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:

 $\ds a^{k + 1} b$ $=$ $\ds a \paren {a^k b}$ $\ds$ $=$ $\ds a \paren {b a^{-k} }$ Induction Hypothesis $\ds$ $=$ $\ds \paren {a b} a^{-k}$ $\ds$ $=$ $\ds \paren {b a^{-1} } a^{-k}$ Basis for the Induction $\ds$ $=$ $\ds b \paren {a^{-1} a^{-k} }$ $\ds$ $=$ $\ds b a^{-\paren {k + 1} }$

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

 $\ds a b$ $=$ $\ds b a^{-1}$ $\ds \leadstoandfrom \ \$ $\ds a b a$ $=$ $\ds b$ $\ds \leadstoandfrom \ \$ $\ds b a$ $=$ $\ds a^{-1} b$

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 $n > 0$.

$\blacksquare$