Product of Generating Elements of Dihedral Group
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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:
- $\forall k \in \Z_{\ge 0}: \beta \alpha^k = \alpha^{n - k} \beta$
$\blacksquare$