Algebra Solutions: Every finite group of even order contains an element of order 2

Let $G$ be a finite group of even order. Prove that $G$ contains an element of order 2.

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.

Algebra Solutions

Every finite group of even order contains an element of order 2

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