I have a map $g : D_{2n} \rightarrow GL_2(\mathbb{R})$ with
$g(r) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}$,
where $\theta = \frac{2 \pi}{n}$, and
$g(s) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$,
and I am asked to extend it to a homomorphism.
Leaving out the messy details, I was able to show that there must exist a unique homomorphism $f : D_{2n} \rightarrow GL_2(\mathbb{R})$ that maps $r$ to $\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}$ and $s$ to $\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$, because the two matrices satisfy the same relations as $s$ and $r$--i.e.., $r^n = s^2 = e$ and $sr = r^{-1}s$.
I guess I am confused about extending $g$ to a homomorphism. What exactly does that mean and how would I do it in this case? I know I can't just conclude that $g=f$?
EDIT: Okay, I took $g(s^i r^j) := f(s^i) f(r^j)$, and this insures $g$ is a homomorphism. Now I am trying to show that it is also injective, but am having difficulty. By way of contradiction, suppose $g(s^i r^j) = g(s^k r^\ell)$, yet $s^i r^j \neq s^k r^\ell$. Then we get
$B^{i-k} A^j = A^\ell$
If $i-k=0$, then
$A^{j-\ell} = I$
This feels like contradiction because I believe $j - \ell < n = |A|$, but I am not sure.
EDIT: I think I figured it out. All I need to do is show that the kernel is trivial. I will flesh out the details some time later.
Well, first you need to really check that both matrices fulfill the very same relations as $\;s,r\;$ in the dihedral group $\;D_{2n}\;$. For this you could work out the following (basic trigonometry needed and induction):
$$A:=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\;,\;\;A^2=\begin{pmatrix}\cos2\theta&-\sin2\theta\\ \sin2\theta&\cos2\theta\end{pmatrix},\ldots,A^n=\begin{pmatrix}\cos n\theta&-\sin n\theta\\ \sin n\theta&\cos n\theta\end{pmatrix}$$
Taking into account that $\;\theta=\cfrac{2\pi}n\;$ shall be enough for you to complete the forementioned checking.
Now, "Extending the homomorphism means" that any element in that dihedral group can be expressed as $\;x=s^\epsilon r^k\;,\;\;\epsilon\in\{0,1\}\;,\;\;k\in\{1,2,...,n\}\;$ , and then you extend
$$g(x):=B^\epsilon A^k\;,\;\;\text{with}\;\;B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$
Finally, $\;f\;$ is just the inverse map to $\;g\;$ , i.e. $\;f=g^{-1}\;$