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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.

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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, [tex] [ab]_K [/tex] is equal to [tex] [ba]_K [/tex], 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, [tex] [ab]_H = [ba]_H [/tex] . Since a and b were generic elements of H, then H/G is abelian.

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