Question

If G is a group of even order, prove that it has an element 'a' which is not equal to 'e' satisfying a^2=e. e is identity element.

Answer 1

Let $|G|=2n$. If $a^2=e$, then $a^2a^{-1}=ea^{-1}$, and consequently $a=a^{-1}$. By analogy the equality $a=a^{-1}$ implies $aa=aa^{-1}$, and thus $a^2=e$. So $a^2=e$ if and only if $a^{-1}=a$. Suppose that $a^2\neq e$ for any $a\neq e$. Then $|\{a,a^{-1}\}|=2$ for any $a\neq e$, and consequently the set $G\setminus\{e\}$ is partitioned into pairs of inverses elements. This imlies that the set $G\setminus\{e\}$ has even order. Therefore, the group $G$ has odd order. This contradiction proves the statement,