Given a ring $R$, let $\text{GL}(n, R)$ be the group of invertible $n \times n$ matrices with entries in $R$. I know the easy counting argument that shows that $\text{GL}(n, \mathbb{F}_q)$ has $(q^n - 1)(q^n - q) \cdots (q^n - q^{n-1})$ elements, where $\mathbb{F}_q$ is a finite field with $q$ elements.
However, I have no idea how to find a formula for the cardinality of $\text{GL}(n, \mathbb{Z}/m\mathbb{Z})$ for a positive integer $m$. Can somebody give me an hint or even a reference? Thanks!
On
Write $m=\prod_{i=1}^s p_i^{r_i}$ with $p_i$'s distinct prime numbers. Then $\text{GL}(n, \mathbb{Z} / m\mathbb{Z})$ is simply the automorphism group of the additive group $\mathbb{Z}_m^n\simeq\oplus_{i=1}^s\mathbb{Z}_{p_i^{r_i}}^n$. First notice that all automorphisms of $\oplus_{i=1}^s\mathbb{Z}_{p_i^{r_i}}^n$ fix every $\mathbb{Z}_{p_i^{r_i}}^n$. Thus we get $\text{GL}(n, \mathbb{Z} / m\mathbb{Z})\simeq\text{Aut}(\mathbb{Z}_m^n)\simeq\text{Aut}(\oplus_{i=1}^s\mathbb{Z}_{p_i^{r_i}}^n)\simeq\oplus_{i=1}^s\text{Aut}(\mathbb{Z}_{p_i^{r_i}}^n)$. Thus the question is reduced to the case $m=p^r$ with a certain prime number $p$.
The calculation the cardinality of $\text{GL}(n, \mathbb{Z} / p^r\mathbb{Z})$ is very similar to the case of $\text{GL}(n, \mathbb{F}_p)$. Let $\{e_1,e_2,...,e_n\}$ be a basis. View $\mathbb Z_{p^r}^n$ as a free $\mathbb Z_{p^r}$-module. For an endomorphisme $\varphi$ over $\mathbb Z_{p^r}^n$ to be a automorphism. $\varphi(e_1)$ can be arbitrary element of order $p^n$, so there are $p^{nr}-p^{n(r-1)}$ many choices; then $\varphi(e_2)$ must be independent with $\varphi(e_1)$, so there $(p^{nr}-p^{n(r-1)})-(p^r-p^{(r-1)})$ many choices, and so on. Thus cardinality equals to $[p^{nr}-p^{n(r-1)}][(p^{nr}-p^{n(r-1)})-(p^r-p^{(r-1)})][(p^{nr}-p^{n(r-1)})-(p^{2r}-p^{2(r-1)})]\cdots[(p^{nr}-p^{n(r-1)})-(p^{(n-1)r}-p^{(n-1)(r-1)})]$.
Here's just a hint (I don't actually know the solution, and have no time to hunt it down, but this is the direction I would take). One needs to count the "bases" of $(\Bbb Z/m\Bbb Z)^n$, subsets for which every element is uniquely a $\Bbb Z/m\Bbb Z$-linear combination.
The fundamental problem is to count the number of elements of $(\Bbb Z/m\Bbb Z)^n$ that span a subgroup of order$~m$ that is a direct factor of the whole group. That depends on some arithmetic property; I think that maybe it is sufficient just to have order$~m$ to be a direct factor (this happens if the $\gcd$ of all coordinates and $m$ together is$~1$). Once this is done, you can quotient by the subgroup and solve the problem of choosing a basis for the quotient recursively. Coming back from the recursion, don't forget that all basis elements for the quotient can independently be lifted to $m$ different representatives in the original group, and that all these give different bases for the original problem (so you get a power of $m$ in the result).