In a finite group of odd order, no nontrivial element is conjugate to its inverse

If G is a group of odd order, prove that for any nonidentity element x \in G, x and x^{-1} are not conjugate.


Suppose to the contrary that axa^{-1} = x^{-1} for some a \in G. Note then that ax = x^{-1}a and that ax^{-1} = xa.

We now prove a few facts.

Fact 1: If m is odd, then a^mx = x^{-1}a^m. Proof: If m = 1, then clearly ax = x^{-1}a. Suppose that the conclusion holds for some odd m \geq 1. Then a^{m+2}x = a^2a^mx = aa x^{-1} a^m = axaa^m = x^{-1}a^{m+2}. By induction, the conclusion holds for all odd positive m. \square

Fact 2: If m is even, then a^mx = xa^m. Proof: If m = 0, then clearly x = x. Suppose that the conclusion holds for some even m \geq 0. Then a^{m+2}x = a^2a^mx = aaxa^m = ax^{-1}aa^m = xa^{m+2}. By induction, the conclusion holds for all even nonnegative m. \square.

Fact 3: If m is odd, then (ax)^m = a^mx. Proof: If m = 1, we have ax = ax. Now suppose the conclusion holds for some odd m \geq 1; we have (ax)^{m+2} = axax(ax)^m = axx^{-1}aa^mx = a^{m+2}x. By induction, the conclusion holds for all odd positive m. \square

Fact 4: If m is even, then (ax)^m = a^m. Proof: If m = 0, we have 1 = 1. Now suppose the conclusion holds for some even m \geq 0. Then (ax)^{m+2} = axax(ax)^m = axx^{-1}aa^m = a^{m+2}. By induction, the conclusion holds for all even nonnegative m. \square

Suppose |a| = m; now m is finite and moreover odd since |G| is odd.

Note that (ax)^{2m} = ((ax)^2)^m = (a^2)^m = (a^m)^2 = 1^2 = 1. On the other hand, (ax)^{2m} = ((ax)^m)^2 = (a^mx)^2 = x^2. Thus |x| = 2, a contradiction.

Hence x and x^{-1} are never conjugate in a finite group G of odd order.

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