$D_6$ as permutation group 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?
 A: 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
$$
s\mapsto
\left(\begin{array}{cccccc}
0&1&0&0&0&0\\
1&0&0&0&0&0\\
0&0&0&1&0&0\\
0&0&1&0&0&0\\
0&0&0&0&0&1\\
0&0&0&0&1&0\\
\end{array}\right).
$$
A: 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.
A: 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 $s \mapsto (2 \ 6)(3 \ 5)$.
