Question

g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and such that G/K is an abelian group. Prove that G/H is an abelian group also.

Answer 1

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab,
`[ab]_K` is equal to
`[ba]_K`, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus,
`[ab]_H = [ba]_H` . Since a and b were generic elements of H, then H/G is abelian.