Generation a subgroup of $GL_2(\mathbb{C})$

Question

Let $G$ be a group and let $S$ be a nonempty subset of a group $G$. The subgroup of $G$ generated by $S$, which is denoted by $\langle S\rangle$, is equal to the set of all elements of $G$ that can be written as products $s_1s_2\ldots s_k$, for some integer $k\geq1$, where each $s_i$ is in $S\cup S^{-1}$. Equivalently, a subgroup generated by $S$ is the intersection of all subgroups which contain $S$.

Clearly, if $S$ consists of one element $S=\{s\}$, then $\langle S\rangle=\langle s\rangle=\{s^k\mid k\in\mathbb{Z}\}$. But what about if $|S|>1$.

Is there any rule for computing the subgroup $H$ generated by $S$ when $S$ is a finite set?

For example, it seems a bit difficult to determine which subgroup of
$GL_2(\mathbb{C})$ is $$ \left< \begin{pmatrix} e^{\frac{\pi i}{2^{n-2}}} & 0 \\ 0 & e^{-\frac{\pi i}{2^{n-2}}} \end{pmatrix} , \begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}\right>. $$ Is there any algorithm or something else that could help?

Answer 1

If you are interested in what elements this subgroup is made up of, it looks like it would be something like this: $$ \left< \begin{pmatrix} e^{\frac{\pi i}{2^{n-2}}} & 0 \\ 0 & e^{-\frac{\pi i}{2^{n-2}}} \end{pmatrix} , \begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}\right> $$ $$= \left\{ \begin{pmatrix} 0 & \pm\alpha^k \\ \pm\bar\alpha^k & 0 \end{pmatrix}, \begin{pmatrix} \pm\alpha^k & 0\\ 0 &\pm\bar\alpha^k \end{pmatrix}\mid k\in\mathbb{Z} \right\}, $$ where $\alpha=e^{\frac{\pi i}{2^{n-2}}}$.

Supplement.

Answer to the question: How did I come to this conclusion?

1. It is checked directly that $$ H=\left<\ \begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}\right>= \left\{ \begin{pmatrix} 0 & \pm1 \\ \pm1 & 0 \end{pmatrix}, \begin{pmatrix} \pm1 & 0\\ 0 &\pm1 \end{pmatrix} \right\}. $$

2. Let $$ F=\left<\ \begin{pmatrix} \alpha & 0 \\0 & \bar\alpha \end{pmatrix} \right>= \left\{ \begin{pmatrix} \alpha^k & 0 \\0 & \bar\alpha^k \end{pmatrix}\mid k\in\mathbb{Z} \right\}. $$

3. It remains to note that $FH=HF$. Here it is important that $\alpha^{-1}=\bar\alpha$ or equivalently $|\alpha|=1$.