Matrix Group generated by Roots of Unity The question is essentially as follows: let $W$ be a primitive $n$-th root of unity where $n$ is an odd integer. Let $G$ be the subgroup (of $GL(n)$) of all 2x2 matrices generated by matrices 
$$\begin{bmatrix}0&-1\\1&0\end{bmatrix} \ \ \text{and} \ \  \begin{bmatrix} W&0\\0&W^{-1}\end{bmatrix}.$$
Prove that $G$ has order $4n.$
 A: A plan: Keep multiplying those matrices together. Observe that you get matrices of the following types ($k$ is an integer parameter)
$$
\left(\begin{array}{cc}W^k&0\\0&W^{-k}\end{array}\right),
$$
$$
\left(\begin{array}{cc}-W^k&0\\0&-W^{-k}\end{array}\right),
$$
$$
\left(\begin{array}{cc}0&-W^k\\W^{-k}&0\end{array}\right),
$$
$$
\left(\begin{array}{cc}0&W^k\\-W^{-k}&0\end{array}\right).
$$
Prove that


*

*There are a total of $4n$ matrices of these types.

*All of them are in the group generated by those two matrices.

*They form a group (a subgroup of $GL_2(\Bbb{C})$).


For full credit answer the following:


*Where did you use the assumption that $n$ is odd? Was it necessary?

A: You probably know the group isomorphism: 
$$\begin{bmatrix}a&b\\c&d\end{bmatrix} \ \leftrightarrow \ \ M(z)=\dfrac{az+b}{cz+d}.$$
between $SL(2,C)$ ("special linear" group of matrices with complex coefficients and determinant $1$) for matrix multiplication and the so-called group of Möbius transformations for functions' composition. 
(see for example this).
In particular, we have:
$$\begin{bmatrix}0&-1\\1& \ \ 0\end{bmatrix} \ \leftrightarrow \ S(z)=\dfrac{0z-1}{1z+0}=-\dfrac{1}{z} \ \ \text{and} \ \ \begin{bmatrix}W&0\\0&W^{-1}\end{bmatrix} \ \leftrightarrow \ R(z)= \dfrac{Wz+0}{0z+W^{-1}}=W^2z$$
It is possible to classify the different possible compositions ($R\circ R \circ S$ etc...) one can do with $R$ and $S$ in two types:
$$z\mapsto  W^{\pm2k}z \ \ \ \ \ \text{or} \ \ \ \ \ z\mapsto  - W^{\pm2k}\dfrac{1}{z}, \ \ k=0,1,...(n-1)$$
(By an immediate recurrence on the number of compositions operations) yielding $2 \times 2n = 4n$ different functions.
One can object that some of these functions could be counted twice, by the fact that $W^{2k}=W^{2k'}$ could occur for some different $k$ and $k' \in [0,n-1]$. This in fact not possible since it would imply $W^{2(k-k')}=1$ which happens if and only if $2(k-k')=0 \ $ modulo $\ n$. As $n$ is odd, it implies that $k-k'=0 \ $ modulo $\ n$.
