2

I need to find an isomorphism between $D_n$ (all symmetries of an $n-gon$) and a subgroup of $S_n$.
I know that Cayley's theorem gives a nice isomorphism that shows that $D_n$ is isomorphic to a subgroup of $S_{2n}$, but this isn't the case here, since I need an isomorphism to $S_n$.
Now, I have a general idea on how to accomplish this; numbering all vertices of the $n-gon$ by $1,2,...,n$, and assign each symmetry in $D_n$ with the permutation of the vertices in respect to the initial symmetry $e$.

For example, if $n=4$, then the element $r^2$ will correspond to the permutation: $$\begin{pmatrix} 1 & 2 & 3 & 4\\ 3 & 4 & 1 & 2 \end{pmatrix}$$

My problem is, I'm having difficulty defining the map itself.
I know that it suffices to define a homomorphism $\varphi$ on the $D_n$ generators, $r$ and $s$, and I noticed that $\varphi(r)$ always equals to the permutation $(n (n-1) (n-2)...321)$ (written in cycle nonation), but I have no idea how to define $\varphi$ on $s$ and since I also need to prove that $\varphi$ is one-to-one homomorphism, I need to reconsider whether this is the optimal way of defining it, or maybe there is better way to define it, in such that it'll be easier for me to prove it.

Any suggestions?

so.very.tired
  • 2,065
  • 1
    You need the rotation to map to $(1,2\ldots n)$ and the reflection to map to some product of disjoint transpositions. If $n$ is even, then it should go to $\prod_{i=1}^{n/2} (i,(n+1-i))$ and if $n$ is odd, almost that same thing.

    It's clear that the image of $r$ and the image of $s$ have the appropriate orders, but you need to check that they commute properly, i.e. $srs=r^{-1}$.

     – Ian Coley Jan 14 '14 at 19:01
  • Any symmetry of an n-gon permutes the vertices. This gives a map to the permutation group. One can also see that the symmetry is actually completely determined by what it does to the vertices, so this map is an injection. – Cheerful Parsnip Jan 14 '14 at 19:25
  • @IanColey, Is it enough to show that this definition of $\varphi$ preserves the $D_n$ group relations, in order to conclude that these two are isomorphic? What about showing that $\forall a,b\in D_n:\varphi (ab)=\varphi (a)\varphi (b)$? and what about the one-to-one property of $\varphi$? – so.very.tired Jan 14 '14 at 22:05

1 Answers1

2

The solution, i.e., an explicit embedding of $D_n$ into $S_n$ is given on the first page here. An explicit definition of a one-to-one map is given. It needs perhaps filling in some details, but should be very helpful.

Edit: In the meantime the link was removed, but also there appeared new posts with an answer, e.g., here:

Defining dihedral groups $\{\sigma \in S_n: $ something $\}$

Dietrich Burde
  • 130,978
  • Thank you. that was very helpful. – so.very.tired Jan 14 '14 at 21:50
  • @DietrichBurde, this is long after the fact, but could you say what reference was being cited here? The link terminus seems to have changed since posting. – Dennis Muhonen Sep 13 '20 at 20:38
  • @DennisMuhonen Thank you. The bachelor thesis has been moved. It is better to link to mathematics stackexchange, you are right. Fortunately there are many posts here with an answer. – Dietrich Burde Sep 14 '20 at 08:35