Show that if $G$ is a finite group of even order, there must exist an element $a\in G, a\ne e$ s.t. $a^2=e$

Question

As order of group is even then $G$ can contain an element of order 2 as 2|even but how it is must?

Answer 1

Let $e$ be the the neutral element of $G$. For $a\in G$ we have $a^2=e\Leftrightarrow a=a^{-1}$. Write as a disjoint union $$G=\{a\in G \mid a\neq a^{-1}\}\dot{\cup} \{e\} \dot{\cup} \{a\in G\mid a\neq e, a=a^{-1}\}.$$ $|G|$ and $|\{a\in G \mid a\neq a^{-1}\}|$ is even $|\{e\}|=1$ so $|\{a\in G\mid a\neq e, a=a^{-1}\}|$ is odd und there is a nontrivial element $a\in G$ with $a^2=e$.

Answer 2

It follows from Cauchy's theorem, which states that if $p$ is prime dividing the order of a group, then there is an element of order $p$.

In this case, you can prove it as follows:

Let $X$ be the set of all pairs $(g, g^{-1})$ as $g$ ranges over $G$. This set is of size $|G|$.

I claim that $C_2 = \{ e, h \}$, the cyclic group of order 2 with $h^2 = e$, acts on $X$ by $h(g, g^{-1}) = (g^{-1}, g)$. (Prove yourself that this is an action.)

Then by the orbit-stabiliser theorem, the orbit of arbitrary $(g, g^{-1})$ has size dividing 2, so it's of size 1 or 2. We're done if we can find a non-identity element $(g, g^{-1})$ whose orbit has size 1, because then $h(g, g^{-1}) = (g^{-1}, g) = (g, g^{-1})$ and so $g = g^{-1}$.

The orbit of $(e \in G, e)$ is of size 1; all the orbits together partition $X$, of size $|G|$ divisible by 2, so we have a sum of (orbits which are of size 2) and (orbits which are of size 1) together making $X$ of even size. That means the number of orbits of size 1 must be even. But we've identified one of them already - $(e, e)$ - so there must be another one, $(g, g^{-1})$ such that $h(g, g^{-1}) = (g^{-1}, g) = (g, g^{-1})$.

Therefore we've found $g$ of order 2.