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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 $4n.$

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    Thank you a lot for the edits (@JeanMarie) – 2017-01-31
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    If first matrix is a, and second is b, then $$ really seems to be a semi-direct product of cyclic groups of order 4 and n. It is sufficient to see at $a^{-1}ba$. – 2017-01-31
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    I haven't really learnt about semi-direct products yet although I know of them. Is there another way to solve this problem? – 2017-01-31
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    Generalization in (http://mathoverflow.net/q/16026) – 2017-02-01

2 Answers 2

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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

  1. There are a total of $4n$ matrices of these types.
  2. All of them are in the group generated by those two matrices.
  3. They form a group (a subgroup of $GL_2(\Bbb{C})$).

For full credit answer the following:

  1. Where did you use the assumption that $n$ is odd? Was it necessary?
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    thank you very much for your simple set of steps to solve the problem in (but still allowing me to do the problem myself). I understand how to all steps apart from 4. Could you help me on that one please? – 2017-02-02
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    You're most welcome @P-S.D! Good job getting there. The catch is that if $n$ were even, then $-1=W^{n/2}$ would be a power of $W$, and you would overcount. – 2017-02-02
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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 $\ n$.

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    Ok, but too complicated ? – 2017-02-01
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    @user1952009 There is almost surely a simpler solution. But simplicity is not the unique criteria for a proof, especially in higher education. I think that for a mathematics student, learning to use (rich) isomorphisms, learning to use geometrical arguments, is a good practice. Besides, I would be interested by a an alternative "simpler" or not solution (btw, I am not at all specialist in group theory). – 2017-02-01
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    @user1952009 I have simplified my presentation. – 2017-02-01