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Prove that groups of even order contain at least one element (which is not the identity) that squares to the identity. 

In case of a cyclic group, it is easy. Such group consists of {e, g, g2, g3, ..., gn-1} and contains element h = gn/2. This element, h2 = (gn/2)2 = gn = e.

But how to prove it when the group is not cyclic?

P.S. Oh, got it. Just count the pairs, element and its inverse.

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