Order of $GL(n,\mathbb{Z}/m \mathbb{Z})$ If $p$ is a prime, then the order of the general linear group $GL(n,F_p)$ is given by
$$(p^n-1)(p^n-p) \cdots (p^n-p^{n-1}).$$
But what if we consider the general linear group over the rings $\mathbb{Z}/m\mathbb{Z}$ where $m$ is any integer? What will be its order?
 A: If $m = \prod_{i=1}^n p_i^{m_i}$ is the prime factorization of $m$, then by the Chinese remainder theorem, you have a ring isomomorphism
$$
\mathbb{Z}/m\mathbb{Z} \to \prod_{i=1}^n \mathbb{Z}/p_i^{m_i}\mathbb{Z}
$$
This translate to an isomorphism between
$$
M_k(\mathbb{Z}/m\mathbb{Z}) \to \prod_{i=1}^n M_k(\mathbb{Z}/p_i^{m_i}\mathbb{Z})
$$
and hence to an isomorphism
$$
GL(k,\mathbb{Z}/m\mathbb{Z}) \to \prod_{i=1}^n GL(k,\mathbb{Z}/p_i^{m_i}\mathbb{Z}) \qquad (1)
$$
so you only need to worry about the case where $m = p^{\ell}$ for some prime $p$. For this case, you have a natural surjective ring homomorphism
$$
\varphi : \mathbb{Z}/p^{\ell}\mathbb{Z} \to \mathbb{Z}/p^{\ell-1}\mathbb{Z} \text{ given by } a + p^{\ell}\mathbb{Z} \mapsto a + p^{\ell-1}\mathbb{Z} 
$$
which induces a surjection (as above)
$$
\varphi_k : GL(k,\mathbb{Z}/p^{\ell}\mathbb{Z}) \to GL(k,\mathbb{Z}/p^{\ell-1}\mathbb{Z})
$$
because if $\overline{a} := \det(A)$ for some $A \in GL(k,\mathbb{Z}/p^{\ell}\mathbb{Z})$, then you can use the fact that
$$
\overline{a} \in (\mathbb{Z}/p^{\ell}\mathbb{Z})^{\ast} \Rightarrow \varphi(\overline{a}) \in (\mathbb{Z}/p^{\ell}\mathbb{Z})^{\ast}
$$
to conclude that $\varphi_k(A) \in GL(k,\mathbb{Z}/p^{\ell-1}\mathbb{Z})$. Hence,
$$
|GL(k,\mathbb{Z}/p^{\ell}\mathbb{Z})| = |\ker(\varphi_k)||GL(k,\mathbb{Z}/p^{\ell-1}\mathbb{Z})| \qquad (2)
$$
Hence, the problem reduces to computing the order of $\ker(\varphi_k)$. Now $A = (a_{i,j}) \in \ker(\varphi_k)$ if and only if
$$
a_{i,i} \equiv 1 \pmod{p^{\ell-1}} \text{ and } a_{i,j} \equiv 0 \pmod{p^{\ell-1}} \text{ if } i\neq j
$$
For any $i,j$, there are precisely as many choices for $a_{i,j}$ as there are elements in
$$
(\mathbb{Z}/p^{\ell}\mathbb{Z})/(\mathbb{Z}/p^{\ell-1}\mathbb{Z})
$$
ie. There are $p$ choices for each $a_{i,j}$. Hence,
$$
|\ker(\varphi_k)| = p^{k^2} \qquad (3)
$$
So, by (2) and (3),
$$
|GL(k,\mathbb{Z}/p^{\ell}\mathbb{Z})| = p^{(\ell-1)k^2}|GL(k,\mathbb{Z}/p\mathbb{Z})| \qquad (4)
$$
Now (1) and (4) together answer your question.
