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 $n > 0$.
$\blacksquare$