Prove that G is abelian given that each element of G has order at most 2.

**feyomi** · March 1st 2011, 02:37 AM

Prove that, if G is a group and each element of G has order at most 2, then G is abelian.

Thanks.

**DrSteve** · March 1st 2011, 03:16 AM

Since $a^2=e$ , $a=a^{-1}$ for all $a$ . Now look at the equation $(ab)^2=e$

**JJMC89** · March 2nd 2011, 07:14 PM

$e$ is the only element of order $1$ .

if $g,h \in G$ then $gh \in G$
so $(gh)^2 = e$
and $(gh)^2 = ghgh$
so $ghgh = e$
$ghghh = h$
$ghg = h$
$ghgg = hg$
$gh = hg$
Therefore $G$ such that all elements have order of at most $2$ is Abelian.

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Prove that G is abelian given that each element of G has order at most 2.

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