# 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:

\(\displaystyle a^{k+1} b\) | \(=\) | \(\displaystyle a \left({a^k b}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \left({b a^{-k} }\right)\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({a b}\right) a^{-k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({b a^{-1} }\right) a^{-k}\) | Basis for the Induction | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b \left({a^{-1} a^{-k} }\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b a^{-\left({k+1}\right)}\) |

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 $a^{-1} b = b a$.

\(\displaystyle a b\) | \(=\) | \(\displaystyle b a^{-1}\) | |||||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle a b a\) | \(=\) | \(\displaystyle b\) | ||||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle b a\) | \(=\) | \(\displaystyle a^{-1} b\) |

... 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 $n > 0$.

$\blacksquare$