Question

Let A and B be normal subgroups of a group G such that G/A and G/B are solvable. Prove that $\large\rm { G/( A \cap B) }$ is solvable.

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Answer 1

Answer:

4. Let G = A1 × A2 × · · · × An and for each i let Bi C Ai

. Prove that B1 × B2 × · · · × Bn C G and that

(A1 × A2 × · · · × An)/(B1 × B2 × · · · × Bn) ' (A1/B1) × (A2/B2) × · · · × (An/Bn).

Solution: Define

φ : G −→ (A1/B1) × (A2/B2) × · · · × (An/Bn)

by (a1, a2, . . . , an) 7→ (a1B1, a2B2, . . . , anBn). Since for each i we have Bi C Ai

, it follows that φ is a

homomorphism, and is obviously surjective. Moreover, it is clear that ker(φ) = B1 × B2 × . . . × Bn so that

B1 × B2 × . . . × Bn C G and we have

(A1 × A2 × · · · × An)/(B1 × B2 × · · · × Bn) ' (A1/B1) × (A2/B2) × · · · × (An/Bn),

as required.

Book problems §5.4:

4. Find the commutator subgroups of S4 and A4.

Solution: First we claim that the only normal subgroups of A4 are A4, V4, and {1}, where V4 is the

klein four group. This follows easily from the fact that any normal subgroup must be a union of conjugacy

classes, and the fact that A4 has conjugacy classes of sizes 1, 4, 4, 3 corresponding to the identity, two classes

of 3-cycles, and the class of 2 × 2 cycles.

Let G = S4, H = A4 and let G0 be the commutator subgroup of G. Since H CG and G/H ' C2 is abelian,

G0 6 H. Moreover, since G0 C G, we have trivially G0 C H. Now G0

cannot be trivial since G is non-abelian.

Thus G0 = H or V4. Observe that V4 C G (since it consists of all 2 × 2 cycles in G and elements of G are

conjugate if and only if they have the same cycle type). Pick any copy of S3 6 G. Then S3 ∩ V4 = {1} so

that (since V4 C G), we see that S3V4 is a subgroup of G of size 24 and therefore coincides with G. The

Second Isomorphism Theorem gives

G/V4 = S3V4/V4 ' S3/(V4 ∩ S3) ' S3

so that G/V4 is not abelian. It follows that G0 = H. On the other hand, H/V4 ' C3 (since, for example, it

clearly has order 3 and there is only one group with this order) is abelian so that if H0

is the commutator

subgroup of H, we must have H0 6 V4. Again, H0 C H, so that by our above remarks, if H0 6= V4 then

H0 = {1}, which cannot be since H is non-abelian. Thus H0 = V4.

5. Prove that An is the commutator subgroup of Sn for all n ≥ 5.

Solution: For n ≥ 5, let G = Sn, H = An and let G0 be the commutator subgroup of G. Since G/H ' C2

is abelian, G0 6 H. On the other hand, G0 C G and so trivially G0 C H. But H is simple since n ≥ 5 so that

G0 = H or {1}. Since G is non-abelian, it follows that G0 = H.

15. If A, B are normal subgroups of a group G with G/A and G/B abelian, prove that G/(A∩B) is abelian.

Solution: Let G0 be the commutator subgroup of G. Recall that G/G0

is the maximal abelian quotient

of G in the sense that if H C G, then G/H is abelian if and only if G0 6 H. Thus, since G/A, G/B are

abelian, we have G0 6 A and G0 6 B. Thus, G0 6 A ∩ B so that G/(A ∩ B) is abelian.

Alternatively, define φ : G → (G/A) × (G/B) by φ(g) = (gA, gB). It is not difficult to see that φ is

a group homomorphism. Moreover, g 7→ (A, B) if and only if g ∈ (A ∩ B), so that ker φ = (A ∩ B) and

φ descends to an injective group homomorphism φ : G/(A ∩ B) → (G/A) × (G/B) so that G/(A ∩ B) is

isomorphic to a subgroup of (G/A) × (G/B). But (G/A) × (G/B) is abelian since G/A and G/B are, and

any subgroup of an abelian group is abelian. Therefore, G/(A ∩ B) is abelian.

1. If H, K are normal subgroups of a group G with H ∩ K = {1} and H.K = G, show hk = kh for all k ∈ K

and h ∈ H. Conclude that H × K ' G.

Solution: Observe that for any k ∈ K and h ∈ H, we have khk−1h

−1 = (khk−1

)h

−1 = h

0h

−1 =

k(hkh−1

) = kk0

since K, H are normal in G. Thus, khk−1h

−1 ∈ K ∩ H so that khk−1h

−1 =

Answer 2

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