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According to Wikipedia the Dihedral group $D_n \cong \; \langle r,s \mid r^n = 1, s^2 = 1, s^{-1}rs = r^{-1}\rangle$. But why does this apply?

As far as I understand the group presentation means that I take $\langle r,s\rangle$ and remove all elements that do not satisfy the provided relations. The Dihedral group $D_n$ consists of $n$ rotations and $n$ reflections. If considering $s$ as reflection, $s^2 = 1$ means that repeating a reflection leads to the identity. But what about the other relations, and why does this apply in general?

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    $\begingroup$ You don't "remove" elements---in a group representation $\langle S | R\rangle$, the relators in $R$ are simply a declaration of how the generators in $S$ are related to one another. $\endgroup$ – Travis Willse Dec 9 '14 at 10:59
  • $\begingroup$ This is more accurately called a presentation of a group. Calling it a representation makes it easy to confuse with the large subject of group representations, which you are apparently not asking about. $\endgroup$ – rschwieb Dec 9 '14 at 11:31
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Well, you can take $r = \left(\begin{array}{clcr} \cos(\frac{2 \pi }{n}) & -\sin(\frac{2 \pi }{n})\\ \sin(\frac{2 \pi }{n}) & \cos(\frac{2 \pi }{n}) \end{array} \right) $ and $s = \left( \begin{array}{clcr} 0 & 1\\ 1 & 0 \end{array} \right) $ to see that $D_{n}$ satisfies the relations in that presentation. This means that $D_{n}$ is a homomorphic image of the presented group. On the other hand, notice that every element of the presented group is expressible in the form $r^{j} s^{i}$ with $ 0 \leq j \leq n$ and $0 \leq i \leq 1.$ Hence the presented group has order $2n,$ so the homomorphism onto $D_{n}$ must be injective, hence an isomorphism.

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  • $\begingroup$ thank you for your answer. I tried to check it with the relations, and it didn't work for me. Please look at this calculation for $s^{-1}rs$, which leads to a strange result. Multiplied with $r$ I don't get the identity matrix. Did I make a mistake? $\endgroup$ – muffel Dec 9 '14 at 20:07
  • $\begingroup$ Oops,no, I did. I wrote down an incorrect matrix for the rotation of order $n$. I will correct now. $\endgroup$ – Geoff Robinson Dec 9 '14 at 20:09
  • $\begingroup$ Thanks! One last question: How do you follow that all elements of the presented group (it is obvious for $D_n$) can be expressed in the form $r^{j}s^{i}$ ? $\endgroup$ – muffel Dec 9 '14 at 20:15
  • $\begingroup$ Hint: Consider $r^{j}s^{i}r^{k}s^{\ell}.$ If $i=0,$ there is nothing to do. If $i=1,$ rewrite as $r^{j}sr^{k}s^{-1}s s^{\ell}$ and use the relations. $\endgroup$ – Geoff Robinson Dec 9 '14 at 20:23
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means that I take $\langle r,s\rangle$ and remove all elements that do not satisfy the provided relations.

As already mentioned, you're not removing elements, but identifying different elements of $\langle r,s\rangle$. The right picture is that you're working with $\langle r,s\rangle/N$ where $N$ is the smallest normal subgroup containing $r^n$, $s^2$ and $rs^{-1}rs$.

It looks an awful lot like you might be reading thus group generator-relation notation as if it were set-builder notation, but they are two different things.

But what about the other relations, and why does this apply in general?

Have you actually played around with a few examples of n gons yet? Thus should make things obvious. The rotation r can be thought of as a rotation (clockwise, say) that moves a vertex onto its neighbor. The relation $r^n=1$ means that after n rotations, the polygon has returned to its starting position.

For the second one, I prefer to think of it as $sr=r^{-1}s$. Just try it out on some regular polygons: if you flip then rotate clockwise one increment, it's the same as rotating counterclockwise by one increment and the flipping.

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  • $\begingroup$ thank you for your answer and your remarks! So how do I get to $N$? What would $N$ be in the current example? As far as I understood, all those relations provided are equal to the neutral element $e$, so $N$ would only contain $e$, which doesn't make any sense, does it? $\endgroup$ – muffel Dec 9 '14 at 20:24
  • $\begingroup$ @muffel the part about "would only contain $e$" is the wrong part. It would be right to say that $r^n\equiv e$ mod $N$, $s^2\equiv e$ mod $N$, and $rs^{-1}rs\equiv e$ mod $N$. The last one could be rewritten as $rs\equiv s^{-1}r$ mod $N$. $\endgroup$ – rschwieb Dec 9 '14 at 21:49
  • $\begingroup$ @muffel Since $N$ is infinite, I'm not sure if it makes much sense to try to write its elements out. I'm just saying that modding out by that normal subgroup gets you the group with these relations. $\endgroup$ – rschwieb Dec 9 '14 at 21:51
  • $\begingroup$ ok, I think I'm starting to understand, thank you for your explanation! $\endgroup$ – muffel Dec 9 '14 at 22:26

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