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$G$ - group, $|G|=2k$, $k\in \Bbb N$

Does there exist $a\in G; a \neq e: a^2 =e$?

I think I should somehow use the fact, that there is odd number of elements of $G$ which are not $e$.

Kuba
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    Yes, using that fact is a good idea. Group the elements pairwise in a convenient way, except for those that don't have a partner. – Daniel Fischer Jan 28 '14 at 13:09

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if $a^2 \ne e$, then $a \ne a^{-1}$.

there are $2k-1$ elements of $G$ which are not $e$, so these exist odd number of elements of $G$ which satisfy $a^2=e$ but $a\ne e$

ziang chen
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