A group of $2\times 2$-matrices which is isomorphic to $\left(\mathcal{O}/24\mathcal{O}\right)^{\times}$ where $\mathcal{O}=\mathbb{Z}[\sqrt{-3}]$ Is there a subgroup $G\leq\mathrm{Gl}_2(\mathbb{Z}_n)$ such that $G\cong\left(\mathcal{O}/n\mathcal{O}\right)^{\times}$ where $\mathcal{O}=\mathbb{Z}[\sqrt{-3}]$ and for a ring $R$, $R^{\times}$ denotes the group of units of $R$?
If yes (which seems to be the case, intuitionally), how does one construct it?
Some context: I am particularly looking for a group of $2\times 2$-matrices which is isomorphic to $\left(\mathcal{O}/24\mathcal{O}\right)^{\times}$.
I would very much appreciate it if somebody could point it out to me.
 A: Let $M$ be any $2$-by-$2$ matrix with entries in $\mathbb{Z}_n$ for which $M^2=-3I$, but for which $M$ and $I$ do not have a nonzero common scalar multiple.
Then, there is a map $\left(\mathcal{O}/n\mathcal{O}\right)^{\times} \to \mathrm{Gl}_2(\mathbb{Z}_n)$ that sends the coset of $a+b\sqrt{-3}$ to $aI+bM$. Using the fact that $M^2=-3I$, this map is easily seen to be a well-defined group homomorphism. It is also injective, since if $aI+bM=I$, then $bM=(1-a)I$ is a common scalar multiple of $M$ and $I$, so $a \equiv 1 \pmod n$ (since $1-a \equiv 0 \pmod n$) and $b \equiv 0 \pmod n$, which means that $a+b\sqrt{-3} \equiv 1 \pmod n$. So, the image is a subgroup of $\mathrm{Gl}_2(\mathbb{Z}_n)$ isomorphic to $\left(\mathcal{O}/n\mathcal{O}\right)^{\times}$.
Such an $M$ is easy to construct for any $n$ (namely, consider $\begin{pmatrix}1&1\\-4&-1\end{pmatrix}$).
A: $GL_2(\Bbb{Z}/n\Bbb{Z})=Aut((\Bbb{Z}/n\Bbb{Z})^2)$. Obviously $\Bbb{Z}[\sqrt{-3}]/(n)$ is isomorphic to $(\Bbb{Z}/n\Bbb{Z})^2$ as a group and choosing such an isomorphism multiplication by elements of $\Bbb{Z}[\sqrt{-3}]/(n)^\times$ give automorphisms $\in Aut((\Bbb{Z}/n\Bbb{Z})^2)$.
