Recently I attempted the question

Let $G$ be the dihedral group defined as the set of all formal symbols $x^{i}y^{j}$, $i=0,1,\ j=0,1,2,\ldots,n-1$ where $x^{2}=e,\ y^{n}=e,\ xy=y^{-1}x$.


(a) The subgroup $N=\{e,y,y^{2},\ldots,y^{n-1}\}$ is normal in $G$.

(b) $G/N \approx W$, where $W = \{1,-1\}$ is the group under multiplication of the real numbers.

It was simple enough to show the first part, and then the second by defining a mapping $ ϕ: G \rightarrow W $ by $$\phi(x^{i}y^{j})= (-1)^{i} $$ , noting that it is a homomorphism with kernel $N$ so by 1st isomorphism theorem we get the result.

But what I don't think I understand is what this represents geometrically. I know $D_{n}$ can be thought of as the group of transformations of a regular $n$-gon in the plane such that the $n$-gon is invariant (we permute its vertices) i.e. x is a reflection in 1 axis and y is a rotation of $ \frac{2 \pi}{n} $.

The index of $N$ in $G$ is 2 as only cosets of $N$ are $eN=N$ and $xN$. If we were to label the vertices of our $n$-gon anticlockwise, then perform a transformation from one coset, they will still be labelled anticlockwise, but the other group of transformations (with reflection) will cause them to be clockwise.

So my question is is this a correct way of interpreting the result, and is this the only way to view it geometrically?

Also, are there any other particularly nice results from isomorphisms that can be pictured geometrically which are non standard?

Any help would be appreciated.

  • 2
    $\begingroup$ Now that the subgroup of proper rotation is taken as identity in $G/N$, the only non-trivial operation left is reflection. $\endgroup$ – Cave Johnson Jul 30 '16 at 14:38
  • 1
    $\begingroup$ I once demonstrated this to a group of students using regular hexagon made of cardboard. One size was white the other red. Doing a rotation did not change the color of the side visible to the students, but a reflection did change it. But your interpretation (clockwise/counterclockwise) is very good, too. It is related to the handedness of the image of a basis under the transformation. Which in turn ( you probably remember this from linear algebra) is visible in the sign of the determinant of the matrix of the transformation. Do write down those matrices, and check their determinants! $\endgroup$ – Jyrki Lahtonen Jul 30 '16 at 22:16
  • $\begingroup$ Basically the map is $\det:G\to\{\pm1\}$. Place the origin at the center of the polygon to make the symmetries all linear transformations. $\endgroup$ – Jyrki Lahtonen Jul 30 '16 at 22:16
  • $\begingroup$ @JyrkiLahtonen ah yes thank you, that makes much more sense! $\endgroup$ – Isomorphism Jul 31 '16 at 8:18
  • $\begingroup$ @CaveJohnson and yep thanks, i will write/remember that explicitly next time :) $\endgroup$ – Isomorphism Jul 31 '16 at 8:19

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