# 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.

• Do you mean the general statement or just for this case? – Elliot G Sep 15 '19 at 22:35
• Either is fine. – Benjamin Sep 15 '19 at 23:42

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.

• Oh, man, you're good! :) (Seriously). – Wlod AA Sep 15 '19 at 23:16
• Regarding whether $\colon \operatorname{Gl}_n(R)\to \operatorname{Gl}_n(R/m)$ is surjective. Call the maps $q\colon R\to R/m$ and $q_*$ for the matrices. Take a candidate $A\in M_n(R)$. It's easy to see that $q\det(A)=\det(q_*A)$, so $x+m$ is invertible in $R/m$. There exists $y\in R$ s.t. $xy-1\in m$, so if $R$ is local, $xy$,is a unit, and thus $x$ is too. – Elliot G Sep 16 '19 at 0:12

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! :)

• love how different this answer is – Elliot G Sep 15 '19 at 23:33
• Why is that "order is the number of the 2×2-matrices which have an odd determinant"? I know that in $\mathbb{Z}_4$, 2 does not have an inverse. But I don't know how to proceed from there. – Benjamin Sep 15 '19 at 23:48
• An element of $\ x\in\mathbb Z/4\$ is invertible iff $\ x=\pm 1,\$ and there are no other odd elements. – Wlod AA Sep 15 '19 at 23:53
• I assumed that the general textbook/wiki material is allowed. I may add something like this in my own words :) about the general linear group, especially in the given case -- a commutative group like $\ \mathbb Z/4,\$ if that's what you expect me to do. (?) – Wlod AA Sep 16 '19 at 0:09

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.