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:
 * $\forall n \in \Z: a^n b = b a^{-n}$

Proof
Proof by induction:

For all $n \in \Z$, 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 $b$:
 * $a b = b a^{-1}$

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

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

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 $n \in \Z: n < 0$ then follows by induction, similarly to the proof for $n > 0$.