1. Let $G = D_6$ be the dihedral group of the rigid motions of a regular hexagon and $X = \{H \subseteq G \mid H \leq G\}$. Consider the action of $G$ on $X$ by conjugation, that is, for $g \in G$, $H \in X$: $$ g \cdot H = g H g^{-1}. $$ a) Find the stabilizer $G_H$ of $H$, for all $H \in X$.
    b) Find the orbit $G(H)$ of $H$, for all $H \in X$.
    c) Find the set of all orbits $X/G$.

I'm working on this problem (note $D_6 = D_{12}$ in the more common notation). So far I've:

  1. Found all the subgroups of $D_{12}$ (all 16, I hope).
  2. Found the normalizer of each subgroup to get the stabilizers for each subgroup. (Is this correct since the action is conjugation?)
  3. Found the orbits via brute force.
  4. Found all the cosets of each $H$ and started constructing the orbits.

However, all this is taking an ENORMOUSLY long time. So much so that I feel I'm missing a much more clever solution.

Is there one?


1 Answer 1


$1$.) I count 16 as well:

$D_{12}$ itself and the identity (trivial) subgroup.

$\langle r\rangle,\langle r^2\rangle, \langle r^3\rangle$

Two isomorphs of $S_3$: $\langle r^2,s\rangle$ and $\langle r^2,rs\rangle$.

Three isomorphs of $V$: $\langle r^3,s\rangle$, $\langle r^3,rs\rangle$ and $\langle r^3,r^2s\rangle$.

Six subgroups generated by a single reflection $\langle r^ks\rangle$ for $k = 0,1,2,3,4,5$.

$2$.) Yes, you want to find the normalizers in $D_{12}$. This is the entire group (as these are all normal) for the first seven subgroups listed, which greatly simplifies things. The isomorphs of $V$ are non-normal but of prime index, so they must be their own normalizers. It is clear that the normalizers of the reflection-generated subgroups of order 2 must be an isomorph of $V$ (the isomorphs of $V$ are abelian so they normalize any subgroup), as $S_3$ has no normal subgroups of order 2 (we are using the fact that the normalizer of a subgroup $H$ in a group $G$ is the largest subgroup of $G$ containing $H,$ in which $H$ is normal).

$3$.) Finding the orbits by brute force is not so bad: the seven normal subgroups will all have a single element of just themselves in their orbits, because they are stabilized by all of $D_{12}$. From Sylow theory (or by inspection), we see the isomorphs of $V$ are all conjugate, so there is another orbit (remember conjugation preserves order of subgroups), and we are left with finding the orbits of the reflection-generated subgroups.

By the orbit-stabilizer theorem (or by examining their generators' (ordinary) conjugacy classes) we have that these must occur in two orbits of three: $\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\}$ and $\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}$

Thus: $G/X = \{\{\{1\}\},\{\langle r\rangle\},\{\langle r^2\rangle\}, \{\langle r^3\rangle\},\{D_{12}\},\{\langle r^2,s\rangle\},\{\langle r^2,rs\rangle\},\{\langle r^3,s\rangle, \langle r^3,rs\rangle,\langle r^3,r^2s\rangle\},\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\},\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}\}.$

(This is a set with 10 elements).


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