Choice of generator in dihedral group

Question

I'm about to begin studying group representation theory, and I want to get more familiar with the symmetric group $\mathfrak{S}_n$ (and its subgroups) first. In particular, I'd like to better understand the dihedral group $D_n$ (of order $2n$).

The generators of $D_n$ are given by $r, s \in \mathfrak{S}_n$ such that $r^n = id, s^2 = id,$ and $srs=r^{-1}$. My question is: will the cycle $r = (123...n)$ always be a suitable element that satisfies this relation? And if so, how does one go about choosing the value of $s$? Is it just trial and error? I know that it can't just be any order two cycle. For instance, $r = (1234)$ and $s = (13)$ generates the dihedral group of order 4 (permutations of a square), but $r = (1234)$ and $s = (12)$ does not.

Answer 1

Yes, the cycle $r = (1,2,3,...,n)$ is always a suitable element.

To find an element $s$ or order $2$ such that $srs=r^{-1}$, you can use this trick:

$D_n$ is the group of isometries of a regular $n$-agon: if you draw a $n$-agon and you label the vertices clockwise with numbers $\{1,2,...,n\}$, you can think that your permutation $r$ is the rotation of $\frac{2\pi}{n}$. Now you can choose any symmetries of this $n$-agon and, with the rotation, they generate $D_n$. If you now represent the symmetry in terms of permutations of vertices, you will find the elements $s$ you are looking for.

I also give you an example:

• if $n=2k$, you can take the symmetry $s=(1,2k)(2,2k-1)...(k,k+1)$ representing the symmetry along the axis of the segment with vertex $1$ and $2k$.
• if $n=2k+1$, you can take the symmetry $s=(2,2k+1)(3,2k)...(k+1,k+2)$ representing the symmetry along the axis passes through the vertex $1$.