Question

2.4. If G is a group of even order, prove that it has an element $a\ne e$ satisfying a² = e.

Answer 1

Define a relation on $G$ by $g\sim h$ if and only if $g=h$ or $g=h^{-1}$ for all $g,h\in G$.

It is easy to see that this is an equivalence relation. The equivalence class containing $g$ is $\{g,g^{-1}\}$ and contains exactly $2$ elements if and only if $g^2\ne e$. Let $C_1,C_2,\dots, C_k$ be the equivalence classes of $G$ with respect to $\sim$. Then $|G|=|C_1|+|C_2|+\dots+|C_k|$.

Since each $|C_i|\in \{1,2\}$ and $|G|$ is even the number of equivalence classes $C_i$, with $|C_i|=1$ is even. Since the equivalence class containing $\{e\}$ has just one element, there must exist another equivalence class with eactly one element say $\{a\}$. Then $e\ne a$ and $a^{-1}=a$ i.e. $a^2=e$.