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.
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 $n$.
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.