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.

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

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    $\begingroup$ Oh, man, you're good! :) (Seriously). $\endgroup$ – Wlod AA Sep 15 '19 at 23:16
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    $\begingroup$ 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. $\endgroup$ – 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! :)

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    $\begingroup$ love how different this answer is $\endgroup$ – Elliot G Sep 15 '19 at 23:33
  • $\begingroup$ 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. $\endgroup$ – Benjamin Sep 15 '19 at 23:48
  • $\begingroup$ An element of $\ x\in\mathbb Z/4\ $ is invertible iff $\ x=\pm 1,\ $ and there are no other odd elements. $\endgroup$ – Wlod AA Sep 15 '19 at 23:53
  • $\begingroup$ 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. (?) $\endgroup$ – 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.


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