Algebra Solutions: 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 $|ab| = |a^{-1}aba| = |ba|$ . $\blacksquare$

Algebra Solutions

Conjugation preserves order

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