Product of Generating Elements of Dihedral Group

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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 $k \in \Z_{\ge 0}$:

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


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.


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