If G is a group of even order, then show that there exists an element a != e, the identity in G, such that a^2 = e.

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There exists a != e in G with a^2 = e (an element of order 2). An element of order 2 is one that is its own inverse but is not the identity, since a = a^{-1} is the same as a^2 = e. The clever counting idea is to pair up every element with its inverse. Elements that differ from their own inverse come in neat two-element pairs {a, a^{-1}}, so together they account for an even number of elements. Because the whole group has an even number of elements and the identity is always its own inverse, the count of self-inverse elements must be even - and since it is at least one (the identity), it is at least two, giving a non-identity self-inverse element.