Why is this group of matrices isomorphic to the dihedral group?

I am reading Abstract Algebra, Theory and Applications by Judson and in exercise $13$ chapter $9$, Isomorphisms, I need to prove that the set of matrices $$A=\pmatrix{ \omega & 0 \\ 0 & \omega ^{-1}} \qquad B=\pmatrix{ 0 & 1 \\ 1 & 0}$$ Where $\omega = e^{2\pi i /n}$ form a group isomorphic to $D_n$ (dihedral group). I know how matrices work to reflect and rotate objects in a dihedral group, and I know that $$R=e^{2 \pi i /n} =\pmatrix{ \cos (\frac{2\pi i}{n}) & -\sin (\frac{2\pi i}{n}) \\ \sin (\frac{2\pi i}{n}) & \cos (\frac{2\pi i}{n})}$$ I can also see that $A^k$ will yield all the rotations I need, reaching the identity rotation with $A^n$, and I can see that multiplying by $B$ allows us to reach the reflections. It is clear that $A$ and $R$ are related, and even though it seems very logical, I cannot prove it. How are these two groups isomorphic?

• What description of $D_{n}$ is available to you? Do you have a group presentation for it, or do you understand it purely in terms of physically-performed manipulations of the regular polygon of $n$ vertices inscribed in the unit disc? – Sinister Cutlass Dec 4 '15 at 21:44
• i have a group presentation for it in terms of $R$ and $B$ (matrices in my post) – Guacho Perez Dec 4 '15 at 22:23
• Well, it looks like you ought to make your homomorphism so that it maps $A$ directly onto $R$ and maps $B$ directly onto $B$. – Sinister Cutlass Dec 5 '15 at 0:16
• Hmm, yes, I can map $\omega$ to $R$ but I can't map $A$ to $R$ – Guacho Perez Dec 5 '15 at 5:21

One way to show that the group $\langle A,B\rangle$ is isomorphic to $D_n$ is, to show that it has the presentation $$\langle A,B\mid A^n=B^2=I, AB=BA^{-1}\rangle,$$ which is just the presentation of the dihedral group $D_n$. A short computation shows that this is the case.