ON CAYLEY'S THEOREMhomepages.math.uic.edu/~rtakloo/papers/group-paper/group5.pdf · n ∈ Z and x...

37
α α α G G S n n = |G| S n n |G| G d(G) n G S n α(G)= d(G) |G| α(G) G d(G) G

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ON CAYLEY'S THEOREM

BY BEN ELIAS, LIOR SILBERMAN, AND RAMIN TAKLOO-BIGHASH

Abstract. In this paper we prove various results about the min-imal faithful permutation representations of �nite groups.

Contents

1. Introduction: Mostly questions and some results 12. Preliminaries 62.1. Subgroups, direct products, and examples 73. Distribution of α 103.1. Limit Points 103.2. The Average of α 134. P-groups 164.1. P-groups and the socle 164.2. Anticovers 184.3. Well-indexed, regular, and antimaximal collections 204.4. An algorithm to determine α for p-groups 274.5. Computational results 29Appendix A. MAGMA Code 32References 36

1. Introduction: Mostly questions and some results

It is a classical theorem of Cayley that given any group G there isa monomorphism G → Sn for n = |G| where here and elsewhere Sn isthe group of permutations on n letters. A natural question is whetherthe bound |G| can be improved. In order to quantify this question wemake the following de�nition:

De�nition. Given a group G, let d(G) be the smallest natural numbern such that there is an injective homomorphism from G to the sym-metric group Sn. Let α(G) = d(G)

|G| . We call α(G) the Cayley Constant

of G. The natural number d(G) is called the degree of G.1

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ON CAYLEY'S THEOREM 2

Calculating α(G) is intimately related to the structure of the groupG. Let us make a de�nition:

De�nition. A faithful collection I = {Hi}mi=1 of G is a set of subgroups

Hi ≤ G such that⋂m

i=1 Hi contains no normal subgroups of G. We call

d(I) :=∑m

i=1|G||H| the degree of I, and α(I) :=

∑mi=1

1|H| the Cayley

Constant of I. The number m is the size of I.

It is easily shown that �nding α(G) is equivalent to �nding thefaithful collection I of G such that α(I) is minimal, and in this caseα(I) = α(G). We call such a collection a minimal faithful collection.We will present details and various examples in Section 2.It is not easy to �nd α(G) for a general group G. In general, the only

statement one can make is that 0 < α(G) ≤ 1; this is the statement ofCayley's theorem! A �rst natural question is what numbers are α(G)for some �nite group G. We set

X ={

x ∈ (0, 1] |x = α(G) for some �nite group G}

,

Xp ={

x ∈ (0, 1] |x = α(G) for some p− group G}

,

Xsol ={

x ∈ (0, 1] |x = α(G) for some solvable group G}

.

(1.1)

What do the sets X, Xp, Xsol look like? What are their limit points?Chapter 3.1 addresses the second problem, and attains some prelim-inary results. If Xp = {x | α(G) = x for some p-group G} then it re-mains unknown what the limit points of Xp are. However, we have thefollowing theorems and conjecture.

Theorem 1.1. Any number of the form 1n, n > 1, is a limit point of

X, Xp for all primes p, and Xsol. Furthermore, 12is the largest limit

point of any of these sets.

Theorem 1.2. All the limit points of Xsol are of the form xn, where

n ∈ Z and x is a limit point of Xp for some p.

Question 1.3. Is it true that all non-zero limit points of X are of theform 1

nfor some natural number n?

For the theorems see 3.1. What can one say about the structure ofthe set X ∩ (1

2, 1)? A related result is the following:

Theorem 1.4. If for a group G we have α(G) > 56, then α(G) = 1.

See 3.1 for a proof. It seems plausible that one may be able tocompletely determine the image of α in (1

2, 1]. At any rate, it follows

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ON CAYLEY'S THEOREM 3

from the above theorems that for any ε > 0, |X ∩ (12

+ ε, 1)| < ∞, andtends to∞ as ε → 0; is it possible to determine the rate of divergence?Typically one expects that α(G) is a small number. This is con�rmed

by the following result from 3.2:

Theorem 1.5.

(1.2) limM→∞

1

M

M∑n=1

∑|G|=n α(G)∑|G|=n 1

= 0.

Question 1.6. Is it true that

(1.3) limM→∞

∑|G|≤M α(G)∑|G|≤M 1

= 0?

In light of the main result of [1] that almost all groups are solvablethe answer to this question should follow from a similar statement forsolvable groups. A widely believed conjecture states that almost allgroups are nilpotent. If one assumes this conjecture then the solutionof the last question is naturally linked to understanding the behaviorof α for p-groups.As pointed out earlier, it is in general a hard problem to calculate

α of a group G. For this reason it is a healthy practice to restrictattention to smaller classes of groups. One of the aims of this paperis to provide an algorithm to �nd the α of a given p-group P . Thealgorithm is presented in 4.4. Let us state the main theorem thatmakes the algorithm possible. Recall that the socle of a group G is theproduct of the minimal normal subgroups. When P is a p-group, thesocle M is the set of all elements of the center which have order p, andis elementary abelian of order pk, where k is the minimal number ofgenerators of the center.

Theorem 1.7. If I = {Hi}ki=1 is a collection of subgroups of P of size

k, such that for all x ∈ M there is some Hi ∈ I with x /∈ Hi andHi maximal with respect to that property, then I is a minimal faithfulcollection. Any p-group P will have such a minimal faithful collection.

For details of the proof and other useful information regarding thealgorithm see Section 4. In that section, especially in 4.2 and 4.3,we collect and prove many of the �ner properties of minimal faithfulcollections for p-groups not all of which will be used in justifying thealgorithm. These results, however, provide insight into the structuretheory of permutation representations for p-groups.This algorithm is much more e�cient than brute force algorithms,

and takes an amount of time and space proportional to the size of

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ON CAYLEY'S THEOREM 4

the subgroup lattice of G. The algorithm was even fortuitous, sinceMAGMA, the algebraic language in which it was coded, already pos-sessed a function which returned an array all the subgroups sorted inorder of size. We could not design an algorithm that did not examinethe subgroup lattice of G, although it seems likely to us that therecould be an algorithm which runs in time proportional to log n, wheren is the number of subgroups of G. This would certainly require a fastway to retrieve the subgroup lattice as a graph and not as an array,and some way to traverse the lattice without creating the whole graph.At a more theoretical level, using the material in Section 4 one can

answer various interesting questions already posed in literature. Forexample, in our terminology, Johnson poses the following questions onpage 864-5 of [6]:

Problem 1.8. Let Gp be the set of p-groups G with a minimal faithfulcollection I = {Hi}m

i=1 such that J = {Z(G) ∩ Hi}mi=1 is a minimal

faithful collection of Z(G). What is the extent of Gp?

Problem 1.9. For which G ∈ Gp is the following true? If Z(G) =Z1×Z2× . . .×Zn is any decomposition into non-trivial cyclic groups,and Hi is any subgroup for which Hi∩Z(G) = Z1× . . .×Zi−1×Zi+1×. . . × Zn and for which Hi is maximal with respect to this property,then I = {Hi}n

i=1 is a minimal faithful collection.

The following theorem answers both questions:

Theorem 1.10. Every p-group is in Gp, and every p-group satis�esthe conditions of the second problem.

See Remark 4.35 for more on these problems.Even with the possibly suboptimal algorithm above, many p-groups

of order p5 could be analyzed, as well as some for p6. The results canbe found in 4.5. Limited though they were, the values of α that werefound pointed to interesting questions which will mention below. Firsta de�nition:

De�nition. Given a prime number p and an integer n, set

(1.4) g(n, p) =∑|G|=pn

α(G).

Then numerical computations have led to the following conjectures:

Conjecture 1.11.

g(4, p) = 1 +5

p+

11

p2+

9

p3for p > 3.

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ON CAYLEY'S THEOREM 5

g(5, p) = 1+7

p+

33 + 2gcd(p− 1, 3) + gcd(p− 1, 4)

p2+

54

p3+

24

p4for p > 3.

One can easily compute g(1, p), g(2, p), and g(3, p).

De�nition. A function f(n) is polynomial on residue classes, or PORC,if f(n) can be expressed as a polynomial of n, whose coe�cients dependon the value of n mod k for some integer k.

In 1959, Graham Higman proposed a conjecture about counting p-groups, which has since been known as Higman's PORC conjecture[5].

Conjecture 1.12. [Higman's PORC Conjecture] The function f(n, p),which is the number of groups of order pn, when considered as a func-tion of p for n �xed, is PORC for all but �nitely many p.

Almost half a century has elapsed since the conjecture was proposed,and although most people currently believe it to be true, it remains aconjecture. Higman was only able to prove it for certain classes ofp-groups; others have shown it to be true for n ≤ 6 [14]. Recent�ndings have renewed interest in the conjecture, and shown connectionsbetween it and modular varieties [12]. However, our knowledge of p-groups for n > 6 is still quite limited, and even assuming the correctnessof the PORC conjecture we do not know how to enumerate all the p-groups. Here we would like to ask the following question:

Question 1.13. [PORC-α] Fix n. Is the function g(n, p), when con-sidered as a function of p, PORC as a function of 1

pfor all but �nitely

many p?

It is unlikely that the PORC-α could be attacked without provingHigman's PORC conjecture, although it is hoped that the same meth-ods which may eventually succeed in proving the PORC conjecturecould be applied to answer the PORC-α. Our numerical results alsosuggest that given any integral polynomial in 1

p, say Q, the number of

p-groups P with α(P ) = Q(1p) is also PORC.

We can also prove several results regarding nilpotent and solvablegroups. We only include the following:

Theorem 1.14. Let G be a solvable group, and let p1, . . . , pr be thedistinct prime factors of |G|. If Pi are the Sylow pi-subgroups of Gthen

α(G) ≤ 1

|G|

r∑i=1

α(Pi)|Pi|.

If G is nilpotent, this is an equality.

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ON CAYLEY'S THEOREM 6

For the proof of this theorem, along with other useful information,see [3].The only published paper we could �nd that discussed the topic in

depth in a fashion similar to ours is one by D. L. Johnson in 1971 [6],which is restricted to a discussion of d(G). Johnson's paper includesthe computation of d for abelian groups, a discussion on the size of theminimal faithful collection for nilpotent groups, and even a method forp-groups of size ≤ p4, as well as some other restricted p-groups. Bythe time we discovered Johnson's paper, we had already proven all butone of its theorems (the result mentioned in Remark 2.15). Johnson'spaper also proposed four questions for further study on the topic ofp-groups, two of which were already solved in our work as mentionedearlier.The research that led to the writing of this paper was instigated in

the fall of 2003 when Andre Kornell, then a Princeton sophomore en-rolled in MAT323, asked the third named author who was teaching thecourse whether Cayley's theorem can be improved. The third namedauthor wishes to thank the students of that class for their enthusiasmand patience while the ideas of this work shaped. Thanks are due toAndre Kornell for asking the questions, Evan Haas who worked outthe �rst examples, Aaron Silberstein for useful lunch discussions, JohnConway for being around, and Calderbank for reading a �rst draft ofthe paper.

2. Preliminaries

In order to study the number d(G), we start by the following. LetH be a subgroup of G. Then G has a permutation action π on thecollection of cosets X = G/H given by

π(g)(xH) = gxH,

for g ∈ G and xH ∈ G/H. Then we have the following simple lemma:

Lemma 2.1. The action π is faithful if and only if the largest normalsubgroup of G contained in H is the trivial group.

Proof. Suppose the action is not faithful. This means that there is ag ∈ G, such that π(g)(xH) = xH for all x. This implies that x−1gx ∈H for all x, which in turn gives

g ∈⋂x∈H

xHx−1.

The conclusion of the lemma is now obvious. �

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ON CAYLEY'S THEOREM 7

Let I = {H1, H2, . . . , Hr} be a collection of subgroups of G. We letXI be the disjoint union of the sets G/Hi for 1 ≤ i ≤ i. The groupG has a natural action πI on the set XI . We observe that any actionπ : G → SX of the group G is obtained as a πI for some choice of theset I. In order to see this, we write the set X as the union of disjointorbits. Then the action of G on each orbit is clearly isomorphic toG/H where H is the stabilizer of any element of the orbit. Now wehave the following generalization of the above lemma whose proof weleave to the reader.

Lemma 2.2. The action πI is faithful if and only if the group HI givenby

HI =r⋂

i=1

Hi

does not contain any normal subgroups of G other than the trivial group.

We call a collection I as above faithful if the corresponding action πI

is faithful. The following result is now obvious:

Proposition 2.3. We have

α(G) = minI faithful

∑H∈I

1

|H|.

2.1. Subgroups, direct products, and examples. Listed below aresome simple results about the relationship of the Cayley Constant of agroup to that of its subgroups.

Proposition 2.4. If H ≤ G, then d(H) ≤ d(G).

Proof. If ρ : G → Sn is injective, then the restriction of ρ to H is stillinjective, so d(H) ≤ n. Choose n such that n = d(G). �

Proposition 2.5. If H ≤ G and N is a faithful subgroup of H, thenN is a faithful subgroup of G.

Proof. Let X be any non-trivial subgroup of N . Then, since N isfaithful in H, X is not normal in H, so ∃h ∈ H such that hXh−1 /∈ X.But h ∈ G, so X is not normal in G. Thus no non-trivial subgroup ofN is normal in G, and N is faithful in G. �

Corollary 2.6. If H ≤ G and I is a faithful collection of H, then I isa faithful collection of G.

Proof. I is a faithful collection of H if⋂

N∈I N is faithful in H. Butthen

⋂N∈I N is faithful in G, so I is faithful in G. �

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ON CAYLEY'S THEOREM 8

Corollary 2.7. If H ≤ G, then α(G) ≤ α(H) ≤ α(G)[G : H].

Proof. d(H) ≤ d(G), meaning that α(H)|H| ≤ α(G)|G|, so α(H) ≤α(G)[G : H]. Now if we have a minimal faithful collection I of H, thenit is still faithful in G, so α(G) ≤

∑1|Hi| = α(H). �

Corollary 2.8. If |G| = n then α(G) ≥ 1(n−1)!

.

Proof. As shown by the Cayley Theorem, every group G is isomorphicto a subgroup of Sn. Thus α(G) ≥ α(Sn) = 1

(n−1)!. �

Now we have a few results about direct products of groups.

Proposition 2.9. If G and H are two groups, then d(G×H) ≤ d(G)+d(H).

Proof. If we have injective homomorphisms ρ : G → Sd(G) and γ :H → Sd(H) then we have the obvious injective homomorphism fromG×H → Sd(G)+d(H) given by ρ× γ : (g, h) 7→ ρ(g)× γ(h) ∈ Sd(G)+d(H),where ρ(g) permutes the �rst d(G) numbers and γ(h) permutes the lastd(H) numbers. �

Corollary 2.10. If G and H are two groups, then α(G×H) ≤ α(G)|H| +

α(H)|G| .

Proof. Again, using the basic de�nitions, α(G×H) = d(G×H)|G||H| ≤

d(G)+d(H)|G||H| =

α(G)|H| + α(H)

|G| . �

Corollary 2.11. If G and H are non-trivial, then α(G×H) < 1 unlessG = H = Z/2Z.

Proof. Both G and H have order ≥ 2, and unless they are both Z2 thenone has order ≥ 3. Thus

α(G×H) ≤ α(G)

|H|+

α(H)

|G|≤ 1

|H|+

1

|G|≤ 1

2+

1

3=

5

6< 1.

Example 2.12. Cyclic groups of prime power order are of Cayley type.If P =< x > and |x| = pn then any subgroup of P is < xpk

> for some

k ≤ n. If H1 =< xpk1 > and H2 =< xpk2 > then H1 ∩ H2 = <xpl>

where l = max(k1, k2). Thus if I is any collection of subgroups, then

∩H∈IH =< xpk> for k the maximal value in H, and < xpk

> is normaland non-trivial whenever k < n. Thus if I is faithful then k = n, so Icontains < xpn

>= 1. Finally, 1 ≥ α(P ) = minI faithful∑

1|Hi| ≥

11

= 1.

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ON CAYLEY'S THEOREM 9

Example 2.13. The generalized Quaternion 2-group

Q2n+1 =< h, x| h2n

= x4 = 1, x−1hx = h−1, h2n−1

= x2 >

is of Cayley type. This is because, once again, any non-trivial subgroupof Q2n+1 must contain x2, and < x2 > is normal in Q2n+1 . It shouldbe noted for later that the Quaternion 2-group and the Cyclic 2-groupare the only groups of order 2n+1 for n > 1 that have a unique elementof order 2 [6]. In particular, the group Q8 is of Cayley type.

Example 2.14. The Klein 4-group V = Z/2Z × Z/2Z is of CayleyType. This is true because if there were an injective homomorphismfrom V to Sn then V ∼= Im, the image of the homomorphism in Sn,and so |V | divides |Sn|. But 4 does not divide n! for n < 4, so theminimal n is 4.

Remark 2.15. It is a theorem in [6] that every group of Cayley type isof the form described in the previous three examples.

Example 2.16. The following fact regarding abelian groups is con-tained in [6]. For any abelian group G =< x1 > × . . .× < xr > with|xi| = qli

i for primes qi not necessarily distinct,

α(G) =

∑ri=1 qli

i∏ri=1 qli

i

=

∑ri=1 qli

i

|G|.

It follows from this result that among of all groups of order pn the onewith the smallest α is the elementary abelian p-group.

Theorem 2.17. α(D2n) = 12α(Z/nZ).

This theorem in the form presented here was discovered by EvanHaas. We present two proofs for this theorem.

First proof. This proof uses Theorem 1.14. First, we know that d(D2n) ≥d(Z/nZ) and thus α(D2n) = d(D2n)

|D2n| ≥d(Z/nZ)

2n= 1

2α(Z/nZ).

Let n = pa11 pa2

2 . . . parr . First, suppose n is odd. Then 2 6= pi. Since

D2n is solvable we have

α(D2n) ≤ 1

2n

r∑i=1

α(Pi)paii +

1

2n[2α(Z/2Z)] =

1

2α(Z/nZ) +

1

n,

where the 1nterm comes from the Sylow 2-complement in the faithful

subgroup I, which in this case is Z/nZ. However, since < σ > is notnormal in D2n, there is no minimal normal subgroup which is of order2, and that subgroup is unnecessary. Thus α(D2n) ≤ 1

2α(Z/nZ).

Suppose n is even. Then the Sylow 2-subgroup of D2n is D2k for k =2m. Now α(Z/kZ) = 1 since Z/kZ is a cyclic 2-group, and α(D2k) ≥

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ON CAYLEY'S THEOREM 10

12α(Z/kZ) = 1

2. But < σ > is a faithful collection of D2k since it is not

normal, so α(D2k) ≤ 1|<σ>| = 1

2. Thus α(D2k) = 1

2α(Z/kZ), and

α(D2n) ≤ 1

2n

∑pi 6=2

α(Pi)paii +

1

2n[2m+1α(D2k)]

=1

2[1

n

∑pi 6=2

α(Pi)paii +

1

n2mα(Z/2mZ)] =

1

2α(Z/nZ)

since Z/2mZ is the Sylow 2-subgroup of Z/nZ.Thus 1

2α(Z/nZ) ≤ α(D2n) ≤ 1

2α(Z/nZ) and α(D2n) = 1

2α(Z/nZ).

Second proof (due to Andre Kornell). This proof is independent of The-orem 1.14. The claim is obviously equivalent to d(D2n) = d(Z/nZ).Since any group that contains a subgroup isomorphic to D2n automat-ically contains the group Z/nZ, the statement that requires proving isthat the smallest symmetric group that contains Z/nZ also containsthe dihedral group. The statement is trivial for a prime power ordercyclic group, since the smallest symmetric group that contains an ele-ment of prime power order is of that prime power order c.f. Example2.12. The symmetric group must contain the dihedral group since itcontains all possible permutations of n points. We now generalize forarbitrary n. By Example 2.16 the generating element of Z/nZ takesthe form of disjoint cycles of prime power order. We now need to showthe existence of an element with the appropriate commutation proper-ties. This element is simply the product of the corresponding elementsfor each cycle in the decomposition, the existence of which was shownfor groups of prime power order. �

3. Distribution of α

3.1. Limit Points. Let Hp = Z/nZ × Z/pZ for a �xed n > 1. Forp > n, (n, p) = 1, and then Example 2.16 implies that

α(Hp) =p +

∑ri=1 qli

i

pn=

1

n+

1

p

∑ri=1 qli

i

n=

1

n+

1

pα(Z/nZ)

where∏

qlii = n is the prime factorization of n. But then as p → ∞,

α(Hp) → 1n. We now have an in�nite sequence of groups whose Cayley

Constant converges to 1n.

Let X = {α(G) |G is a �nite group}. Then 1nis a limit point of X!

This raises the natural question of whether or not we can categorize allthe non-zero limit points of X.

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ON CAYLEY'S THEOREM 11

Proposition 3.1. The largest limit point of X is 12.

Proof. Once again, if |G| = n and n is not a prime power, then any twoSylow subgroups will form a faithful collection. Thus α(G) ≤ 1

q+ 1

r

where q and r are any two prime power divisors of n. Although thereare certain small values for n which will allow α(G) > 1

2, it is clear that

after a point the only n for which there are not two prime divisors qand r with 1

q+ 1

r≤ 1

2will be of the form n = 2pa for p an odd prime.

Thus the only groups we need be concerned with are either of the formn = 2pa or of the form n = pa.Any group G of order 2pa will have a Sylow p-group S of order pa,

and once again by the subgroup rule, α(S) ≥ α(G) > 12. If α(S) = 1

then S must be cyclic, so G is either cyclic or the Dihedral group. Inthe former case α(G) = 1 and in the latter α(G) = 1

2α(Z/nZ) ≤ 1

2.

If α(S) 6= 1, and if we can show that the largest limit point of α forgroups of order pa is 1

2, then clearly the largest limit point for G will

be 12. So we need only check limit points of prime power groups.

Let |G| = pa. If 1 > α(G) > 12then we still have that α(Z(G)) > 1

2

and Z(G) is not cyclic. If Z(G) has two generators then some minimalfaithful collection has 2 elements, so α(G) = 1

pn + 1pm with n,m ≥ 1,

but the largest limit point the set { 1pn + 1

pm} is 12. But if Z(G) has k

generators of order ai then α(Z(G)) =∑k

i=1 pai

pa , so the only possibility

where α(Z(G)) > 12and k ≥ 3 is when Z(G) = Z/2Z× Z/2Z× Z/2Z.

Once again, some minimal faithful collection has 3 elements and is well-indexed, so if H is in the collection then |H ∩Z(G)| = 4, implying that|H| > 4. Hence α(G) = 1

2m + 12n + 1

2o for m,n, o ≥ 2, and once again,the largest limit point of these numbers is 1

2. �

In fact, this �nal argument can be extended to any prime p:

Proposition 3.2. There is no limit point of α(P ) for p-groups Pgreater than 1

p.

Proof. We know the result is true when p = 2, so let p ≥ 3. Let|G| = pa and let Z(G) have k generators for k > 1 (clearly when k = 1,α(G) ≤ 1

por α(G) = 1). Then any minimal faithful collection has k

elements (assuming p ≥ 3), and each must be of size at least pk−1,since that is the size of an element's intersection with the socle. Thusα(G) = 1

pk−1 (∑k

i=11

pni) for some ni ≥ 0. This achieves its maximum

value when ni = 0 for all i, which is when α(G) = kpk−1 . But k

pk−1 >1p⇒ logp k > k − 2, which has no solutions for k > 2 when p ≥ 3, so

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ON CAYLEY'S THEOREM 12

k ≥ 3 ⇒ α(G) ≤ 1p. When k = 2, α(G) = 1

pn + 1pm where m, n ≥ 1, and

the largest limit point of such numbers is 1p. �

In fact, this also shows that for p > 3, either α(P ) ≤ 2por α(P ) = 1.

And now we have the �nal theorems of this chapter.

Theorem 3.3. Let Y be the set of all positive integers that only containprime factors from among the set π = {p1, p2, . . . , pk}. Then the onlynon-zero limit points of W = {α(G) s.t. |G| ∈ Y and G is solvable}are of the form x

mwhere m ∈ Y and x is a limit point of α(P ) for

p-groups P with p ∈ π.

Proof. We have shown that for solvable groups of order n = pa11 pa2

2 . . . pakk ,

α(G) ≤ 1

|G|

k∑i=1

α(Pi)|Pi| =k∑

i=1

α(Pi)∏j 6=i p

aj

j

where Pi is a Sylow pi-subgroup of G. This inequality is true evenwhen ai = 0, since the extra term merely increases the right side ofthe inequality. Now, any in�nite sequence of groups in W whose αconverges must have orders which go to in�nity, so there is at leastone pt such that at is unbounded. If α of the sequence is convergentthen any α of any subsequence converges to the same number, so wechoose a subsequence where at →∞. If within that subsequence someother as is unbounded, then we choose a further subsequence whereat → ∞ and as → ∞. Otherwise,

∏j 6=t p

aj

j will be bounded withinthe subsequence, so choose a further subsequence where that productconverges as well (i.e. it is constant since the choices are discrete). Inthat case let m =

∏j 6=i p

aj

j , m ∈ Y .

Since α(Pi) ≤ 1, we get that the term α(Pi)∏j 6=i p

ajj

will become negligible

whenever i 6= t, since patt is in the denominator. If as →∞ then every

term in the sum becomes negligible, and the limit of α is 0. Otherwise,the only remaining term in the limit sum is α(Pt)

m, and the limit of α of

the sequence must be xmwhere x is a limit point of α(Pt). �

Theorem 3.4. If W is a sequence of solvable groups where α converges,and there is no �nite set π of primes such that all the prime factorsof |G| are in π for all G ∈ W , then limG∈W α(G) = 0 or 1

nfor some

n ∈ Z.

Proof. Once again, we use α(G) ≤∑k

i=1α(Pi)∏j 6=i p

ajj

. Let pt be the max-

imal prime in a given group G ∈ W . If {pt} is bounded then π ={p s.t. p is prime and p ≤ pt} will violate the requirements of this

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ON CAYLEY'S THEOREM 13

theorem. So without loss of generality (by looking at subsequences)

pt →∞. Again, for any i 6= t, α(Pi)∏j 6=i p

ajj

will be negligible. More impor-

tantly, if there are more than 2 primes then∑i6=t

α(Pi)∏j 6=i p

aj

j

≤ 1

patt

∑i6=t

1∏j 6=i,t p

aj

j

≤ 1

pt

∑k−1

1

2k−2=

1

pt

k − 1

2k−2<

2

pt

is negligible, so the only remaining term in the sum is α(Pt)∏j 6=t p

ajj

. However,

once pt > 3, either α(Pt) = 1 or α(Pt) ≤ 2pt. So either there is a

subsequence where α(Pt) = 1, and lim α(G) = lim 1∏j 6=t p

ajj

= { 1nor 0},

or α(Pt) → 0 and lim α(G) = 0. Note that the former case is possible,since we can let Gi = Z/2Z×Z/piZ where pi is the i-th prime, and aswe have already shown, lim α(Gi) = 1

2. �

Corollary 3.5. The only non-zero limit points of {α(G) |G is solvable}are of the form x

n, where n ∈ Z and x is a limit point of {α(P ) |P is a p-group}.

It seem likely that one maybe able to completely determine the imageof α in the interval (1

2, 1]. We only include the following proposition:

Proposition 3.6. If α(G) > 56, then α(G) = 1.

Proof. Let |G| = n, and let α(G) > 56. Suppose that n has more than

one prime factor. Then let H1 and H2 be Sylow p-groups of G fordi�erent primes p. Since H1 ∩ H2 = 1, we see that I = {H1, H2} is afaithful subgroup with µ(I) = 1

|H1| +1|H2| ≤

12

+ 13

= 56. So α(G) ≤ 5

6, a

contradiction; hence, n is a prime power.So G is a p-group. Since Z(G) is a subgroup of G, we have that

α(G) ≤ α(Z(G)) and thus α(Z(G)) > 56. However, Theorem 2.16

implies that Z(G) is cyclic or Z(G) = Z/2Z × Z/2Z, and either way,α(Z(G)) = 1. If Z(G) is cyclic then we know that some minimalfaithful collection of G will have exactly 1 element, and thus eitherα(G) ≤ 1

2or α(G) = 1. If Z(G) = Z/2Z × Z/2Z then some minimal

faithful collection will have exactly 2 elements, but p = 2, and thuseither α(G) = 1 or α(G) ≤ 1

2+ 1

4= 3

4. So if α(G) > 5

6then α(G) =

1. �

This proposition showed that 1 is an isolated point of X, and thusnot a limit point.

3.2. The Average of α. The purpose of this paragraph is to show thata certain average of α over all groups tends to 0. More speci�cally,

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ON CAYLEY'S THEOREM 14

Theorem 3.7.

1

M

M∑n=1

∑|G|=n α(G)∑|G|=n 1

→ 0

as M →∞.

Proof. It is important to note that at all points in this proof, we willbe using the bound α(H) ≤ 1 for all groups H of order |H| = pa forsome a. Now let G be a solvable group of order n = pa1

1 pa22 . . . pak

k .Philip Hall's Theorem, in one form, states that G is solvable i� everySylow group has a complement (p. 105 of [2]). In other words, forall subgroups Pi of order pai

i there exists a subgroup Bi (called theSylow pi-complement) of order [G : Pi] = n

paii

such that PiBi = G. Let

I = {Bi}ki=1. Then HI =

⋂Bi must be the trivial subgroup, since if

x ∈ HI then |x| divides |Bi| for each i, but this means that |x| can nothave pi as a factor for each i, and thus |x| = 1. Hence we have that Iis a faithful collection of subgroups of G, and

α(G) ≤k∑

i=1

1

|Hi|=

k∑i=1

paii

n=

1

n

k∑i=1

paii .

Since this will be true for all solvable G of size n, the average ofα(G) over solvable groups of size n will be less than 1

n

∑ki=1 pai

i . Soover solvable groups

1

M

M∑n=1

∑|G|=n α(G)∑|G|=n 1

≤ 1

M

M∑n=1

1

n

k∑i=1

paii ,

and we seek a bound for the latter sum.For prime powers n, where k = 1 and pa1

1 = n, the bound α(G) ≤1n

∑ki=1 pai

i is precisely α(G) ≤ 1, which we knew before. Luckily, the

number of prime powers less than M grows as Mlog M

according to thePrime Number Theorem, so we can consider these separately. We de-note the set of natural numbers ≤ M that are not prime powers AM .Now it is useful to introduce the notation that n = mip

aii , so that

|Bi| = mi, and∑

n∈AM

1n

∑ki=1 pai

i =∑

n∈AM

∑ki=1

1mi. We estimate

this by looking at the set of all mi for all n ∈ AM . Now if m is anatural number such that there is a p and an r satisfying(1) (m, p) = 1 and(2) m · pr ≤ M (or pr ≤ M

m),

then n = m · pr ∈ AM , and m will be one of the mi for that n. Itis obvious from these criteria that m ≤ M

2, since pr ≥ 2. Instead of

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ON CAYLEY'S THEOREM 15

summing over all n ∈ AM and over all mi, we can instead sum over allm ≤ M

2and all pr that satisfy the criteria. So

∑n∈AM

k∑i=1

1

mi

=∑

m≤M2

∑pr≤M

mand (m,p)=1

1

m

∑m≤M

2

1

m

∑pr≤M

m

1

∑m≤M

2

1

m(

Mm

log Mm

)

= M∑

m≤M2

1

m2 log Mm

.

Since we will later be dividing by M , we want a bound for the sumwhich goes to 0 as M → ∞. To do this, choose δ ∈ (0, 1) such that1 < M δ < M

2, and split the sum into two ranges, where m ≤ M δ and

where m > M δ. For the �rst,

∑m≤Mδ

1

m2 log Mm

≤∑

m≤Mδ

1

m2

1

log M − log M δ

<1

(1− δ) log M

∑m

1

m2=

ζ(2)

(1− δ) log M.

For the second,

∑Mδ<m≤M

2

1

m2 log Mm

≤ 1

log 2

∑Mδ<m≤M

2

1

m2<

1

log 2

∑m>Mδ

1

m2

<1

log 2

∫ ∞

1

x2dx =

1

M δ log 2.

And so �nally, we have

1

M

∑n∈AM

∑|G|=n α(G)∑|G|=n 1

� ζ(2)

(1− δ) log M+

1

M δ log 2.

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ON CAYLEY'S THEOREM 16

This means that for all solvable groups,

1

M

M∑n=1

∑|G|=n α(G)∑|G|=n 1

≤ 1

M

∑prime powers ≤M

1 +1

M

∑n∈AM

∑|G|=n α(G)∑|G|=n 1

� 1

M

M

log M+

1

(1− δ) log M+

1

M δ� 1

(1− δ) log M+

1

M δ.

For any 0 < δ < 1 this sum will go to 0 as M →∞!If we wish to clarify the asymptotics by eliminating dependence on

δ, we see that the restriction on δ is that M δ < M2, implying that

δ < log M−log 2log M

→ 1 as M →∞. Hence we can let δ be arbitrarily closeto 1 and the asymptotics given by this estimation on the average are

1log M

+ 1M. �

4. P-groups

4.1. P-groups and the socle. Here we collect various elementaryfacts for ease of reference.

De�nition 4.1. The socle of G is the product of all its minimal normalsubgroups.

The intersection of two normal subgroups is another normal sub-group. Hence, it is obvious that any two minimal normal subgroupsmust intersect trivially, or else their intersection would be a smaller,normal subgroup. Recall the important fact (which we will use often)that if H E G and K E G and H ∩K = 1 then HK ∼= H ×K.

Theorem 4.2. (p. 87, [11]). Suppose that K1, . . . , Kn are minimalnormal subgroups of a group G and K = K1K2 . . . Kn. Then there issome subset of the groups, indexed by i1 to im such that K = Ki1 ×Ki2 × . . .×Kim.

Proof. Let X denote all the non-empty subsets {i1, . . . , im} of {1, . . . , n}such that Ki1Ki2 . . . Kim = Ki1×Ki2×. . .×Kim . Single element subsets{j} trivially fall into X. Now choose x ∈ X with m as large as possible,and let L = Ki1 ×Ki2 × . . .×Kim . Clearly L E G and L ≤ K.Now suppose K 6= L. Then there exists some Kj such that Kj is not

a subgroup of L. Then Kj ∩ L = 1 and thus KjL = Kj × L. But nowone can see that

Ki1Ki2 . . . KimKj = LKj = L×Kj = Ki1 ×Ki2 × . . .×Kim ×Kj,

contradicting the maximality of m. Therefore K = L. �

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ON CAYLEY'S THEOREM 17

Corollary 4.3. If M is the socle of G, then M is isomorphic to thedirect product of some minimal normal subgroups.

Now we turn to p-groups.

Theorem 4.4. (p. 188, [2]) Let P be a p-group of order pn, and H bea subgroup of P. Then(1) Z(P ) 6= 1.(2) If H E P and H 6= 1 then H ∩ Z(P ) 6= 1.(3) If H E P and H 6= 1 then there exists K < H, |K| = p, K E P .

Proof. Let a1, . . . , ar be representatives of the non-central conjugacyclasses of P , such that a1, . . . , ak ∈ H and ak+1, . . . , ar /∈ H. Then

by the class equation |P | = |Z(P )|+∑r

i=1|P |

|CP (ai)| . However, p divides

|P | and p divides |P ||CP (ai)| for all i, so since |Z(P )| ≥ 1, p must divide

|Z(P )|. Thus Z(P ) 6= 1, proving (1). Let H 6= 1 be normal in P . Nowany conjugacy class C must either be contained within H or disjointwith H, since H is normal. So restricting the class equation to H

we get that |H| = |H ∩ Z(P )| +∑k

i=1|P |

|CP (ai)| . Once again, p divides

every term other than |H ∩ Z(P )|, so p must divide |H ∩ Z(P )| andH ∩ Z(P ) 6= 1, proving (2). By (2), H ∩ Z(P ) is non-trivial, so itcontains an element of order p which generates a subgroup K of orderp. Clearly K is normal since it is central, proving (3). �

Corollary 4.5. The minimal normal subgroups of a p-group P are thesubgroups of Z(P ) of order p.

Proof. If |H| > p and H is normal then by (3) there is a smallernormal set contained in it, so H is not minimal. If |H| = p then by (2)|H ∩Z(P )| is at least of order p, so H = H ∩Z(P ) and H ≤ Z(P ). Soa minimal normal subgroup is of order p and is contained in Z(P ).If H ≤ Z(P ) and |H| = p then clearly H is normal and minimal. �

Corollary 4.6. The socle M of a p-group P is the set of all elementsof Z(P ) of order p.

Proof. If z ∈ Z(P ) and |z| = p then < z > is a minimal normal sub-group, so < z >≤ M . If m ∈ M , then m is the product of commutingelements of Z(P ) of order p, so m ∈ Z(P ) and |m| = p. �

Corollary 4.7. Every non-trivial x ∈ M generates a minimal normalsubgroup of P .

Proof. Every x ∈ M is central and of order p, so it generates a centralsubgroup of order p, which is minimal and normal. �

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ON CAYLEY'S THEOREM 18

Theorem 4.8. The socle M of a p-group P is elementary abelian oforder pk, where k is the minimal number of generators of Z(P ).

Recall Theorem 4.2, which showed that the socle is the direct productof some of the minimal normal subgroups. This, or Corollary 4.6, issu�cient to show that M is elementary abelian. It is also simple toshow that |M | = pk, but I shall use the process below because it playsan important role in the p-group algorithm to come. It functions alongthe same lines as the proof of Theorem 4.2.

Lemma 4.9. Let B be an elementary abelian p-group and let C < Bbe a proper subgroup of size pa. If b ∈ B but b /∈ C, then C < b >∼=C× < b > is a subgroup of B of size pa+1.

Proof. The proof is immediate from the fact that if H,K E B andH ∩K = 1 then HK ∼= H ×K. �

Lemma 4.10. Let P be a p-group and let M be its socle. Let K0 = 1.For 1 ≤ i ≤ n, let bi be any element of M not contained in Ki−1, andlet Ki = Ki−1 < bi >. Then for any 1 ≤ i ≤ n, Ki is elementaryabelian of order pi. Furthermore, n ≤ l, the number of generators ofM, and the process can always be continued until n = l.

Proof. Everything said follows immediately from the previous lemma,since M is elementary abelian and thus of order pl. �

This Lemma is important, because it provides us an inductive wayto build the socle and �nd a set of its minimal generators. And now,to complete the proof of Theorem 4.8.

Proof of Theorem 4.8. Let {a1, . . . , ak} be a set of minimal generators

of Z(P ), so Z(P ) ∼=< a1 > × . . .× < ak >. Let bi = a|ai|/pi , so that bi is

an element of order p in < ai > (in fact, it generates all the elements oforder p in < ai >). Applying the method in the lemma above, we getthat Ki−1 ∼=< b1 > × . . .× < bi−1 >, which clearly does not containbi. This process yields Kk such that Kk ≤ M , and Kk is elementaryabelian of order pk. But Kk is the largest elementary abelian groupcontained in Z(P ) (if there is an elementary abelian subgroup of orderpk+1 within Z(P ) then Z(P ) must have at least k + 1 generators), sobecause M is also an elementary abelian subgroup of Z(P ), |M | ≤ |Kk|and Kk = M . �

4.2. Anticovers. The following statement while trivial is extremelyuseful:

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ON CAYLEY'S THEOREM 19

Proposition 4.11. A collection of subgroups I is faithful in G if, forevery minimal normal subgroup N ≤ G, there is some H ∈ I such thatN is not a subgroup of H.

To �nd a faithful collection one need only choose a set of subgroupssuch that each minimal normal subgroup is not contained in one ofthem. To simplify the notation a bit, we introduce the following de�-nition:

De�nition 4.12. An anticover of a subgroup H ≤ G (or an elementx ∈ G) is a subgroup K ≤ G such that H � K (or x /∈ K). In thisinstance, we say that K anticovers H. A maximal anticover of H (orx, etc.) is an anticover of H such that there is no other anticover of Hof larger size. Then, by Proposition 4.11, a collection I is faithful i� itcontains an anticover for every minimal normal subgroup.

Proposition 4.13. A collection I of a p-group P is faithful i� it con-tains an anticover for every non-trivial x ∈ M , the socle.

Proof. This follows immediately from the fact, proven in the previoussection, that every non-trivial x ∈ M generates a minimal normalsubgroup. �

It is obvious that every non-trivial element of G has a anticover ofmaximal size, even if that maximal anticover is the trivial subgroup.Note that each group H may have multiple maximal anticovers of thesame size, and that a given set can be a maximal anticover for morethan one minimal normal subgroup. One viable method to �nd a faith-ful collection would be to �nd all the minimal normal subgroups, foreach �nd the maximal anticovers, and choose a collection from amongthose anticovers which contains at least one anticover for each minimalnormal subgroup. However, this is far more complex than is required.For instance:

Note 4.14. If Z(P ) has k generators then there will be 1+p+ . . .+pk−1

minimal normal subgroups.

Proof. M will be of order pk, and will be the set-union of t di�erentminimal normal subgroups, all of which have size p and intersect in theidentity. Thus

pk = |M | = 1 +∑

minimal normal subgroups N

(|N | − 1) = 1 + t(p− 1)

and t = pk−1p−1

= 1 + p + . . . + pk−1. �

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ON CAYLEY'S THEOREM 20

4.3. Well-indexed, regular, and antimaximal collections.Throughout the rest of this chapter, we will let P denote a prime

power group for prime p, M the socle of P , and k the minimal numberof generators of M . All collections are assumed to be collections ofsubgroups of P . We now de�ne three kinds of faithful collections, andendeavor to compare them.

De�nition 4.15. Let K be a subgroup of P . We say that K is well-indexed if, for all subgroups B of the socle such that B is not containedin K, [B : B ∩ K] = p. In other words, for all B ≤ M either [B :B ∩K] = p or [B : B ∩K] = 1. We say a faithful collection I = {Ki}is well-indexed if Ki is well-indexed for all i.

This is the same as saying that if K is an anticover for some x ∈ Bthen [B : B ∩K] = p.

De�nition 4.16. We say that a faithful collection I = {Ki} of ap-group P is regular if it is of size k, and there are some {xi}k

i=1,generators of M , such that Ki a maximal anticover of xi for all i, andxi ∈ Kj for all j 6= i. We call {xi} the regular generators of I.

And �nally,

De�nition 4.17. We say that a faithful collection I = {Ki} of a p-group P is antimaximal if for every nontrivial x ∈ M , there is someKi ∈ I such that Ki is a maximal anticover for x.

First, we show some basic facts about well-indexed collections.

Theorem 4.18. [Well-Indexed Theorem] Any p-group P has a minimalfaithful collection which is well-indexed. If p 6= 2 then every minimalfaithful collection is well-indexed.

Proof. We must show that if B is an elementary abelian p-group con-tained in the socle, and Lj is an anticover for some x ∈ B, then[B : B ∩ Lj] = p. Suppose that [B : B ∩ Lj] ≥ p2. Following Lemma4.9, we can �nd x, y ∈ B such that x /∈ Lj, y /∈ Lj < x >, and then(Lj ∩M) < x >< y >∼= (Lj ∩M)× < x > × < y >. But in that case,let L1 = Lj < x > and L2 = Lj < y >. Clearly Lj ∩M ≤ L1∩L2∩M .If z ∈ L1∩L2∩M then z = lxi = l′yj for some l, l′ ∈ Lj, so yjx−i ∈ Lj.So yj ∈ Lj < x >, but < y > ∩(Lj < x >) = 1, so yj = 1. Thenx−i ∈ Lj, but < x > ∩Lj = 1 so x−i = 1. So yj = xi = 1, and z ∈ Lj.Thus L1∩L2∩M ≤ Lj ∩M , so both sides are equal. If z ∈ M , z /∈ Lj,then either z /∈ L1 or z /∈ L2.Every element anticovered by Lj is anticovered by either L1 or L2.

Thus J ′ = {Li}i6=j ∪ {L1, L2} is a faithful collection of P . L1 is bigger

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ON CAYLEY'S THEOREM 21

than Lj, so |L1| ≥ p|Lj|; the same is true of L2. Then

µ(J ′) = µ(J)− 1

|Lj|+

1

|L1|+

1

|L2|≤ µ(J)− 1

|Lj|+

1

p|Lj|+

1

p|Lj|

= µ(J) +1

|Lj|(2

p− 1) ≤ µ(J),

where the inequality is strict when p > 2. For p > 2 this contradictsthe minimality of J , proving that J is well-indexed.For p = 2, this will still contradict the minimality of J unless |L1| =

|L2| = 2|Lj|. We know that p ≤ [B : B ∩ L1] < [B : B ∩ Lj], andthe same is true of L2. For any other subgroup C of M we havethat [C : C ∩ L1] ≤ [C : C ∩ Lj], because Lj ≤ L1. Thus replacing{Lj} with {L1, L2} will not change µ but will lower the sum of theindices [B : B ∩ L] over all L in the collection, and will not increase[C : C ∩ L] for any C 6= B. We can repeat the process on L1 and L2

if necessary, or on other Li, until [B : B ∩ L] = p or 1 for all L in thecollection. Then, we can repeat the process again for other subgroupsC ≤ M , until we arrive at a �nal collection where, for all subgroups C,[C : C ∩ L] = p or 1 for all L. At every step the Cayley sum µ of thecollection is unchanged, so the �nal result will be a collection which isminimal and well-indexed. �

We already know of a minimal faithful collection which is not well-indexed: the trivial collection of the 4-group. If G =< x1 > × < x2 >with x2

1 = x22 = 1, then as we have seen, α(G) = 1. Thus {1} and

{< x1 >,< x2 >} are two distinct minimal faithful collections. Theformer is not well-indexed, since [G : G ∩ 1] = 4 6= 2. We are onlyinterested in collections that are well-indexed, like the latter, becausethey have several nice properties.

Lemma 4.19. [Minimal Collection Index Lemma, or MCIL] Let J ={Li}m

i=1 be a well-indexed minimal faithful collection. If {i1, i2, . . . , ir}is some subset of {1, 2, . . . ,m} not containing j, then

[M ∩ Li1 ∩ Li2 ∩ . . . ∩ Lir : M ∩ Li1 ∩ Li2 ∩ . . . ∩ Lir ∩ Lj] = p.

Proof. Suppose �rst that

[M ∩ Li1 ∩ Li2 ∩ . . . ∩ Lir : M ∩ Li1 ∩ Li2 ∩ . . . ∩ Lir ∩ Lj] = 1.

If x ∈ M has Lj as an anticover, then it is not in M ∩Li1 ∩Li2 ∩ . . .∩Lir ∩Lj, so it is not in M ∩Li1∩Li2∩ . . .∩Lir either. Then some Lim isan anticover of x as well. Thus Lj adds no new anticovered elements,and J ′ = {Li}i6=j still has an anticover for every element of M , so it

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ON CAYLEY'S THEOREM 22

is faithful. Since µ(J ′) = µ(J)− 1|Lj | we have that J is not minimal, a

contradiction. Thus

[M ∩ Li1 ∩ Li2 ∩ . . . ∩ Lir : M ∩ Li1 ∩ Li2 ∩ . . . ∩ Lir ∩ Lj] 6= 1,

and some x ∈ M ∩ Li1 ∩ Li2 ∩ . . . ∩ Lir is not contained in Lj. Then,by the de�nition of well-indexed, the index must be p. �

Corollary 4.20. Let J = {Li}mi=1 be a well-indexed minimal faithful

collection. Then |M ∩ La1 ∩ . . . ∩ Lar | = pk−r for all subsets {ai}ri=1 of

{1, . . . , k}.

Proof. This follows immediately from repeated application of the MCIL.�

Corollary 4.21. Let J = {L1, . . . , Ln} be a well-indexed minimal faith-ful collection. Then it has size k, the number of generators of the socle.

Proof. Since |M ∩L1 ∩ . . .∩Li| = |M |/pi, if n < k the collection is notfaithful because |M ∩ L1 ∩ . . . ∩ Ln| > 1, and if n > k then Lk+1 cannot have index p in M ∩ L1 ∩ . . . ∩ Lk = 1. �

Corollary 4.22. If J is a minimal faithful collection of size k, then Jis well-indexed.

Proof. We already showed that if J were minimal but not well-indexedthen we could create a larger well-indexed minimal faithful collection.However, any well-indexed minimal faithful collection has size k, so nolarger such collection exists. Thus J is well-indexed. �

The following lemma provides the link between well-indexed collec-tions and maximal anticovers.

Lemma 4.23. [Anticover Index Lemma, or AIL] If B is an elementaryabelian p-group contained in the socle, and K is a maximal anticoverof x ∈ B, then [B : B ∩K] = p.

Proof. Both B and B ∩K are elementary abelian p-groups, and B ∩Kis smaller than B because it does not contain x, so [B : B ∩ K] ≥ p.By Lemma 4.9 we have that (B ∩ K) < x > is a larger elementaryabelian p-group, and if [B : B ∩K] > p then there is some remainingelement y ∈ B, y /∈ (B ∩ K) < x >. Let Y =< y >. In thatcase, KY will not contain x. For suppose x = ky0 with k ∈ K andy0 ∈ Y non-trivial, then k = xy−1

0 ∈ B, so k ∈ B ∩ K. But thisis impossible, since then y0 = k−1x ∈ (B ∩ K) < x >, which is falsebecause y ∈ <y0>. Therefore KY is also an anticover of x, and is largerthan K, contradicting the maximality of K. Thus [B : B∩K] = p. �

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ON CAYLEY'S THEOREM 23

Corollary 4.24. If B is as above, K1 is a maximal anticover of x ∈ B,and K1, . . . , Kj are such that Ki is a maximal anticover for x ∈ B ∩K1 ∩ . . . ∩Ki−1, then |B ∩K1 ∩ . . . ∩Kj| = |B|/pj.

Proof. This follows immediately from repeated application of the abovelemma. �

Lemma 4.25. Let J = {L1, . . . , Lk} be a well-indexed minimal faithfulcollection. Then it is regular.

Proof. First we �nd a set of generators for which Li is an anticoverfor xi and Lj is not. For all i, let xi be an arbitrary element in C =M ∩ L1 ∩ . . . ∩ Li−1 ∩ Li+1 ∩ . . . ∩ Lk. xi /∈ Li, since C ∩ Li = 1 by theMCIL. If {a1, a2, . . . , ar} is some subset of {1, 2, . . . , k} not containingj, then xj /∈< xa1 >< xa2 > . . . < xar > because each xam is containedin Lj, so any product of them must be contained in Lj, and xj is not.From this it immediately follows from induction that

< x1 > . . . < xi >∼=< x1 > × . . .× < xi >

and is elementary abelian of order pi for 1 ≤ i ≤ k. When i = k theresult is elementary abelian of order pk and is contained in M , andtherefore it is M . Thus {xi} is the desired set of generators.Suppose that Lj is not a maximal anticover for xj. Let Kj be a

maximal anticover for xj. C as above is a subgroup of the socle of orderp, by the MCIL, and since xj ∈ C by construction, the AIL impliesthat [C : C ∩ Kj] = p. Thus C ∩ Kj = 1 and J ′ = {Li}i6=j ∪ {Kj}is a faithful collection. But µ(J ′) = µ(J) − 1

|Lj | + 1|Kj | < µ(J) since

|Kj| > |Lj|. This contradicts the minimality of J , and thus Li is amaximal anticover for xi. �

Remark 4.26. It should be noted that, although we have not found acounterexample, we do not know yet if every set of generators {xi}necessarily corresponds to some regular faithful collection. There maybe no maximal anticover of xi which contains all of xj for j 6= i.

Lemma 4.27. Let J = {Li} be a well-indexed minimal faithful collec-tion, and {xi}k

i=1 its regular generators. Then M ∩ La1 ∩ La2 ∩ . . . ∩Lam =< xb1 >< xb2 > . . . < xbn > where {xai

}mi=1 and {xbi

}ni=1 are

disjoint subsets of {xi}ki=1, and n + m = k.

Proof. This follows from the fact that xi ∈ Lj for all j 6= i. LettingC = M ∩ La1 ∩ . . . ∩ Lam , and letting {xbi

} be the set of all remaininggenerators, then D =< xb1 > . . . < xbn > is contained in C, since eachxbi

∈ C. But |D| = |C| since both are elementary abelian of orderpn = pk−m by the MCIL, so D ≤ C ⇒ D = C. �

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ON CAYLEY'S THEOREM 24

Lemma 4.28. Let J = {Li} be a regular faithful collection (NOTassuming minimal), with {xi}k

i=1 the regular generators, such that M ∩La1 ∩ . . . ∩ Lam =< xb1 > . . . < xbn > as in the preceding lemma. Forany subset {xbi

}ni=1, if x ∈< xb1 > . . . < xbn > is non-trivial then some

L ∈ {Lb1 , . . . , Lbn} is a maximal anticover for x.

Proof. We prove the statement by induction on n. If n = 1 and xis non-trivial, then x ∈< xb1 >⇒ x /∈ Lb1 , as desired. Now assumethat for any combination of generators of length n − 1, the statementis true, and suppose without loss of generality that there is a non-trivial x ∈< x1 > . . . < xn > such that {Li}n

i=1 does not contain anymaximal anticovers of x. Now x = xa1

1 . . . xann . If ai = 0 for any i, then

x ∈ <x1> . . . < xi−1 >< xi+1 > . . . < xn >, which is a set of generatorsof length n − 1. Then one of {L1, . . . , Li−1, Li+1, . . . , Ln} must be amaximal anticover for x by the inductive hypothesis, a contradiction.So we can assume that ai 6= 0 for all i. Now x /∈ Li for all 1 ≤ i ≤ n,since M ∩ Li =< x1 > . . . < xi−1 >< xi+1 > . . . < xk > by the abovelemma, and x is not contained in that product.Let K be a maximal anticover of x, such that |K| > |Li| for all

1 ≤ i ≤ n, since they are all anticovers of x but are not maximal. Ifxi /∈ K for any 1 ≤ i ≤ n, then K would be a larger anticover than Li,which is a maximal anticover of xi; thus xi ∈ K for all 1 ≤ i ≤ n. Butthis implies that x = xa1

1 . . . xann is also in K, a contradiction! So some

Li is a maximal anticover for x, and the induction is valid. �

Corollary 4.29. A well-indexed minimal faithful collection J is regularand antimaximal.

Proof. If x ∈ M then x ∈< x1 > . . . < xk > so some L ∈ J is a maximalanticover for x, by the above lemma. Thus J is antimaximal. �

Corollary 4.30. Every p-group has a well-indexed, regular, antimaxi-mal minimal faithful collection.

Now we investigate the properties of the last of our descriptors, theantimaximal collections.

Lemma 4.31. A faithful antimaximal collection J must have at leastk elements.

Proof. There is some non-trivial x1 ∈ M , so it must have some maximalanticover K1 ∈ J , and by the AIL, |M ∩K1| = pk−1. Then, if k > 1,there is some non-trivial x2 ∈ M ∩K1, so it must have some maximalanticover K2 ∈ J , with |M ∩ K1 ∩ K2| = pk−2 . It is obvious that,numbering some elements of J in this fashion, that Ki 6= Kj for i < j,since Kj is an anticover for some x ∈ Ki. Repeating this argument,

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ON CAYLEY'S THEOREM 25

there must be at least k distinct elements of J for |M∩K1∩. . .∩Kk| = 1,which is required for J to be faithful. �

Lemma 4.32. If a faithful antimaximal collection J has exactly k el-ements, then:(1) If {a1, . . . , ar} is any subset of {1, . . . , k} not containing j then

Kj is a maximal anticover for some x ∈ M ∩Ka1 ∩ . . . ∩Kar ;(2) If {a1, . . . , ar} is any subset of {1, . . . , k} then |M ∩Ka1 ∩ . . . ∩

Kar | = pk−r.

Proof. We prove the two statements by a joint induction. Denote by(1[n]) and (2[n]) the respective statements above when r = n. Denoteby (1[0]) the statement that Kj is a maximal anticover for some x ∈M for all j, and by (2[0]) the statement that |M | = pk. It is clearthat (1[n]) and (2[n]) ⇒ (2[n + 1]), because if |

⋂ni=1 Kai

∩M | = pk−n

and Kan+1 is a maximal anticover for some x ∈⋂n

i=1 Kai∩ M then

[⋂n

i=1 Kai∩M :

⋂ni=1 Kai

∩M ∩Kan+1 ] = p by the AIL, so |⋂n+1

i=1 Kai∩

M | = pk−n−1. This applies equally to the case when n = 0. We nowneed only show that (2[n]) ⇒ (1[n]). Since we know that (2[0]) is true,these two implications will force (1[n]) and (2[n]) to be true for alln ≤ k.Assume (2[r]) for r < k. Let {ai}r

i=1 be any subset of {1, . . . , k} andj ≤ k be not contained in {ai}. Let {b1, . . . , bk−r−1} be {1, . . . , k} −{a1, . . . , ar} − {j}, or the empty set if r = k − 1; it must be true that|{bi}| = k− r− 1. Let C = M ∩Ka1 ∩ . . .∩Kar , and suppose Kj is nota maximal anticover for any x ∈ C. We know by (2[r]) that |C| = pk−r.Repeating the argument of the previous lemma, there is a non-trivialx ∈ C, which must have a maximal anticover in J , and this maximalanticover is not Kai

for any i, or Kj by supposition, so it must be Kbi

for some i (and without loss of generality, it is Kb1). Then by the AIL|C ∩Kb1| = pk−r−1, so there is some non-trivial x ∈ C ∩Kb1 which hasa maximal anticover, let it be Kb2 . Repeating this process we get that|C∩Kb1 ∩ . . .∩Kbk−r−1

| = p, so there is some non-trivial x contained init which must have a maximal anticover. But this maximal anticovermay not be any Kai

, or any Kbi, or Kj, and as such, it has no maximal

anticover in J , a contradiction! Thus Kj is a maximal anticover forsome x ∈ C, and (1[r]) is proven. �

We now come to �nal two theorems of this section, the �rst of whichsums up what is known so far, and the second of which proves onefurther equivalence.

Theorem 4.33. For minimal faithful collections J , the conditions ofbeing well-indexed, regular, and antimaximal are all equivalent.

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ON CAYLEY'S THEOREM 26

Proof. We have already shown that well-indexed faithful collectionsare both regular and antimaximal. We have also shown that minimalfaithful collections of size k are well-indexed, and so regular collections,which must be of size k, are both well-indexed and antimaximal. Fi-nally, a minimal antimaximal collection must have at least size k byLemma 4.31, and we can �nd k elements such that M∩K1∩. . .∩Kk = 1.If there were more than k elements, than any after the �rst k could beeliminated and J would still be faithful but would have a smaller Cay-ley sum, which is impossible if J is minimal. Thus J has exactly kelements, implying that it is both well-indexed and regular. �

Theorem 4.34. Let I = {Ki} be an antimaximal faithful collectionwhich has exactly k elements. Then I is minimal.

Proof. Let J = {Li} be a well-indexed minimal faithful collection. Weseek to show that µ(I) = µ(J). Since J is minimal, we know thatµ(J) ≤ µ(I), so we need only show that µ(J) < µ(I) is impossible. Sosuppose that µ(J) < µ(I).First, renumber the indices of the collections from largest to smallest.

In other words, without loss of generality we let |K1| ≥ |K2| ≥ . . . and|L1| ≥ |L2| ≥ . . .. We have already shown that both collections areof size k. There must be some i such that |Li| > |Ki|. Otherwise,

|Li| ≤ |Ki| for all i, and µ(I) =∑k

i=11|Ki| ≤

∑ki=1

1|Li| = µ(J). Let n

be the least integer such that |Ln| > |Kn|, so that |Lj| ≤ |Kj| for allj < n.Let C = M ∩ L1 ∩ . . . ∩ Ln and let D = M ∩ K1 ∩ . . . ∩ Kn−1 if

n > 1, or let D = M if n = 1. If x ∈ D is non-trivial then someK ∈ I is its maximal anticover, and it can not be Ki for i ≤ n − 1.So if Kj is a maximal anticover of x then |Kj| ≤ |Kn|, because j ≥ n.Thus |Kj| < |Ln|, and due to the ordering, |Kj| < |Li| for all i ≤ n.This implies that x ∈ Li for all i ≤ n, or else Li would be a largeranticover than Kj. Thus D ⊂ C. However, |D| = pk−n+1 by Lemma4.32, whereas |C| = pk−n by the MCIL. Thus D can not be containedin C, because D is bigger. We have derived a contradiction. �

Remark 4.35. It should be noted that by now the two questions posedby Johnson in the introduction have been solved. Every p-group Psatis�es the criterion of Problem 1.9, since the collection I de�nedthere will be regular and by Lemma 4.28 it will also by antimaximal.Since it has k elements, it will be minimal. Finally, I will clearly alsobe a minimal faithful collection when restricted to Z(P ), again beingregular and antimaximal, and thus every group G is in Gp, solvingProblem 1.8.

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ON CAYLEY'S THEOREM 27

4.4. An algorithm to determine α for p-groups.

Algorithm 4.36. [The P-group Algorithm] The following algorithmwill correctly �nd the Cayley Constant and a minimal faithful collectionfor any p-group P.Construct M , the socle of P .Let K1 be a maximal anticover of M .Let K2 be a maximal anticover of M ∩K1.Let Ki be a maximal anticover of M ∩K1 ∩ . . . ∩Ki−1.Repeat until M ∩K1 ∩K2 ∩ . . . ∩Km = 1.Then I = {K1, K2, . . . , Km} will be a faithful collection of P with

minimal Cayley sum, i.e. α(I) = α(P ).

It should be noted that this algorithm is easy to implement if oneis given a list of subgroups of P sorted by size. In that case, one canmerely traverse the list from largest to smallest, choosing Ki wheneverKi does not contain C = M ∩ K1 ∩ . . . ∩ Ki−1. Then Ki will bea maximal anticover of C since every larger set L has already beenvisited, and either was selected to be Kj for j < i (implying that Lis not an anticover of C) or was not chosen at all (implying that Lcontained M ∩K1 ∩ . . . ∩Kj for some j < i, and thus L contains C).Although much of the machinery of the previous section is unneeded

for the proof that this algorithm is correct (although needed to answerJohnson's questions), it played a vital role in the actual process ofinventing this algorithm. After having seen that antimaximal faithfulcollections with k elements are minimal, it became merely a task of�nding such a collection. The proof follows.

Lemma 4.37. If M has k generators (i.e. |M | = pk) then the algo-rithm will terminate after k steps, yielding a collection I = {K1, K2, . . . , Kk}with k elements.

Proof. This is obvious, since M ∩ K1 ∩ . . . ∩ Ki has order pk−i byCorollary 4.24, and the algorithm stops when this order is equal to1. �

Corollary 4.38. The result I = {K1, K2, . . . , Kk} of the algorithm isa faithful collection of P .

Proof. Once again by the AIL, M ∩ K1 ∩ . . . ∩ Kk = 1, thus there isno non-trivial x ∈ M such that x ∈ Ki for all i. This implies that Icontains an anticover for every single non-trivial member of M . �

Theorem 4.39. The result I = {K1, K2, . . . , Kk} of this algorithm isantimaximal.

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ON CAYLEY'S THEOREM 28

Proof. Let x ∈ M . Since M ∩ K1 ∩ . . . ∩ Kk = 1, there is some isuch that x /∈ Ki, so let Kn be the �rst anticover of x. Then x ∈M ∩K1 ∩ . . .∩Kn−1. If Kn is not a maximal anticover of x then thereis some L, a larger anticover of x. But then L is a larger anticover ofM ∩K1 ∩ . . . ∩Kn−1, so L would have been chosen by the algorithmand not Kn. Hence Kn is a maximal anticover of x. �

Theorem 4.40. If I = {K1, K2, . . . , Km} is the result of the algorithm,then I is a minimal faithful collection, and α(P ) = α(I).

Proof. By Theorem 4.34, an antimaximal faithful collection with k el-ements is minimal. �

Remark 4.41. It seems a lot of work only to �nd an algorithm for p-groups, but as [3] shows we can extend our knowledge of p-groups andthe methods used here to apply to nilpotent and solvable groups.

We can use the algorithm to �nd a regular set of generators:

Theorem 4.42. Let I = {K1, K2, . . . , Kk} be the result of the algo-rithm. For each 1 ≤ i ≤ k let xi be a non-trivial element of M ∩K1 ∩. . . ∩Ki−1 ∩Ki+1 ∩ . . . ∩Kk. Then {xi} are the regular generators ofI. Furthermore, if x ∈ M and x = xa1

1 . . . xakk , then Ki is a maximal

anticover of x for the �rst i for which ai 6= 0.

Proof. Clearly xi ∈ Kj for j 6= i, and Ki must be a maximal anticoverof xi since no other element of I could be. Following our knowledge ofhow to construct a socle, all we need to do to show that {xi} generateM is that xi /∈< x1 > × < x2 > × . . .× < xi−1 >, since we knowthere are k of them. Suppose that xi ∈< x1 > × . . . × <xi−1> = B,for the least such i ≤ k (clearly i > 1). If this is the least such i then< x1 > × . . .× < xi−1 > is elementary abelian of order pi−1. By theAIL, we get that B ∩K1 ∩ . . . ∩Ki−1 = 1. But since xi ∈ B, xi mustbe trivial, a contradiction. Thus there is no such i ≤ k, and {xi}k

i=1

generate the socle M .Let x = xa1

1 . . . xakk . When we proved that I is antimaximal we also

proved that if Kn is the element of I of smallest index which is ananticover of x, then Kn is a maximal anticover of x. If ai = 0 thenx is in the subgroup generated by {xj}j 6=i, and since each xj ∈ Ki byde�nition of regularity, we have that x ∈ Ki. However, if ai 6= 0 thenx is not in the subgroup generated by {xj}j 6=i, which equals M ∩ Ki

since the former is contained in the latter and both are of size pk−1. Sothen x /∈ Ki. Thus, the �rst n such that x /∈ Kn is the �rst n such thatan 6= 0. �

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ON CAYLEY'S THEOREM 29

4.5. Computational results. Using the algebraic programming lan-guage MAGMA [7], we implemented the P-group Algorithm and testedit on p-groups of order pn for n ≤ 6 and small p, since every such groupis in MAGMA's group database [9]. The goal, as described in the intro-duction, was to �nd some polynomial in 1

pthat would describe the sum

of α over all groups of a given prime power order. We chose to restrictour analysis to odd primes, since 2-groups tend to be anomalous. Thealgorithm could analyze many groups of large size, but sadly, for manygroups (those with large subgroup lattices) the algorithm exhaustedthe memory of the computer. The set of prime powers for which wecould analyze every group of that order was smaller than originally an-ticipated; however, some encouraging results were found. The resultsare recorded in detail in [3], but are summarized here.Recall the PORC-α from the introduction. The answer to PORC-α

is almost trivially positive for n ≤ 3. There is only one group of orderp1 for a given p, and it is cyclic and has α(G) = 1, so Q1(x) = 1. Thereare only two groups of order p2, one cyclic and the other elementaryabelian, so

∑α(G) = 1 + 2

p, or Q2(x) = 1 + 2x. There are only �ve

groups of order p3: one cyclic with α = 1; one elementary abelian withα = 3

p2 ; one abelian with a generator of order p2, having α = 1p+ 1

p2 ; and

two non-abelian groups both having α = 1p. Hence

∑α(G) = 1+ 3

p+ 4

p2 ,

and Q3(x) = 1 + 3x + 4x2. These values of α come directly from theclassi�cation of groups of order pn for n ≤ 3, and the simple fact thata non-abelian group of order p3 must have α = 1

p. This is because we

must have |Z(G)| = p, and if {H} is the minimal faithful collection and|H| = p2 then G = Z(G)H ∼= Z(G) ×H, which is impossible becauseH has a non-trivial center which would be central in G. As such, webegin our discussion of results and investigation at n = 4.For any prime p ≥ 3, there are always exactly 15 groups of order p4,

and these can be enumerated and described. Thus, it is not surprisingthat α for these groups is very neatly de�ned. The results are shown inthis table, where the value on the left is the value of α, and the value

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ON CAYLEY'S THEOREM 30

on the right is the number of groups with that α:

4p3 11p2 2

1p2 + 1

p3 21p2 + 2

p3 12p2 31p

41p

+ 1p3 1

1 1

This has been checked for all 3 < p < 50, as well as several large valuesof p (≈ 1000). Thus, we have the conjecture:

Conjecture 4.43. For p > 3,

∑|G|=p4, p>3

α(G) = 1 +5

p+

11

p2+

9

p3.

Although we do not have a proof of this proposition, it is likely thatsomeone more well-versed in groups of order p4 could prove this. Inhis paper on faithful actions, Johnson says that groups of order p4 canbe analyzed because of the restrictions on the structure of the center(Example 3, [6]).When p = 3 the results are similar, although there is one more group

with α = 13, and one fewer with α = 1

32 . Although this di�erence seemssmall, it becomes more pronounced when looking at groups of orderp5, for which again p = 3 is an anomaly. This is not surprising at all,because for n ≥ 5, the number of subgroups of order 3n does not evenfollow the standard PORC for that n.The results for groups of order p5, with p ≥ 5, are summarized in

this table:

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ON CAYLEY'S THEOREM 31

5p4 11p3 2

1p3 + 1

p4 21p3 + 2

p4 21p3 + 3

p4 12p3 12 + p

2p3 + 1

p4 33p3 11p2 XXXX

1p2 + 1

p4 41p2 + 2

p4 11p2 + 1

p3 142p2 1 + p1p

41p

+ 1p4 1

1 1

These results were true for all primes 5 ≤ p < 20 (which was the upperlimit before the computer ran out of memory). The number XXXXabove was the only number which did not follow an obvious pattern,for it was:

p XXXX5 177 1911 1513 2117 1719 19

However, we know the number of groups of order p5 is 61 + 2p +2gcd(p−1, 3)+gcd(p−1, 4), so it is clear that x = 11+2gcd(p−1, 3)+gcd(p− 1, 4), and the results are consistent with this [14].Classifying groups by their Cayley constant, we see that the types

of groups whose structure is a�ected by the prime p seems to be onlythose with α = 2

p3 ,2p2 ,

1p2 , and the only groups a�ected by the residue

class e�ect had α = 1p2 . Now we can make a conjecture for p5:

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ON CAYLEY'S THEOREM 32

Conjecture 4.44. For p > 3,∑|G|=p5

α(G) = 1 +7

p+

34 + 2gcd(p− 1, 3) + gcd(p− 1, 4)

p2+

54

p3+

24

p4.

Sadly, some of the groups of order 76 were too large for the computerto handle, and thus the only complete data on n = 6 is for groups oforder 36 and 56. Since we already suspect that 36 is an anomaly, onedata point is not enough to �nd the relevant polynomial for p6.What is important to note about these results, aside from the fact

that they clearly support the PORC-α conjecture, is that they supportan even stronger claim: not only is the sum of all α a PORC, butthe number of groups with any given α is a PORC! It is even possible,although one can not tell without investigating groups of higher orders,that one could show, for any given n, the following claim: for certainα, the number of groups with that α is a constant function of p; forother types of α, like 2

pm for some m ≤ n− 2 (perhaps), the number ofgroups with that α is a polynomial function of p; and for other types,like 1

p2 in this case, the number of groups will be a PORC of p. Thiswould certainly help enumerate all groups of order pn!

Appendix A. MAGMA Code

Printed here is the �nal version of the code the �rst author usedin [3] to calculate α for p-groups. Because MAGMA is an obscureprogramming language, we have placed comments next to the code toexplain its function, when not obvious. Some observations will be madeafterwards.

function getSocle(G)// returns product of minimal normal subgroups of G// calls it K.Z:=Centre(G);gens:=Generators(Z);// gens is set of generators of Z, see notes belowgenK:=[]; // To become the generators of Kwhile (IsEmpty(gens) eq false) do

// These two lines are essentially// a for loop over all generatorsExtractRep(~gens,~gen);ord:=Order(gen);// with the generator called genp:=Factorization(ord)[1][1];n:=Factorization(ord)[1][2]; // |gen|=p^n

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ON CAYLEY'S THEOREM 33

gen2:=gen^(p^(n-1));// gen2 is the element of order p in <gen>Append(~genK,gen2);// and as such is a generator of K

end while;K:=sub<G|genK>;// The subgroup of G generated by genKreturn K;end function; //getSocle

function getMinimalFaithfulCollection(G)// This is the algorithm used.// First it finds K, the socle.// Then choose as K_1 the largest subgroup// of G which does not contain K.// Then choose as K_2 the largest subgroup of G// which does not contain K meet K_1.// Then choose as K_i the largest subgroup of G// which does not contain K meet ... meet K_i-1.// repeat until K meet K_1 meet ... meet K_k = 1.//Inter:=getSocle(G);// Inter is, at any point, K meet K_1 ... meet K_isubs:=Subgroups(G);// The Subgroups function returns records of all the// subgroups of G, already sorted.F:=[]; // To become the faithful collectionfor j:=#subs to 1 by -1 do

// Going through the subgroups from largest...// to smallest, calling them sub.sub:=subs[j]`subgroup;if((Inter subset sub) eq false) then

//If sub is an anticover of Inter...Append(~F,sub); // Add it to the collectionInter:=Inter meet sub; // maintain Inter

if(IsTrivial(Inter)) then break; end if;// And stop if Inter = 1end if;

end for;return F;end function; //getMinimalFaithfulCollection

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ON CAYLEY'S THEOREM 34

function mu(F)// This is the Cayley sum of a collection.// Returns the sum of 1/Order(x) for all x in F.alph:=0;for i:=1 to #F do

alph +:= 1/Order(F[i]);end for;return alph;end function;function abelianFactors(G)// If G is abelian then G=G_1 X G_2 X ... X G_k// where |G_i|=p^n and cyclic for some prime p,// The return value is an array of the orders of G_i.orders:=[];gens:=Generators(G); // see notes belowfor i in gens do

Append(~orders,Order(i));end for;return orders;end function;function alphaAbelian(G)// Returns alpha for an abelian group.orders:=abelianFactors(G);return (&+orders)/(&*orders);//The sum of all orders over their product.end function;

function Alpha(G)// Goes through the process, return alpha for G.if IsAbelian(G) then return alphaAbelian(G);

// see noteselse

F:=getMinimalFaithfulCollection(G);return mu(F);

end if;end function;

function SumAlpha(Glist)

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ON CAYLEY'S THEOREM 35

// If Glist is a list of groups (for instance,// all groups of order p^n)// then SumAlpha finds alpha of each group,// and prints it in the form// "1/3 + 1/9 + 2/9 + 2/27 = 20/27"s:=0;for i:= 1 to #Glist do

a := Alpha(Glist[i]);s +:= a;if i eq 1 then output:=Sprintf("%o",a);// Concatenate stringselse output:=Sprintf("%o + %o",output,a);end if;

end for;output:= Sprintf("%o =",output);return output,s;end function;

SumAlpha(SmallGroups(3^5));SumAlpha(SmallGroups(7^6));// SmallGroups(n) returns a list of all// groups of order n. See notes below.

The �rst few things to note, pertaining to the correctness of the code,are that the Generators(G) function is guaranteed to provide a minimalset of generators when the group G is a p-group, according to [8], andthat the SmallGroups(n) function will provide all groups of order nwhen n = pk for any prime p, and k ≤ 6, according to [9].The solution provided by the algorithm for abelian p-groups is the

same as the solution given by Theorem 2.16, but the latter was cho-sen for computational reasons. Abelian groups tend to have the mostsubgroups and certainly have centers with the highest number of gen-erators, and as such, the algorithm runs slowest on them. However,the special algorithm for abelian groups runs much faster, since nosubgroups need to be investigated, so it is preferred.This algorithm is the fastest of the algorithms which the �rst author

devised for this problem, but it is still quite slow on some p-groups.Most of the time spent calculating α for each group is in fact spentwithin MAGMA's built-in Subgroups function, which �nds all the sub-groups in sorted order. Unless there is a fast way to �nd maximal

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ON CAYLEY'S THEOREM 36

anticovers for subgroups of the socle without looking at all the sub-groups of G, we doubt this time expenditure can be eliminated whilestill using the MAGMA language. It should be noted that the limit-ing factor when calculating α for all groups of order pn was not time,but memory; all 2GB of available memory was used up. The only sig-ni�cant usage of memory in the program is, again, storing the resultof the Subgroups function, so it appears that this algorithm, as it iscurrently implemented, can not calculate α for certain p-groups withlarge subgroup lattices. Any attempt to improve the algorithm wouldhave to �nd a way to traverse the subgroup graph from top to bottomwithout actually creating the whole graph, or to �nd more economicalways to store the data.

Remark A.1. Before our algorithm was devised, we were examininga subtly di�erent algorithm: Choose x1 ∈ M , �nd K1 the maximalanticover, then choose x2 ∈ M ∩K1 and K2 its maximal anticover, andso forth. Once again, we can easily show that the algorithm terminatesafter k steps, and that {xi} are a set of generators of the socle, but wewere unable to show that {xi} were regular generators, or that {Ki}was in fact an antimaximal collection. This algorithm is inferior tothe current one, since it requires k passes through the subgroup arrayinstead of 1, but it did produce identical results, making us believethat it is also correct, and that there may be some simpli�cations tobe found in the future.

References

[1] A. R. Camina, G. R. Everest and T. M. Gagen, �Enumerating Non-SolubleGroups - A Conjecture of John G. Thompson,� Bulletins of the London Math-

ematical Society, vol 18 (1986), pp. 265-268.[2] David S. Dummit and Richard M. Foote, Abstract Algebra (3rd edition), John

Wiley and Sons Inc., 2004.[3] B. Elias, �Minimally faithful group actions and P -groups,� Senior Thesis,

Princeton University, 2005.[4] G. Higman, 'Enumerating p-groups I,' Procedings of the London Mathematical

Society, Vol 3, No. 10 (1960) pp. 24-30.[5] G. Higman, 'Enumerating p-groups II,' Procedings of the London Mathematical

Society, Vol 3, No. 10 (1960) pp. 566-582.[6] D. L. Johnson, �Minimal Permutation Representations of Finite Groups,�

American Journal of Mathematics, vol. 93, No. 4 (Oct. 1971), pp. 857-866.[7] MAGMA Computational Algebra System, http://magma.maths.usyd.edu.au/

.[8] MAGMA Computational Algebra System Online Documentation,

http://magma.maths.usyd.edu.au/magma/htmlhelp/text314.htm#2261.

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ON CAYLEY'S THEOREM 37

[9] MAGMA Computational Algebra System Online Documentation,http://magma.maths.usyd.edu.au/magma/htmlhelp/text385.htm .

[10] Peter M. Neumann, 'An Enumeration Theorem for Finite Groups,' QuarterlyJournal of Mathematics, Vol 2, No. 20 (1969), pp. 395-401.

[11] John S. Rose, A Course on Group Theory, Cambridge University Press, 1978.[12] Marcus du Sautoy, �Zeta Functions and Counting Finite p-groups,� Electronic

Research Announcements of the American Mathematical Society, Vol 5 (1999),pp. 112-122.

[13] Lior Silberman and Ramin Takloo-Bighash, �On Cayley's Theorem,�12/17/2003.

[14] Brett Witty, �Computational Approach to Higman's PORC Conjecture,�http://wwwmaths.anu.edu.au/~witty/bw_austms04.pdf , Mathematical Sci-ences Institute, Australian National University, 2004.