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I'm trying to prove that if $G$ is a group with $|G|=2m$ and $m$ odd, then $G$ has exactly one element of order two. What I have is that there actually exist at least one element of order two, and that by Sylow theorems every two involutions are conjugate. I also proved that if a group has odd order then there are no elements of order two. I'm trying to glue all this info together with no success.

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    $\begingroup$ This is not true: Take $G = S_3$, then $|G| = 6$, but $G$ has both $(12)$ and $(13)$ which are of order 2. $\endgroup$ – Prahlad Vaidyanathan Sep 24 '15 at 1:45
  • $\begingroup$ This is true if $G$ is abelian, though. $\endgroup$ – pjs36 Sep 24 '15 at 1:51
  • $\begingroup$ @PrahladVaidyanathan I see, thanks. There is then a mistake in my teacher's document. $\endgroup$ – Jose Paternina Sep 24 '15 at 1:56
  • $\begingroup$ As @pjs36 points out, it is likely that $G$ is meant to be abelian. $\endgroup$ – Prahlad Vaidyanathan Sep 24 '15 at 1:58
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This is false, a dihedral group of order $2m$ with $m$ odd provides a counterexample. If $\rho$ is the rotation and $\tau$ is the reflection then $\rho^n\tau$ has order $2$ regardless of the choice of $n$.

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In the abelian case, the proof is a one-liner:

Two different elements of order $2$ generate a subgroup isomorphic to $C_2 \times C_2$, hence $4$ divides the order of the group.

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