Other method(s) to prove "a group cannot have exactly two elements of order $2$" If $a,b$ are elements of a group having order $2$ then, if $a,b$ commute, $ab(\ne a , \ne b)$ has order $2$, and if $a,b$ do not commute, then $aba^{-1}(\ne a , \ne b, \ne e)$ has order $2$. Using this we can deduce "a group cannot have exactly two elements of order $2$".

Is there any other method to prove this statement?


A somewhat complicated way consists in noting that in any group, the subgroup generated by two distinct involutions is dihedral, of order at least $4$ (possibly infinite). And in a dihedral group there are plenty of involutions.
