Power of Product with Inverse

From ProofWiki
Jump to navigation Jump to search


Let $G$ be a group whose identity is $e$.

Let $a, b \in G: a b = b a^{-1}$.


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


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 $n > 0$.