# Product of Generating Elements of Dihedral Group

## Theorem

Let $D_n$ be the dihedral group of order $2 n$.

Let $D_n$ be defined by its group presentation:

$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$

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

$\beta \alpha^k = \alpha^{n - k} \beta$

## Proof

The proof proceeds by induction.

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

$\beta \alpha^k = \alpha^{n - k} \beta$

$\map P 0$ is the case:

 $\ds \beta \alpha^0$ $=$ $\ds \beta e$ $\ds$ $=$ $\ds e \beta$ $\ds$ $=$ $\ds \alpha^n \beta$ $\ds$ $=$ $\ds \alpha^{n - 0} \beta$

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

### Basis for the Induction

We have:

 $\ds \beta \alpha \beta$ $=$ $\ds \alpha^{-1}$ Group Presentation of Dihedral Group $\ds \leadsto \ \$ $\ds \beta \alpha \beta^2$ $=$ $\ds \alpha^{-1} \beta$ $\ds \leadsto \ \$ $\ds \beta \alpha$ $=$ $\ds \alpha^{-1} \beta$

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:

$\beta \alpha^r = \alpha^{n - r} \beta$

from which it is to be shown that:

$\beta \alpha^{r + 1} = \alpha^{n - r - 1} \beta$

### Induction Step

This is the induction step:

 $\ds \beta \alpha^{r + 1}$ $=$ $\ds \alpha^{n - r} \beta \alpha$ $\ds$ $=$ $\ds \alpha^{n - r} \alpha^{-1} \beta$ Basis for the Induction $\ds$ $=$ $\ds \alpha^{n - r - 1} \beta$

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

Therefore:

$\forall k \in \Z_{\ge 0}: \beta \alpha^k = \alpha^{n - k} \beta$

$\blacksquare$