Order of Conjugate Element equals Order of Element

Theorem
Let $$\left({G, \circ}\right)$$ be a group whose identity is $$e$$.

Then
 * $$\forall a, x \in \left({G, \circ}\right): \left|{x \circ a \circ x^{-1}}\right| = \left|{a}\right|$$

where $$\left|{a}\right|$$ is the order of $a$ in $G$.

Corollary

 * $$\forall a, x \in \left({G, \circ}\right): \left|{x \circ a}\right| = \left|{a \circ x}\right|$$

Proof

 * Let $$\left|{a}\right| = k$$.

Then $$a^k = e$$, and $$\forall n \in \N^*: n < k \implies a^n \ne e$$.


 * We have:

$$ $$ $$ $$

Thus $$\left|{x \circ a \circ x^{-1}}\right| \le \left|{a}\right|$$.


 * Now suppose $$a^n = y, y \ne e$$.

Then $$x \circ a^n \circ x^{-1} = x \circ y \circ x^{-1}$$.

If $$x \circ y = e$$, then $$x \circ a^n \circ x^{-1} = x^{-1}$$.

If $$y \circ x^{-1} = e$$, then $$x \circ a^n \circ x^{-1} = x$$.

So $$a^n \ne e \implies x \circ a^n \circ x^{-1} = \left({x \circ a \circ x^{-1}}\right)^n \ne e$$.

Thus $$\left|{x \circ a \circ x^{-1}}\right| \ge \left|{a}\right|$$, and the result follows.

Proof of Corollary
From the main result, putting $$a \circ x$$ for $$a$$:
 * $$\left|{x \circ \left({a \circ x}\right) \circ x^{-1}}\right| = \left|{a \circ x}\right|$$

from which the result follows by $$x \circ x^{-1} = e$$.