# Dihedral Group/Group Presentation

## Theorem

The dihedral group $D_n$ has the group presentation:

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

That is, the dihedral group $D_n$ is generated by two elements $\alpha$ and $\beta$ such that:

$(1): \quad \alpha^n = e$
$(2): \quad \beta^2 = e$
$(3): \quad \beta \alpha = \alpha^{n - 1} \beta$

## Proof

By definition, the dihedral group $D_n$ of order $2 n$ is the group of symmetries of the regular $n$-gon.

So, let $P$ denote a regular polygon with $n$ sides.

Let $\alpha$ be a rotation of $P$ by $\dfrac {2 \pi} n$.

Let $\beta$ be a reflection $P$ whose axis of reflection is the $y$ axis.

It takes $n$ rotations by $\dfrac {2 \pi} n$ to return $P$ to its original position.

Thus $\alpha$ is of order $n$, and so:

$\alpha^n = e$

Two reflections of $P$ in the same axis of reflection also returns $P$ to its original position.

Thus $\beta$ is of order $2$, and so:

$\beta^2 = e$

It remains to be shown that $\beta \alpha \beta = \alpha^{-1}$.

Let the vertices of $P$ be labelled in order anticlockwise from the $y$-axis: $A_1, A_2, \ldots, A_n$.

Let $P$ be arranged so that $A_1$ is on the $y$-axis.

From here, $\beta$ on $P$ exchanges the places of $A_2$ and $A_n$.

From here, $\alpha$ sends $A_2$ to the position of $A_1$, $A_1$ to the original position of $A_2$, and $A_3$ to the original position of $A_n$.

From here, $\beta$ sends exchanges the positions of $A_3$ and $A_1$.

So $\beta \alpha \beta$ on $P$ results in $A_1$ moving to the original position of $A_n$, $A_2$ moving to the original position of $A_1$ and $A_3$ moving to the original position of $A_2$.

That is, $\beta \alpha \beta$ is a rotation of $P$ by $-\dfrac {2 \pi} n$, or $\alpha^{-1}$.

$\blacksquare$