Pair up if possible each element of G with its inverse, and observe that

\(g^{2}\) not equal \(e\Leftrightarrow\) not equal \(g^{−1} \Leftrightarrow\) there exists the pair \((g,g^{−1})\)

Now, there is one element that has no pairing: the identity e (since indeed \(e=e−1 \Leftrightarrow e2=e,e=e −1\) \(\Leftrightarrow e^{2} =e\)), so since the number of elements of G is even there must be at least one element more, say e not equal \(a\in G\) e not equal \(a\in G\), without a pairing, and thus \(a=a−1 \Leftrightarrow a2=e.a=a^{−1} \Leftrightarrow a^{2} =e\).