Genady Posted January 29 Posted January 29 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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