Number of fixed points of elements of a dihedral group

Question

Suppose that $D_{2n}=\langle r,s\vert r^n=s^2=1,\, srs=r^{-1} \rangle=\lbrace1,r,\cdots, r^{n-1}, s,rs,\cdots, r^{n-1}s\rbrace$ denotes the dihedral group of order $2n$. We know that the conjugacy classes of $D_{2n}$ are as follows: If $n$ is odd, then the conjugacy classes are: $\{1\}, \{r,r^{-1}\},…, \{r^{\frac{n-1}{2}},r^{\frac{-(n-1)}{2}}\}$, $\{s, rs,…,r^{(n-1)}s\}.$
If $n=2m$ is even, then the conjugacy classes are: $\{1\}, \{r^{m}\}, \{r,r^{-1}\}$,$…, \{r^{(m-1)},r^{-m+1}\}, \{r^{2j}s,0\leq j\leq m-1\}, \{r^{2j+1}s, 0\leq j\leq m-1\}.$
Consider $D_{2n}$ as a transitive permutation group on a set $\Omega$ of size $n$. Recall that for each element $g$ of a group $G$, fix$(g)=\lbrace \alpha \in \Omega\vert \alpha^g=\alpha\rbrace$ and supp$(g)=\lbrace \alpha \in \Omega\vert \alpha^g\neq \alpha\rbrace$. We know that two elements of the symmetric group $S_n$ are conjugate if and only if they have the same cycle structure. I am interested in finding the cycle structure of elements of $D_{2n}$ as a permutation group on $\Omega=\lbrace 1,\cdots, n\rbrace$. Using GAP, I figured out that when $n$ is odd, then in the cycle structure of elements of the group, all the points $\lbrace 1,\cdots, n\rbrace$ appear, but when $n$ is even, some points of $\Omega$ have been fixed by some elements of order 2. For example, the elements of $D_{12}$ as a permutation group on the set $\lbrace 1,\cdots, 6\rbrace$ have cycle structures of the types: one cycle of length 6 or two cycles of length 3 or two cycles of length 2 or three cycles of length 2. I am looking for a technique to find the cycle structure of elements of the dihedral group $D_{2n}$, when $n=p_1^{a_1}\cdots p_s^{a_s}$ for distinct primes $p_1,\cdots, p_s$. I guess it is related to the conjugacy classes of the group. I don't know how to find the sets fix$(g)$ and supp$(g)$ for every $g\in D_{2n}$.

Answer 1

I deleted my last comment because it was too inaccurate.

Everything below is just a long comment.

What is more accurate is the following. Recall that I am used to the fact that $D_n$ is a dihedral group of order $2n$. Let $r=(1,2,\ldots,n)$, $s=(1,n)(2,n-1)\ldots(n/2,n/2+1)$. If $n=2m$ and $m$ is even, then $r$ is a cycle of even length and hence $r$ is an odd permutation, and $s$ is the product of an even number of transpositions and hence an even permutation. Then from geometrical considerations each permutation of the form $r^is$ has at most two fixed symbols. Hence, it follows that $r^is$ at even $i$ is the product of $m$ independent transpositions and at odd $i$ such transpositions are $m-1$.

The case when $m$ is odd is treated similarly.

If I understand your question correctly.