Order of the group $GL_3(\mathbb{Z_4})$

Question

I want to compute the order of the group $GL_3(\mathbb{Z_4})$ of invertible 3 by 3 matrices with coefficients in $\mathbb Z_4$.

Since this is not a field I don't think the answer here applies.

Answer 1

We can generalize the argument used in this math.SE answer of mine for $GL_2$, as follows. The idea is to consider the quotient map to $GL_3(\mathbb{F}_2)$ and then compute the size of its kernel and image. For the rest of this answer all rings are commutative.

Proposition: Let $R$ be a local ring and $m$ its unique maximal ideal. Then the quotient map $f : GL_n(R) \to GL_n(R/m)$ is surjective.

Here $R = \mathbb{Z}_4$ with maximal ideal $m = (2)$.

Proof. Let $X \in GL_n(R/m)$ and consider any lift of it to a matrix $\widetilde{X} \in M_n(R)$. Its determinant $\det(\widetilde{X}) \in R$ must map to a unit in $R/m$ (since it must be equal to $\det(X) \in (R/m)^{\times}$), and so does not lie in the maximal ideal $m$; by locality it must itself be a unit in $R$, so $\widetilde{X} \in GL_n(R)$ as desired. $\Box$

(In the case of the quotient map $\mathbb{Z}_4 \to \mathbb{Z}_2$ things are very clear and we don't need this level of generality; I'm recording this argument because I'm interested in seeing what can be said in general.)

This implies that if $R$ is finite then $|GL_n(R)|$ is $|GL_n(R/m)|$ times the size of the kernel of the quotient map $f$ above, so we turn to this next.

Proposition: Let $R$ be an Artinian local ring and $m$ its unique maximal ideal. Then the kernel of the quotient map $f : GL_n(R) \to GL_n(R/m)$ is the set $1 + M_n(m)$ of matrices in $M_n(R)$ congruent to $1 \bmod m$.

Proof. The kernel by definition consists of matrices of the form $1 + M_n(m)$ which are also invertible, so we just need to show that all of them are. But since $m$ is nilpotent any matrix of the form $M_n(m)$ is nilpotent so this is clear. $\Box$

Corollary: Let $R$ be a finite local ring with maximal ideal $m$. Then $|GL_n(R)| = |GL_n(R/m)| |m|^{n^2}.$

This gives

$$|GL_3(\mathbb{Z}_4)| = |GL_3(\mathbb{Z}_2)| \cdot 2^4 = (8 - 1)(8 - 2)(8 - 4) \cdot 2^4 = \boxed{2688} = 2^7 \cdot 3 \cdot 7.$$

Every finite ring is a product of finite local rings so this allows us to compute $|GL_n(R)|$ for $R$ any finite ring.