TITLE: A group of order $2n$ has an element of order $2$
QUESTION [6 upvotes]: I'm having problems proving this. The full question is:
"Let $G$ be a group which order is a pair number. Show that $G$ has an element of order $2$".
Can anyone give me a hint?

REPLY [6 votes]: Hint $\ $ Inversion $\rm\:x\to x^{-1}\:$ is an involution $\rm\: (x^{-1})^{-1} = x,$ so the cycles (orbits) of this permutation partition $\rm\:G\:$ into orbits of length $2$ or $1$. Since $\rm\:|G|\:$ is even so too is the number of length $1$ orbits, i.e. fixed points $\rm\:a = a^{-1};\:$ these include $\rm\:a = 1,\:$ hence, having even cardinality, must include at least one other fixed-point, necessarily of order $2$ by $\rm a^{-1}=a\:$ $\Rightarrow$ $\rm\:a^2 = 1$ but $\rm\:a\ne 1$.
For an analogous application of orbit decomposition (without parity) see my prior answer today on Wilson's theorem for groups.