# Subgroups of the dihedral group $D_n$ modulo $Aut(D_n)$

This question is related to this math.se question.

Consider the dihedral group $D_n = \langle r,s \rangle.$

Two subgroups $G, H \leq D_n$ are said to be ''isomorphic'' if there is an $f \in \rm{Aut}(D_n)$ such that $f(G) = H.$

I would like to describe all non-isomorphic subgroups of $D_n.$ I can describe all subgroups of $D_n$ up to conjugation but don't see how to do the final step of checking how these list collapses with respect to outer automorphisms of $D_n.$

What I do know

• Every automorphism $f$ of $D_n$ sends $f(r) = r^k$ and $f(s) = sr^l$ for $1 \le l,k < n$ such that $\gcd(k,l) = 1.$
• The inner automorphisms of $D_n$ are of the form $f(r) = r^{\pm 1}, f(s) = sr^{2k}$

• For $n$ odd the subgroups of $D_n$ are (up to conjugacy): $$\{ \langle r^m, s \rangle \quad m | 2n \quad \mbox{ and } m \equiv 1 \pmod{2} \} \cup \{ \langle r^{m/2} \rangle | m |2n \mbox{ and } m \equiv 0 \pmod{2}\}.$$

• For $n$ even the subgroups of $D_n$ up to conjugation are $$\{ \langle r^m, s \rangle \mid m |2n \mbox{ and } m \equiv 1 \pmod{2}\} \cup \{ \langle r^{m/2} \rangle \mid m \not | n \mbox{ and } m \equiv 0 \pmod{2} \}$$ and $$\{ \langle r^{m/2} \rangle, \langle r^m,s \rangle , \langle r^m,rs\rangle \mid m | n \mbox{ and } m \equiv 0 \pmod{2}\}.$$

My question now is, how to take into account the isomorphisms of $\rm{Out}(D_n)$ and the listed classes of subgroups?

• When $n$ is odd, there is no fusion of classes under ${\rm Aut}(D_n)$. When $n$ is even, the outer automorphism $r \mapsto r$, $s \mapsto rs$ fuses some pairs of classes (you can work out which) and apart from that it is easy to see that no other classes can be fused (they are mostly mutually non-isomorphic). – Derek Holt Apr 6 '15 at 16:52
• @DerekHolt Hm.. Why is there no fusion for $n$ odd and why is it enough to consider the automorphism $r \mapsto r$ and $s \mapsto rs$? – Jernej Apr 7 '15 at 9:59
• For $n$ odd the subgroups consist of a cyclic group of order $m$ and a dihedral group of order $2m$ for each $m$ dividing $n$. So the subgroups are all mutually nonisomorphic, and hence there can be no fusion. For $n$ even, there are sometimes two dihedral groups of the same order, and they are fused by that outer automorphism. There can be no further fusion, because there are no other pairs of isomorphic subgroups (except for the subgroups of order $2$ need a bit more thought, but only one of those is central, so it cannot be fused to any others). – Derek Holt Apr 7 '15 at 19:50
• @DerekHolt Derek, thanks that helped. If you're willing to copy paste your comment into an answer I'll upvote& accept it. – Jernej Apr 13 '15 at 10:43