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 $G$.

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

Then $a^k = e$, and:
 * $\forall n \in \N_{>0}: n < k \implies a^n \ne e$

by definition of the order of $a$ in $G$

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.