A group of order $2n$ has an element of order $2$ 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?
 A: 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.
A: Partition $G = E \cup A \cup B$, where $E=\{e\}$, $A=\{x \in G : x = x^{-1}, x\ne e\}$, $B=\{x \in G : x \ne x^{-1}\}$. Then $E$ has one element and $B$ has an even number of elements because $B$ is invariant under inversion. Since $G$ has an even number of elements, $A$ must have an odd number of elements. In particular, $A$ has at least one element. Every element in $A$ has order 2.
A: If there is no element of order 2 then show that $G$ has odd number of elements.
(Hint: think of elements and inverses)
