Product of Generating Elements of Quaternion Group

Theorem

Let $Q = \Dic 2$ be the quaternion group:

$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$

Then for all $k \in \Z_{\ge 0}$:

$b a^k = a^{-k} b$

Proof

The proof proceeds by induction.

For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition:

$b a^k = a^{-k} b$

$\map P 0$ is the case:

\(\ds b a^0\)

\(=\)

\(\ds b e\)

\(\ds \)

\(=\)

\(\ds e b\)

\(\ds \)

\(=\)

\(\ds a^{-0} b\)

Thus $\map P 0$ is seen to hold.

Basis for the Induction

We have:

\(\ds a b a\)

\(=\)

\(\ds b\)

\(\ds \leadsto \ \ \)

\(\ds b a^1\)

\(=\)

\(\ds a^{-1} b\)

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis

Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:

$b a^r = a^{-r} b$

from which it is to be shown that:

$b a^{r + 1} = a^{-r - 1} b$

Induction Step

This is the induction step:

\(\ds b a^{r + 1}\)

\(=\)

\(\ds b a^r a\)

\(\ds \)

\(=\)

\(\ds a^{-r} b a\)

Product of Generating Elements of Quaternion Group

\(\ds \)

\(=\)

\(\ds a^{-r} \alpha^{-1} b\)

Basis for the Induction

\(\ds \)

\(=\)

\(\ds a^{-r - 1} b\)

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall k \in \Z_{\ge 0}: b a^k = a^{-k} b$

$\blacksquare$