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Let $G$ be an Abelian group and $p$ be a prime, $$G_p:=\{x\in G\mid x\text{ is a $p$-element }\}.$$ Then $G_p$ is a characteristic $p$-subgroup of order $|G|_p$.

My question is a simple one.

At the beginning of the proof, it was mentioned that “For $x,y ∈ G_p$ also $xy$ is a $p$-element; $\color{red}{\text{use } xy = yx}$ and 1.1.2(which is and actually should be 1.1.3 that states any algebraically closed non-empty subset is a subgroup) on page 4”.

I may think it just suffices without using $xy=yx$. What have I missed? Any help would be sincerely appreciated!

PS: It’s on page 45 of my textbook (page 58 of the pdf), The Theory of Finite Groups, An Introduction.

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    $\begingroup$ here’s a counterexample with $p=2$: let $g$ be the mapping $g:\Bbb R\to\Bbb R$ by $g(x)=-x$, and $h$, by $h(x)=1-x$. Both of these are involutions (square is the identity). Then $g\circ h:x\mapsto x-1$ and $h\circ g:x\mapsto x+1$. So $g\circ h\ne h\circ g$, and in fact their composition is of infinite order. $\endgroup$ – Lubin Mar 27 '18 at 4:14
  • $\begingroup$ @Lubin , thanks! It’s so kind of you to give such an illuminating comment! $\endgroup$ – Benny Mar 27 '18 at 4:21

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