Prove that G is abelian if (G:H) is prime, where H is in Z(G)

Question

I'm attempting to prove this, but am quite frankly stuck and haven't made any progress.

I know that $\frac{|G|}{|H|}=p$, and I know by Cauchy's thm, that $\exists x \in G$ such that $x^p$ = e, and that any group of prime order is cyclic which implies Abelian, but that's as many relevant details I can think of. Any hints or tips would be appreciated.

Thanks!

Based on the duplicate and its not entirely helpful answer, it might be an idea to ask explicitly, "why can't $G/Z(G)$ be nontrivial and cyclic?" as a new question. — hmakholm left over Monica, Mar 20 '17 at 01:09
@HenningMakholm That particular question would undoubtedly be rapidly closed as a duplicate to this or one of its numerous duplicates. I didn't grep through all of the dupes, but almost certainly one has pointed out that conclusion... — rschwieb, Mar 20 '17 at 01:36

score 0 · Answer 1 · answered Mar 20 '17 at 04:04

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If the quotient group has prime order then it is cyclic. If the quotient group is cyclic then $G$ is Abelian.

answered Mar 20 '17 at 04:04

AnotherPerson

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Prove that G is abelian if (G:H) is prime, where H is in Z(G)

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