Power of Conjugate equals Conjugate of Power

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

Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$.

That is, let $x$ and $y$ be conjugate.

Then:
 * $\forall n \in \Z: y^n = \paren {a^{-1} \circ x \circ a}^n = a^{-1} \circ x^n \circ a$

It follows directly that:
 * $\exists b \in G: \forall n \in \Z: y^n = b \circ x^n \circ b^{-1}$

In particular:
 * $y^{-1} = \paren {a^{-1} \circ x \circ a}^{-1} = a^{-1} \circ x^{-1} \circ a$

Proof
Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition $y^n = a^{-1} \circ x^n \circ a$.

$\map P 0$ is true, as this just says $e = a^{-1} \circ e \circ a$.

Basis for the Induction
$\map P 1$ is the case $y = a^{-1} \circ x \circ a$, which is how conjugacy is defined for $x$ and $y$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


 * $y^k = a^{-1} \circ x^k \circ a$

Then we need to show:


 * $y^{k + 1} = a^{-1} \circ x^{k + 1} \circ a$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: y^n = a^{-1} \circ x^n \circ a$

Now we need to show that if $\map P n$ holds, then $\map P {-n}$ holds.

That is:
 * $y^{-n} = a^{-1} \circ x^{-n} \circ a$

Let $n \in \N$.

Then:

Thus $\map P n$ has been shown to hold for all $n \in \Z$.

Hence the result.