Solutions to Homework IV

Ans 6: Let g, g′ ∈ G and H be such that Hg = g′H. But

g ∈ Hg ∩ gH ⇒ g′H ∩ gH 6= φ ⇔ Hg = g′H = gH

By definition, this means that g ∈ NG(H).

Ans 8: We know that the order of a subgroup divides the order ofthe group. Thus, in this case, we get

|H ∩K||gcd(|H|, |K|) = 1

Ans 11: We have to show, in general, that, if H ≤ K ≤ G, then

[G : H] = [G : K][K : H]

If both sides are finite, we let x1, ...., xk be the coset representativesfor K in G and y1, ..., yh be the coset representatives for H in K.Then, we can check that the xiyjH form a complete distinct set ofcoset representatives for H in G.

The next part is to show that the LHS is finite if and only if theRHS is. Assume that [G : K] < ∞ and [K : H] < ∞, i.e. the RHSis finite. As before, we let x1, ...., xk be the coset representatives forK in G and y1, ..., yh be the coset representatives for H in K. Then,we can check that the xiyjH form a complete distinct set of cosetrepresentatives for H in G. Thus, [G : H] < ∞.

Now assume that [G : H] < ∞. Then, we can get z1, ..., zh such that

G = ∪hi=1ziH

But H ⊆ K and hence

∪hi=1ziK ⊇ ∪h

i=1ziH = G

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This gives [G : K] < ∞. Furthermore,

K = G ∩K = ∪hi=1(ziH) ∩K

If ziH ∩K 6= φ, then zih ∈ K for some h ∈ H; hence zi ∈ K andziH ⊆ K. Thus,

K = G ∩K = ∪hi=1(ziH) ∩K = ∪ziH∩K 6=φziH ∩K = ∪zi∈KziH

This proves that [K : H] < ∞.

Ans 18 Since N is normal in G, we can form the quotient groupG/N , which has order [G : N ]. Suppose that H is not a subgroupof N , then we can get h ∈ H −N . Then hN is a nontrivial elementof the group G/N . The order of hN in G/N divides [G : N ]. Ifthe order of h in G(and hence in H) is k, then k||H|. Further,(hN)k = hkN = N and hence the order of hN in G/N divides kand hence it also divides |H|. Thus, hN has order 1 in G/N andhence h ∈ N , a contradiction.

Ans 3 Let K be another subgroup of G. Then, if H ( HK, then[G : HK]|[G : H] = p and [G : HK] < [G : H]. Thus, [G : HK] = 1and G = HK. Then, [K : K ∩H] = [HK : H] = [G : H] = p.

Otherwise, we get HK = H. Then K ⊆ H.

Ans 8 Clearly, we have (zz′)p = zpz′p and hence the map φ : z 7→ zp

is a homomorphism. If z is a p-power root of unity, then zpn= 1

for some n ≥ 0. If n > 0, z = φ(zpn−1); if n = 0, z = 1 = φ(1).

The map φ : G → G is surjective and has nontrivial kernel (sincee2πi/p 6= 1 and lies in ker(φ)), G becomes isomorphic to a properquotient G/Ker(φ).

Ans 9: We can check directly that this subgroup of order 4 is

{1, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}

This is isomorphic to V4 ≈ Z/2×Z/2 by the association (1, 2)(3, 4) 7→(1̄, 0) and (1, 3)(2, 4) 7→ (0, 1̄). Let us call this subgroup H.

Note that all 3-cycles are in A4. There are 8 distinct 3-cycles in S4.This brings the total to 4 + 8 = 12 elements and hence exhausts allof A4; thus A4 has a unique subgroup of order 4. If x ∈ S4, thenxHx−1 ⊆ xA4x

−1 = A4 and hence xHx−1 is a subgroup of A4 oforder 4. Thus, xHx−1 = H.

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Ans 10 (a) H acts by multiplication on the left cosets of K in G,taking

(h, xK) 7→ hxK = {hxk|k ∈ K}The orbit of xK under this action is all cosets hxK. If these cosetshave distinct set of representatives x1K,...,xnK, then

HxK = {hxK|h ∈ H, k ∈ K} = ∪ni=1xiK

(b) is the same result for right cosets instead of left cosets.

(c) As said in part (a), HxK is the union of elements in the orbitof xK under the action of H on the left cosets of K. We know thattwo cosets of K are either equal or disjoint; and that two oribits areeither equal or disjoint; combining we see that any two HxK areequal or disjoint. Since the orbits form a partition of the set G : Kand cosets xK form a partition of G, the HxK form a partition ofG.

(d) We see that, if the orbit of xK under the action of H, containsn elements,

|HxK| = n|K|Since H acts on the left cosets of K, if Ox denotes the orbit of xKand Gx denotes the stabilizer of xK,

|H| = |Gx||Ox| = n|Gx|

But

hxK = xK ⇔ x−1hx ∈ K h ∈ xKx−1 ∩H

Thus, |Gx| = |H ∩ xKx−1|. Hence,

|HxK| = n|K| = (|H|/|H ∩ xKx−1)|K| = |K|[H : H ∩ xKx−1]

and we are through.

(e) Similar.

Further Exercises

Ans 5. H and K are two distinct subgroups of G of index p withH ∩K = {e}. Take y, y′ ∈ H −K. Then yK is a nontrivial in G/Kand G/K has order p. Thus, yp ∈ K. But, yp ∈ H. Hence yp = eand y has order p. Further, since G/K has order p, it must be cyclic.

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Thus, there exists n such that (y′K) = (yK)n; hence y′ = ynk forsome k ∈ K. Thus, y′y−n = k. The LHS is in H and the RHS in Kand hence both sides are e. Thus, y′ = yn. Thus, H is cyclic andgenerated by y. Since yp = e, |H| = p. Since H has index p, we seethat |G| = p2.

By similar reasoning, K is also cyclic of order p. But |HK| = p2,since |H ∩K| = 1. Thus, HK = G and all elements in G have orderp; hence G cannot be cyclic of order p2.

Ans 6. We know that the elements g1, ..., gr commute. Thus, thesubgroup of G, say C generated by the elements g1, ..., gr is abelian.We claim that the union

∪x∈GxCx−1 = G

Since the gi are representatives of all the conjugacy classes, for anyg in G, we can find a gi and an x such that g = xgix

−1. Hence,g ∈ xCx−1. Thus, ∪x∈GxCx−1 = G.

This will contradict the result of problem 7 if C is a proper subgroup.Thus, C = G and hence G is abelian.

Ans 7 LetS = ∪x∈GxHx−1

For any x ∈ G, we have |xHx−1| = |H|. Now, note that

xHx−1 = yHy−1 ⇔ y−1xH(y−1x)−1 = H y−1x ∈ N(H)

Thus, if the xi, i = 1, 2, ..., s form a complete set of coset represen-tatives for N(H) in G, we get

S = ∪si=1xiHx−1

i

Since all the |xHx−1| = |H| and the identity lies in all of them,

|S| < s|H| = [G : N(H)]|H| ≤ [G : N(H)]|N(H)| = |G|

Hence S 6= G.

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