Power of Product with Inverse

Theorem
Let $$G$$ be a group whose identity is $$e$$.

Let $$a, b \in G: a b = b a^{-1}$$.

Then $$\forall n \in \mathbb{Z}: a^n b = b a^{-n}$$.

Proof
Proof by induction:

For all $$n \in \mathbb{Z}$$, let $$P \left({n}\right)$$ be the proposition $$a^n b = b a^{-n}$$.


 * $$P(0)$$ is trivially true, as $$a^0 b = e b = b = b e = b a^{-0}$$.

Basis for the Induction

 * $$P(1)$$ is true, as this is the given relation between $$a$$ and $$b$$. 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:

$$a^k b = b a^{-k}$$.

Then we need to show:

$$a^{k+1} b = b a^{-\left({k+1}\right)}$$.

Induction Step
This is our induction step:

So $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \in \mathbb{N}: a^n b = b a^{-n}$$.


 * Now we show that $$P \left({-1}\right)$$ holds, i.e. that $$a^{-1} b = b a$$.

... thus showing that $$P \left({-1}\right)$$ holds.


 * The proof that $$P \left({n}\right)$$ holds for all $$n \in \mathbb{Z}: n < 0$$ then follows by induction, similarly to the proof for $$n > 0$$.