Conjugation preserves order

Let G be a group and let x,g \in G. Prove that |x| = |g^{-1} x g|. Deduce that |ab| = |ba| for all a,b \in G.


First we prove a technical lemma: for all a,b \in G and n \in \mathbb{Z}, (b^{-1} a b)^n = b^{-1} a^n b. The statement is clear for n = 0. We prove the case n > 0 by induction; the base case n = 1 is clear. Now suppose (b^{-1} a b)^n = b^{-1} a^n b for some n \geq 1; then (b^{-1} a b)^{n+1} = (b^{-1} a b) (b^{-1} a b)^n = b^{-1} a b b^{-1} a^n b = b^{-1} a^{n+1} b. By induction the statement holds for all positive n. Now suppose n < 0; we have (b^{-1} a b)^n = ((b^{-1} a b)^{-n})^{-1} = (b^{-1} a^{-n} b)^{-1} = b^{-1} a^n b. Hence, the statement holds for all integers n. square

Now to the main result.

Suppose first that |x| is infinity and that |g^{-1}xg| = n for some positive integer n. Then we have (g^{-1} x g)^n = g^{-1} x^n g = 1, and multiplying on the left by g and on the right by g^{-1} gives us that x^n = 1, a contradiction. Thus if |x| is infinity, so is |g^{-1} x g|. Similarly, if |g^{-1} x g| is infinite and |x| = n, we have (g^{-1} x g)^n = g^{-1} x^n g = g^{-1} g = 1, a contradiction. Hence if |g^{-1} x g| is infinte, so is |x|.

Suppose now that |x| = n and |g^{-1} x g| = m for some positive integers n and m. We have (g^{-1} x g)^n = g^{-1} x^n g = g^{-1} g = 1, so that m \leq n, and (g^{-1} x g)^m = g^{-1} x^m g = 1, so that x^m = 1 and n \leq m. Thus n = m. \square

Now let a and b be arbitrary group elements. Letting x = ab and g = a, we see that |ab| = |a^{-1}aba| = |ba|. \blacksquare

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