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ioda
3 years ago
5

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.
Mathematics
1 answer:
allsm [11]3 years ago
4 0

Answer:

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.

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Which fact would help you to solve for x in the diagram shown?
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