Let $d$ be a divisor of $n$.
Then obviously, $\langle r^d\rangle$ is a normal subgroup, so we may take the quotient$D_n/\langle r^d\rangle$ . It is not hard to show that this is isomorphic to $D_d$.
But what is the geometric meaning of this isomorphism? How do I visualize this? Do we start from an $n$-gon and end up with a $d$-gon?
edit: $D_n = \langle r,s\,\mid\,r^n=s^2=1, rs=sr^-1\rangle$, the dihedral group of order $2n$.
Best Answer
Yes, your intuition is correct, we start with and $n$-gon and end up with a $d$-gon. One way to think of this is to consider just the rotations (subgroup generated by $r$). We can think of $r$ as rotating the $n$-gon by $360/n$ degrees. Of course, that means that applying $r$ $n$-times results a one full rotation of the $n$-gon. When we mod out by $\left<r^d\right>$, we are saying that applying $r$ $d$-times is the identity, ie, it results in one full rotation. So the subgroup generated $r\left<r^d\right>$ has order $d$. This yields the presentation
\begin{equation} D_n/\left<r^d\right> = \left< r\left<r^d\right>, s\left<r^d\right> \;|\; \left(r\left<r^d\right>\right)^d = \left(s\left<r^d\right>\right)^2 = 1,\; \left(r\left<r^d\right>\right)\left(s\left<r^d\right>\right) = \left(s\left<r^d\right>\right)\left(r\left<r^d\right>\right)^{-1} \right> \end{equation} which is a presentation of $D_d$.