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 $b$. 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:

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

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

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

... 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 $n > 0$.

$\blacksquare$