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Homework question:

Let $G$ a finite group with order of $2p$, where $p > 2$ is prime. given that there's $a \in Z(G)$ such that $o(a) = 2$. Prove: $G$ is abelian.

Can you give me some hints and help to solve it? Thanks in advance.

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marked as duplicate by TravisJ, graydad, colormegone, Rebecca J. Stones, Jonas Meyer May 22 '15 at 3:30

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    $\begingroup$ Hint: What will the order of $G/\langle a\rangle$ be? $\endgroup$ – Tobias Kildetoft May 21 '15 at 19:45
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Hint: by Cauchy's theorem there is an element $x$ of order $p$. What does the subgroup generated by $a$ and $x$ look like, and what is its order?

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  • $\begingroup$ Ahh, even simpler than my version. Though Cauchy is not needed, just that there is an element not of order $2$ (and if they are all of order $2$, then the group being abelian is a standard exercise). $\endgroup$ – Tobias Kildetoft May 21 '15 at 19:47
  • $\begingroup$ @TobiasKildetoft true, but you still need Lagrange's theorem with that argument. $\endgroup$ – Matt Samuel May 21 '15 at 19:49
  • $\begingroup$ @TobiasKildetoft Also your hint is a standard exercise, and that's the way I would have solved it. I agree this hints is very nice, and too bad only one wrote an answer as I would have upvoted both. $\endgroup$ – Timbuc May 21 '15 at 19:49
  • $\begingroup$ Sure, but Lagrange is generally covered earlier than Cauchy. $\endgroup$ – Tobias Kildetoft May 21 '15 at 19:50
  • $\begingroup$ @TobiasKildetoft so the idea is to make the G = { $x^i$, $a*x^i$ } for $0 \le i<p$ ? $\endgroup$ – Billie May 21 '15 at 21:45

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