Let $G$ be a finite group of even order. Prove that $G$ contains an element of order 2.
Solution: Let $t(G)$ be the set $\{ g \in G \ |\ g \neq g^{-1} \}$. Note that if $g \in t(G)$, then $g^{-1} \in t(G)$, and by definition $g^{-1} \neq g$.
Now let $A = \{ \{ g, g^{-1} \} \ |\ g \in t(G) \}$. Clearly $t(G) = \bigcup A$, and if $\alpha = \{ g, g^{-1} \}$, $\beta = \{ h, h^{-1} \} \in A$ are distinct then $\alpha \cap \beta = \emptyset$. So $A$ is a partition of $t(G)$, and $|A|$ must be finite. Thus we have $$|t(G)| = \sum_{\alpha \in A} |\alpha| = 2k$$ for some positive integer $k$; in particular, $t(G)$ contains an even number of elements. Moreover, $1 \notin t(G)$ since $1^{-1} = 1$. Now by definition, every nonidentity element of $G \setminus t(G)$ has order 2.
Now we have $|G| = |t(G)| + |G \setminus t(G)|$. Since 2 divides $|G|$ and $|t(G)|$, it must also divide $|G \setminus t(G)|$; hence $|G \setminus t(G)|$ must contain at least two elements, one of them the identity. The other is an element of order 2.
