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 $\map P n$ be the proposition $a^n b = b a^{-n}$.

$\map P 0$ is trivially true, as $a^0 b = e b = b = b e = b a^{-0}$.

Basis for the Induction
$\map P 1$ 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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


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

Then we need to show:


 * $a^{k + 1} b = b a^{-\paren {k + 1} }$

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

Now we show that $\map P {-1}$ holds, that is, $a^{-1} b = b a$.

thus showing that $\map P {-1}$ holds.

The proof that $\map P n$ holds for all $n \in \Z: n < 0$ then follows by induction, similarly to the proof for $n > 0$.