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 $x^{2}=e,\ y^{n}=e,\ xy=y^{-1}x$.
Prove
(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.