Question

Prove that in a group of even number of elements, at least one element, apart from

the identity element, must equal to its identity​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Here’s a contribution to the festival of saying the same thing in different ways. The function [math]f: G \to G[/math] given by [math]f(g)=g^{-1}[/math] is a permutation of [math]G[/math]. Written in cycle notation, [math]f[/math] consists of 1-cycles (fixed points) and 2-cycles (transpositions). The elements of [math]G[/math] appearing in the 2-cycles are even in number, so there must be an even number of 1-cycles if the order of [math]G[/math] is to be even. The identity element of [math]G[/math] gives one 1-cycle, and since there have to be an even number of 1-cycles, there must be a 1-cycle that isn’t the identity of [math]G[/math], and this is an element whose square is the identity.