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I am trying to understand hos one can define the dihedral groups $D_n$. I have seen the "definition" that just says this is the group of symmetries of an $n$-polygon. So you have rotations and reflections. But I feel this definition is a bit vague. I asked around and heard that one can define this using generators and relations. I don't know about that.

I know that one can "realize" for example $D_4$ as a subgroup of $S_4$. For example $$ D_4 =\{(1), (13), (24), (14)(23), (1234), (12)(34), (1432), (13)(24)\}. $$ I do understand that I get these elements from labelling the vertices of the $4$-gon. This is very concrete for me.

Therefore my question is: Is there a nice way to actually define the general dihedral group $D_n$ as a specific concrete subgroup of $S_n$?

So, for example, I am looking for something like $$D_n = \{\sigma \in S_n : \text{something} \}.$$

I am not looking for a vague algorithmic way of defining the groups.

From the example of $D_4$ I am thinking that it should always have $2$-cycles, but I don't think that this is always true. For example with $D_5$.

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  • $\begingroup$ Label the vertices the $n$-gon by $1,\ldots,n$ etc. $\endgroup$ – Lord Shark the Unknown Oct 13 '17 at 17:14
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Sure. For instance, you can use the fact that $D_n$ is generated by a rotation and a reflection. In terms of permuting the vertices, this means you can define $D_n$ as the subgroup of $S_n$ generated by the $n$-cycle $(1\ 2\ 3\ 4\ \dots\ n)$ and the permutation $(2\ n)(3\ n-1)(4\ n-2)\dots$ (that is, the permutation that sends $k$ to $n-k+2$ mod $n$). If the vertices are labelled $1,\dots, n$ going around the $n$-gon counterclockwise, the first generator is the rotation by an angle of $2\pi/n$ and the second generator is the reflection through the line of symmetry passing through vertex $1$.

You can in fact explicitly describe all the elements of $D_n$ similarly. The rotations are permutations of the form $k\mapsto k+m$ mod $n$ for some fixed $m$ (that's a rotation by an angle of $2\pi m/n$). The reflections are permutations of the form $k\mapsto m-k$ mod $n$ for some fixed $m$ (that's the reflection across the line of symmetry passing through vertex $\frac{m}{2}$ if $m$ is even and the line of symmetry passing through the edge between vertices $\frac{m-1}{2}$ and $\frac{m+1}{2}$ if $m$ is odd). So you could explicitly define $D_n$ as the set of permutations $\sigma\in S_n$ which have the form $\sigma(k)=ak+m$ mod $n$ for $a\in\{-1,1\}$ and $m\in\{0,\dots,n-1\}$.

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  • $\begingroup$ Can you elaborate on the second paragraph? $\endgroup$ – John Doe Oct 13 '17 at 17:28
  • $\begingroup$ I don't know what kind of elaboration you want. What don't you understand? $\endgroup$ – Eric Wofsey Oct 13 '17 at 17:29
  • $\begingroup$ For example, when you say $k\mapsto k+m$ mod $n$ where is this function going from and to? $\endgroup$ – John Doe Oct 13 '17 at 17:30
  • $\begingroup$ An element of $S_n$ is by definition a bijection from $\{1,\dots,n\}$ to itself, so all these functions are from $\{1,\dots,n\}$ to $\{1,\dots,n\}$. $\endgroup$ – Eric Wofsey Oct 13 '17 at 17:31
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    $\begingroup$ As you may have discovered by now this works except for $n=2$ because $S_2$ has order $2$ and is thus too small to embed $D_2$ which has order $4$. $\endgroup$ – Stephen Meskin Nov 25 '17 at 2:50
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The dihedral group is symmetries of the $n$-gon, and it is generated by a rotation by $2\pi/n$ and a reflection about one of the vertices. If you label the vertices by $1,2,\dots,n$ counterclockwise, then the rotation can be represented by the cycle $(1\ 2\ \ldots\ n)\in S_n$, and the reflection by $(1)(2\ n)(3\ (n-1))\dots \in S_n$ (where the end is either $(\frac{1}{2}n)$ or $(\frac{1}{2}(n-1)\ \frac{1}{2}(n+1))$ depending on whether $n$ is even or odd, respectively).

Presentation-wise, if $\sigma$ is rotation of the plane by $2\pi /n$ and $\tau$ is the flip about the $x$-axis, then it ends up being that all you care about is that $\sigma^n=1$, $\tau^2=1$, and how they fail to commute: $\tau\sigma=\sigma^{-1}\tau$. The group is $\langle \sigma,\tau\mid \sigma^n=\tau^2=1,\tau\sigma=\sigma^{-1}\tau\rangle$.


In general, Cayley's theorem gives a way to take any finite group and embed it as a subgroup of the symmetric group.

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If $n = 2k$ is even, $$D_n = \langle (12 \cdots n), (1 n)(2, n - 1) \cdots (k, k + 1) \rangle$$

If $n = 2 k + 1$ is odd, $$D_n = \langle (12 \cdots n), (1 n)(2, n - 1) \cdots (k, k + 2) \rangle$$

In terms of generators and relations, $$D_n = \langle x, y \text{ } | \text{ } x^n = 1, y^2 = 1, y x y = x^{-1}\rangle$$

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  • $\begingroup$ This is kinda of what I think I am looking for. Is there a way to write this without the $\dots$? $\endgroup$ – John Doe Oct 13 '17 at 17:24
  • $\begingroup$ @JohnDoe Would you prefer product notation (eg. $\prod_{k = 1}^n$)? $\endgroup$ – Qudit Oct 13 '17 at 17:25
  • $\begingroup$ Is there a way to have it look like $D_n = \{\sigma\in S_n: $ something $\}$? $\endgroup$ – John Doe Oct 13 '17 at 17:26
  • $\begingroup$ Everyone in this thread has given you an answer of the form $D_n = \{ \sigma \in S_n : \text{.....}\}$. $\endgroup$ – Randall Oct 13 '17 at 17:27
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    $\begingroup$ @Randall Not exactly in that form. Perhaps the question is whether there is a simple predicate that can say "this element of $S_n$ is an element of this particular dihedral subgroup." $\endgroup$ – Kyle Miller Oct 13 '17 at 17:29
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Here is an answer in the form you originally requested.

$$D_n =\{\sigma \in S_n|\quad\sigma = (1,2,3,...n)^i \quad or\quad (1,n)(2,n-1)(3,n-2)...(1,2,3,...n)^i\}$$

It should be clear that:
1. We only gets $2n$ distinct $\sigma$s because $(1,2,3,...n)^i=(1,2,3,...n)^j$ iff $i=j$ mod $n$
2. The last transposition in the sequence $(1,n)(2,n-1)(3,n-2)...$ is either $$\big(\frac{n}{2},\frac{n}{2}+1\big)$$ if $n$ is even or $$\big(\frac{n-1}{2},\frac{n+3}{2}\big)$$ if $n$ is odd.

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Yes. Label the vertices of your regular $n$-gon with $1, 2, 3, \ldots, n$. Keep track of where the vertices go after the rotation or reflection is done. The result is a literal permutation on $\{1, 2, 3, \ldots, n\}$ and the operation is still composition, so $D_n$ is a subgroup of $S_n$.

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  • $\begingroup$ if you wish to contact me, I can send you a handout I use in my intro group theory class that exactly does all of this from the ground up. In the end, you end up creating the Cayley table for $D_4$ (for me, of order 8). $\endgroup$ – Randall Oct 13 '17 at 17:15
  • $\begingroup$ Sorry I should have been more clear. I wasn't looking for this algorithmic way of defining the groups. $\endgroup$ – John Doe Oct 13 '17 at 17:24
  • $\begingroup$ Then I do not understand the question at all. $\endgroup$ – Randall Oct 13 '17 at 17:25

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