Every group element of odd order is an odd power of its square

Let G be a finite group and let g \in G be an element of order n. Prove that if n is odd, then x = (x^2)^k for some k.


Since n is odd we have n = 2k - 1 for some integer k. Then x^n = x^{2k-1} = x^{2k} x^{-1} = 1, hence (x^2)^k = x.

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