Order of Conjugate Element equals Order of Element

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Then

$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$

where $\order a$ denotes the order of $a$ in $G$.


Corollary

$\forall a, x \in \struct {G, \circ}: \order {x \circ a} = \order {a \circ x}$


Proof

Let $\order a = k$.

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:

\(\displaystyle \paren {x \circ a \circ x^{-1} }^k\) \(=\) \(\displaystyle x \circ a^k \circ x^{-1}\) Power of Conjugate equals Conjugate of Power
\(\displaystyle \) \(=\) \(\displaystyle x \circ e \circ x^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle x \circ x^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle e\)


Thus $\order {x \circ a \circ x^{-1} } \le \order a$.


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:

$a^n \ne e \implies x \circ a^n \circ x^{-1} = \paren {x \circ a \circ x^{-1} }^n \ne e$

Thus:

$\order {x \circ a \circ x^{-1} } \ge \order a$

and the result follows.

$\blacksquare$


Sources