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 $x \circ x^{-1} = e$.