Order of $GL(2, \mathbb{Z}_4)$ In this wiki page, it is stated that the order for $GL(2, \mathbb{Z}_4)$ is 96. But I don't understand the explanation it gives. It seems that they are using some sort of formula. I never encountered a formula like that before. Can someone clarify this for me? It would be much appreciated if you can tell me how it is proved.
 A: The said order is the number of the $2\times 2$-matrices
which have an odd determinant. A determinant is odd iff
the diagonal products are of different parity.
A diagonal product can be odd in exactly $\ 4 =2\cdot 2\ $
ways hence it can be even in $\ 4\cdot 4-2\cdot 2\ =\ 12\ $
ways. This gives you the answer: the said order is
$$ 4\cdot 12\ +\ 12\cdot 4\ = 96 $$
Great! :)
A: The starting point is that we know a classic formula for the size of $GL_n(F)$ if $F$ is a finite field: namely, if $|F| = q$ then 
$$|GL_n(F)| = (q^n - 1)(q^n - q) \dots (q^n - q^{n-1}).$$
This can be proven e.g. by considering the rows or columns of $GL_n(F)$ one at a time; see this groupprops page. In particular we have
$$|GL_2(F)| = (q^2 - 1)(q^2 - q).$$
Unfortunately, $\mathbb{Z}_4$ is not a field. It is, however, a finite local ring $R$ with unique maximal ideal $m$ such that $R/m$ is a field, namely $\mathbb{F}_2$. With this setup we can hope to calculate the size of $|GL_n(R)$| by considering the map $GL_n(R) \to GL_n(R/m)$, proving that this map is surjective, calculating the size of $GL_n(R/m)$, then calculating the size of the kernel. 
The kernel of the map $GL_n(R) \to GL_n(R/m)$ consists of invertible matrices of the form $I + mX$ where $X \in M_n(R)$. In our setup the maximal ideal $m$ will always be nilpotent, so every such matrix is invertible. That means we just need to count $|M_n(m)| = |m|^{n^2}$. If we know that the quotient map $GL_n(R) \to GL_n(R/m)$ is surjective, this gives
$$|GL_n(R)| = |GL_n(R/m)| |M_n(m)| = |GL_n(R/m)| |m|^{n^2}.$$
Now for this business involving "length," which has to do with computing the size of $m$. I believe that length refers to the smallest positive integer $\ell$ such that $m^{\ell} = 0$ (this should be equivalent to the length of the composition series $R \supseteq m \supseteq m^2 \supseteq m^3 \dots \supseteq (0)$), and that in sufficiently nice cases $m$ has size $q^{\ell-1}$ (this should be because all of the quotients $m^i/m^{i+1}$ have size $q$). This is at least true in our case, and substituting $n = 2$ above, where $q = |R/m|$, gives
$$|GL_2(R)| = (q^2 - 1)(q^2 - q) q^{4 \ell - 4}$$
which is the formula the article is using. I'm confused about the article's use of the term discrete valuation ring, though; with the standard definition of that term, DVRs are always integral domains. In any case everything is fine if either $R = \mathbb{Z}_{p^{\ell}}$ or $R = \mathbb{F}_p[t]/t^{\ell}$, both of which have length $\ell$ and satisfy $q = p, |m| = q^{\ell-1}$. These rings are both quotients of the form $R/m^{\ell}$ where $R$ is a DVR so the article apparently uses a definition which encompasses this case. 
I also don't know off the top of my head when the map $GL_n(R) \to GL_n(R/m)$ is surjective. For this particular calculation we can apparently replace $\mathbb{Z}_4$ by $\mathbb{F}_2[t]/t^2$, according to that article; in that case this map is clearly surjective because the quotient map $\mathbb{F}_2[t]/t^2 \to \mathbb{F}_2$ has a section. 
A: The general formula for $\operatorname {GL}(n,\Bbb Z_p)$ is $(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1})$.  This follows from a simple argument counting linearly independent columns. 
Here we have $\Bbb Z_4$.  Since $4$ is not prime, we can't use the above formula. 
Apparently a correction  factor involving the length of the ring $\Bbb Z_4$ comes into play. I'm not exactly sure how this factor arises;  though it makes sense because the length roughly measures the size of the ring. 
