. Define an action ∗ by:

g ∗ P = gPg−1

for all g ∈ G, P ∈ X. Now we get a homomorphism

ρ : G 7→ Sym(X); ρ(g)(Pr) = g ∗ Pr

Since all Sylow 2-subgroups are conjugate, there exists g ∈ G such thatgP1g

−1 = P2. So g ∗ P1 = P2. Therefore, ρ(g)(P1) = g ∗ P1 = P2, so ρ(g) isnot the identity map. So ρ is not a trivial homomorphism. This means thatits kernel is not equal to the whole group, so | ker(ρ)| < |G| = 12. However| ker(ρ)| must divide the order of the group, so we must have |ker(ρ)| ≤ 6. Also| im(ρ)| must divide the order of Sym(X), which is 3! = 6, so | im(ρ)| ≤ 6. So|G/ ker(ρ)| = | im(ρ)| ≤ 6. This means that | ker(ρ)| ≥ 2.

Now, 2 ≤ | ker(ρ)| ≤ 6. We will show that ker(ρ) is solvable. First notethat ker(ρ) has order dividing the order of G, so along with the constraint

88 A. SOLVABILITY WITHOUT THE TRANSFER

2 ≤ | ker(ρ)| ≤ 6, we have that | ker(ρ)| is 2, 3 or 6. If | ker(ρ)| = 2 or | ker(ρ)| = 3then it has prime order and therefore is Abelian, therefore solvable. If| ker(ρ)| = 6, then we can apply Lemma A.3.1 to determine that ker(ρ) issolvable.

Since | ker(ρ)| is 2, 3 or 6, we also have |G/ ker(ρ)| is equal to 2, 3 or 6. If|G/ ker(ρ)| = 2, 3 then G/ ker(ρ) has prime order and therefore is Abelian. Soker(ρ) is a solvable normal subgroup with G/ ker(ρ) Abelian, so G is solvableby Lemma 1.4.1, and we are done.

So we turn to the case |G/ ker(ρ)| = 6. Then | ker(ρ)| = 2. By Propo-sition A.2.1, there is a normal subgroup N of G/ ker(ρ) with index 2. ByLemma 1.1.1, N = H/ ker(ρ) for some H E G (normal as N E G/ ker(ρ)) andker(ρ) E H. So |H| ≥ 2 (as ker(ρ) E H, | ker(ρ)| = 2). In fact, |H| 6= 2, since if|H| = 2 then

|N | = |H|| ker(ρ)|

=22

= 1

so [G/ ker(ρ) : N ] = |G/ ker(ρ)| = 6, but the index should equal 2. This is acontradiction, therefore |H| 6= 2, so |H| > 2. So |H| = 3, 6. If |H| = 3 then Hhas prime order, therefore is Abelian, therefore solvable. If |H| = 6 then H issolvable by Lemma A.3.1. Note also that |G/H| = 2, 4, and so G/H is Abelian(as it has order either equal to a prime or a prime squared). So G is solvable,by Lemma 1.4.1. �

Now we simply tie all our results together.

Proposition A.3.3. Suppose G is a group of order p2q < 200, where p andq are distinct primes. Then G is solvable.

Proof. Suppose G is a group of order p2q < 200, where p and q are distinctprimes. If G has odd order then we are done, by Theorem A.1.1. So supposeG has even order. Then 2 divides the order of G, so either p = 2 or q = 2,but not both (as they are distinct). If q = 2 then p must be odd (as it is aprime not equal to 2), so the group order has the form 2m where m = p2 isodd. Then G is solvable by Proposition A.2.1. If p = 2 then the order of G isequal to 4q, where q is odd (as it is a prime not equal to 2). Then G is solvableby Proposition A.3.2. �

Classiﬁcation of some groups of order pqr Adam Burley

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