I do recognize that this problem from Dummit&Foote Abstract Algebra has been posted before. However the full completed proof was not shown and there was debate regarding the correctness of the definition in the hint. I am hoping that my attempt can be verified.
PROOF
Let G be a finite group of even order.
Notice that for an element $g$ of $G$, if $g=g^{-1}$, then $g^2=gg=gg^{-1}=e=1$. We have shown previously that this holds if order of $g$, $\vert g\vert$, is $1$ or $2$. Also, we know that only the identity of a group has order $1$.
Define $t(G)=\{g\in G \vert g\ne g^{-1} \}$
By definition, $G$ contains a unique identity $e=1$. Further $e=e^{-1}$, so $e\notin t(G)$.
If $t(G)$ is empty, as $G$ has even order, it follows that $\exists g\in G$, distinct from $e$. Since $g=g^{-1}$ and $g\ne e$, $\vert g\vert = 2$.
Assume $t(G)$ is nonempty. By definition a group contains the inverse of each element. Thus $\forall x\in t(G)$, $x^{-1}\in t(G)$. Hence $t(G)$ contains an even number of elements. But $e\in G-t(G)$. Therefore for $G$ to have even order, $\exists g\in G$, distinct from $e$, such that $g\notin t(G)$. As above $\vert g\vert = 2$.
Thus a finite group of even order contains an element of order 2.
