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I want to prove that a group $G$ of order $10$ with a normal subgroup $H$ of order $2$ is abelian.

It is clear that $H$ is contained in the center of $G$ but I can't go further.

Shaun
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flower7182
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    There are only two groups of order $10$ . – lulu Dec 31 '22 at 19:04
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    Note that $G/H$ is cyclic - this makes it quite easy - you can find the theorem that if $Z$ is the centre and $G/Z$ is cyclic then $G$ is abelian many times on this site. The proof is not difficult and applies in this case. – Mark Bennet Dec 31 '22 at 19:23
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    The better phrasing is that if $N$ is contained in the center and $G/N$ is cyclic, then $G$ is abelian. Here is a proof of the theorem phrased that way. – Arturo Magidin Dec 31 '22 at 19:50
  • @ArturoMagidin Indeed it is better, and it is just as easy to prove as you note. But most sources I've encountered use the less general formulation. Thanks for finding the more general version. – Mark Bennet Dec 31 '22 at 20:10
  • @MarkBennet Yes, the formulation is both extremely common and potentially confusing. A pet peeve of mine, as it happens. – Arturo Magidin Dec 31 '22 at 20:12
  • As you pointed out, $H$ of order $2$ is contained in the center. Then $|Z(G)|=2$ or $|Z(G)|=10$. But the former case is ruled out because then every noncentral element has centralizer of order $2$, and the class equation yields: $10=2+5k$, a contradiction because $5\nmid 8$. No need, then, to consider the quotient $G/H$ as in the other comments. – citadel Jan 15 '23 at 22:16

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