Power of Conjugate equals Conjugate of Power

Theorem
Let $$\left({G, \circ}\right)$$ 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 $$\forall n \in \Z: y^n = \left({a^{-1} \circ x \circ a}\right)^n = a^{-1} \circ x^n \circ a$$.

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

In particular, $$y^{-1} = \left({a^{-1} \circ x \circ a}\right)^{-1} = a^{-1} \circ x^{-1} \circ a$$.

Proof
Proof by induction:

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


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

Basis for the Induction

 * $$P(1)$$ 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 $$P \left({k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ 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 $$P \left({k}\right) \implies P \left({k+1}\right)$$ 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 $$P \left({n}\right)$$ holds, then $$P \left({-n}\right)$$ holds.

That is, $$y^{-n} = a^{-1} \circ x^{-n} \circ a$$.

Let $$n \in \N$$. Then:

$$ $$ $$ $$ $$


 * Thus $$P \left({n}\right)$$ has been shown to hold for all $$n \in \Z$$, hence the result.