0

Recall that the dihedral group $D_{6} \cong \mathbb{Z}_{6} \rtimes _{\phi} \mathbb{Z}_{2},$ where the reflection $\mu$ acts on the rotation $\rho$ by \begin{equation*} \prescript{\mu}{}{\rho} = \rho^{-1}. \end{equation*} Show that we also have: $$D_{6} \cong S_{3} \times \mathbb{Z}_{2}.$$ Hint: Label the vertices of a regular hexagon $1,2,3,4,5,6$ consecutively, and consider the stabilizer of the set $\{ 2,4,6 \}$ by the action of $D_{6}$ on the set of vertices.

My question is:

I do not know how considering the stabilizer of the set $\{ 2,4,6 \}$ by the action of $D_{6}$ on the set of vertices of the regular hexagon will prove that $D_{6} \cong S_{3} \times \mathbb{Z}_{2}.$ Could anyone explain this to me, please?

  • The stabilizer of this set is exactly the symmetries of the triangle formed by those vertices (and note that $S_3\cong D_3$). – Nathan Lowry Sep 26 '20 at 03:45
  • And why we need the symmetries of the triangle formed by those vertices and what is the importance of $S_{3}$ being isomorphic to $D_{3}$? could you clarify more please? –  Oct 03 '20 at 21:44

1 Answers1

1

So $D_6$ has a subgroup isomorphic to $S_3$, namely the stabilizer of $\{2,4,6\}$.

It also has a two element subgroup, isomorphic to $\Bbb Z_2$, whose intersection with $S_3$ is trivial (consider a reflection that doesn't stabilise $\{2,4,6\}$.)

By counting $S_3\Bbb Z_2=D_6$.

Thus $D_6\cong S_3×\Bbb Z_2$.

  • And what is the importance of stating that it is the semidirect product of $\mathbb{Z}{6}$ and $\mathbb{Z}{2}$ ? –  Sep 26 '20 at 12:00
  • 1
    Dihedral groups cannot be written as the direct product of cyclic groups. The following has the intuition for semidirect products: https://math.stackexchange.com/questions/1704410/intuition-about-the-semidirect-product-of-groups/3590049#3590049 – user404974 Sep 26 '20 at 14:25
  • 1
    @Smart20 so we know what group we are talking about –  Sep 26 '20 at 16:51
  • How can I count $S_{3}\mathbb{Z}_{2}$? can you give me an example of counting one element of it please? –  Oct 03 '20 at 21:46
  • 1
    $S_3$ is a normal subgroup, because it is of index two. If follows that $S_3\Bbb Z_2$ is a subgroup. But, because of the presence of the identity in each of $S_3$ and $\Bbb Z_2$, $S_3$ and $\Bbb Z_2$ are proper subgroups of $S_3\Bbb Z_2$. But by Lagrange the order has to divide the order of the group. It follows that $S_3\Bbb Z_2$ is too big to be anything but the whole group. –  Oct 03 '20 at 21:57
  • In this statement "It also has a two element subgroup, isomorphic to $\Bbb Z_2$, whose intersection with $S_3$ is trivial (consider a reflection that doesn't stabilise ${2,4,6}$.) " ..... Why should I consider a reflection and not rotation? –  Oct 04 '20 at 14:53
  • But how should we calculate the left cosets of $S_{3}$ in $D_{6}$? or in our case we should calculate the left cosets of the stabilizer of ${2,4,6}$? –  Oct 04 '20 at 16:28
  • There's one rotation you could use. A reflection is just the first thing I thought of, since they all have order two. No real need for cosets. We are just checking that the definition of a direct product are met. –  Oct 04 '20 at 20:31
  • This is a question about your answer here (if you want to answer it I would appreciate that) –  Oct 06 '20 at 22:30
  • Sure. All I have used is that $S_3$ and $\Bbb Z_2$ are subgroups with $S_3\cap\Bbb Z_2=\emptyset$ and $S_3\Bbb Z_2=D_6$. This means by definition that $D_6\cong S_3\times\Bbb Z_2$. Is it a different part you are stuck on? –  Oct 06 '20 at 22:36
  • https://math.stackexchange.com/questions/3854252/showing-that-d-6-has-a-two-element-subgroup-isomorphic-to-bbb-z-2-whose?noredirect=1#comment7951450_3854252 I am sorry I forgot to put the link :) –  Oct 06 '20 at 22:37
  • Yeah I am speaking about a different part. Also can not we take the center instead of this subgroup of order 2 that is isomorphic to $\mathbb{Z}_{2}?$ –  Oct 06 '20 at 22:39
  • Yes we can. None of the $\Bbb Z_2$'s stabilize ${2,4,6}$. The center is, as you know, generated by the rotation through $\pi$. –  Oct 06 '20 at 22:51
  • How did you know that the center is the rotation by $\pi$? –  Oct 06 '20 at 22:53
  • It's well-known: for odd $n$, the center of $D_{2n}$ is trivial, and for even $n$, it is ${\rho,e}$, where $\rho$ is rotation through $\pi$. –  Oct 06 '20 at 22:56
  • Let us continue this discussion in chat. –  Oct 06 '20 at 23:08