I am trying to solve some exercises for a course in representation theory. We are studying finite groups and I have an exercise about the dihedral groups $D_n = \left<r,s|r^n,s^2,rs=sr^{n-1}\right>$ for $n>2$. After finding all irreducible representations of $D_n$ and computing the character tables, I got to the following question:

$D_6$ may be understood as a permutation group, acting on the vertices of the hexagon. We hence obtain a repr. of $D_6$ on $\mathbb{C}^6$. Decompose this repr. into irreducibles.

My problem is: I don't understand the first part of the question, i.e. how $D_6$ is to be interpreted as permutation group. Anyone has any clue?


Just number the vertices of a regular hexagon. Let's say we number them counterclockwise in the order $1,2,3,4,5,6$. Then the rotation $r$ maps them according to the 6-cycle $r=(123456)$ and (one of the) reflection(s) as a product of three transpositions $s=(12)(36)(45)$. This gives a homomorphism $D_6\to S_6$. You are expected to compose that with the 6-dimensional "natural" representation of $S_6$.

This gives a representation of $D_6$ by 6x6 matrices such that $$ r\mapsto\left(\begin{array}{cccccc} 0&0&0&0&0&1\\ 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ \end{array}\right) $$ and $$ s\mapsto \left(\begin{array}{cccccc} 0&1&0&0&0&0\\ 1&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&0&1&0&0&0\\ 0&0&0&0&0&1\\ 0&0&0&0&1&0\\ \end{array}\right). $$

  • $\begingroup$ By "natural" you mean the regular representation? $\endgroup$ – Daniel Robert-Nicoud Mar 5 '13 at 20:52
  • $\begingroup$ @DanielRobert-Nicoud: No, the regular representation of $S_6$ has dimension $|S_6|=720$. I mean the 6-dimensional representation, where $S_6$ acts by permuting the elements of the the natural basis of $\mathbb{C}^6$. Editing..... $\endgroup$ – Jyrki Lahtonen Mar 5 '13 at 20:55
  • $\begingroup$ Ah, seems reasonable :) Thanks. $\endgroup$ – Daniel Robert-Nicoud Mar 5 '13 at 20:56
  • $\begingroup$ YW. Note that some of the reflecting have fixed points (see the answers by Jim and user58512). Consequently their traces will be different here. No wonder, because the two types of reflections form two distinct conjugacy classes in $D_6$. This is different from $D_n$ with odd $n$. $\endgroup$ – Jyrki Lahtonen Mar 5 '13 at 21:02
  • 1
    $\begingroup$ Yes, actually we have been given a hint about the fact that $D_4$ can be interpreted in two different ways. The difference would be that $\chi(s) = 2$ instead of $0$ and a couple of other things like that. $\endgroup$ – Daniel Robert-Nicoud Mar 5 '13 at 21:44

it's quite simple, imagine a hexagon:

   1   2

6         3

   5   4

then $D_6 = \langle (1\;2\;3\;4\;5\;6), (1\;5)(2\;4)) \rangle$ is generated by the rotations, and flipping over of that hexagon.

  • $\begingroup$ So, it would be "as a subgroup of a permutation group" (as in Cayley's theorem). Anyway shouldn't it be $(1\ 5)(2\ 4)$? $\endgroup$ – Daniel Robert-Nicoud Mar 5 '13 at 20:51
  • $\begingroup$ @DanielRobert-Nicoud, yeah that was a typo - in general permutation group means a pair $(G,X)$ where the group $G$ acts on the set $X$. $\endgroup$ – user58512 Mar 5 '13 at 21:00

Let $r$ be a rotation of $\frac{\pi}{3}$ (so each vertex of the hexagon gets sent to an adjacent vertex) and let $s$ be a reflection about a line passing through two vertices that are opposite of each other.

This realizes $D_6$ as symmetries of a hexagon. We then get an inclusion $D_6 \to S_6$ by considering how those symmetries permute the vertices of that hexagon. Depending on how you number your vertices you'll get that this map is $r \mapsto (1 \ 2 \ 3 \ 4 \ 5 \ 6)$ and $s \mapsto (2 \ 6)(3 \ 5)$.


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