# Subgroup structure of dihedral groups

I am currently looking into structure of dihedral groups; I am interested in their subgroup structure. Dihedral group have two kinds of elements; I will use their geometric meaning and call them rotation and reflection: $r$ and $s$. All dihedral groups can be described in this way: $$D_n = \{r^is^j|0 \le i <n; 0 \le j <2\}$$

Now about the group structure. It is simple that all reflections form subgroups of order 2, because reflection of reflection is unchanged image: $\{\operatorname{id}, r^0s^1 \}, \dots ,\{\operatorname{id}, r^ns^1\}$. It is also simple that rotations create their own subgroup: $\{\operatorname{id}, r^1s^0, \dots, r^ns^0\}$. I am also sure that rotation subgroups will always be normal, because $r_1^{-1}rr_1$ and $s_1^{-1}rs_1$ is always a rotation again. If the $n$ in question isn't a prime, there will also be subgroups of rotations on number of angles equal to divisors of $n$.

I believe I understand nature of subgroups, which consist only of rotations or reflections, but I don't understand subgroups which contain rotations and reflections alike. Is there a simple explanation of their structure and geometric nature? How can I say if they are normal or not without exhaustive search of possibilities?

• You mean, apart from the fact that they are themselves dihedral groups? – M. Vinay Jun 17 '14 at 14:57
• @M.Vinay I think the poster wants to know if there is a general way to tell if a subgroup of a dihderal group is normal. – angryavian Jun 17 '14 at 14:58
• @angryavian That is one of his questions, but another is about their nature in general. I was just wondering whether he had noticed that such subgroups are dihedral. – M. Vinay Jun 17 '14 at 15:00
• >I was just wondering whether he had noticed that such subgroups are dihedral. No, I didn't. – Dark Archon Jun 17 '14 at 15:05

It should be obvious that the rotations form a normal subgroup of index 2 (the fact that it is of index 2 is sufficient to prove normality). If we call the rotation subgroup $R$, and the reflection coset $F$, we have:

$RR = R$

$RF = F$

$FR = F$

$FF = R$.

Subgroups containing only rotations are cyclic, due to the fact that $R$ is cyclic. We thus get exactly one subgroup of order $d$ contained in $R$, for each divisor $d$ of $n$.You should prove any (and all) of these subgroups are normal.

Subgroups containing only reflections and the identity must have order a power of 2. Since a reflection times a reflection is a rotation, with $(r^ks)(r^ms) = r^{k-m}$, it should be clear that any such subgroup is in fact of order 2 (we must have $k = m)$. These subgroups are typically NOT normal, but there is an exception for $n = 2$ ($D_2 = V$ is abelian).

Which brings us to "mixed subgroups", containing at least one rotation, and one reflection. These are going to be of the form $\langle r^k,s\rangle$ where $k|n$, that is, isomorphic to $D_m$, where $m = \dfrac{n}{k}$ . You can think of these as symmetries of an $m$-gon, which are also symmetries of an $n$-gon, since $n$ is a multiple of $m$ (the axes of symmetry of an $n$-gon include all the axes of symmetry of the $m$-gon, plus more).

These "mixed subgroups" aren't, in general, normal, but in some special cases they are: for example if $n$ is even and $k = \dfrac{n}{2}$, or $k = 2$.

Another case worth mentioning is when $n$ is an odd prime; in this case, any subgroup containing more than one reflection, or a reflection and a (non-trivial) rotation, is the entire group, which limits the possibilities for subgroups.

For a more complete analysis, see: https://in.answers.yahoo.com/question/index?qid=20091014113730AA7KJDt