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In a previous question: Prove that G is abelian if (G:H) is prime, where H is in Z(G), I was directed to: If H ≤ Z(G) ≤ G, where G is a finite group,Z(G) is its center, and (G:H) = p for some prime p, then G is abelian.

Here, an answer states that "hence G/Z(G) is a group of prime order, hence G/Z(G) is nontrivial cyclic, which is impossible (well-known exercise)."

However, I'm unable to see why this is impossible.

Any tips or hints would be greatly appreciated.

Thanks!

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chikin
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1 Answers1

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The assumption is $[G:Z(G)]=p$, then we know that $G/Z(G)$ is cyclic.
Hence $G$ is abelian and $G=Z(G)$.
This means that $[G:Z(G)]=1$ which is a contradiction to the assumption.
So $[G:Z(G)]=p$ cannot be happened.

Wang Kah Lun
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