A couple of recent questions on MO have involved the characters or the orders of specific finite groups of the form $G(\mathbb{Z}/n\mathbb{Z})$ for a familiar algebraic group $G$ defined over $\mathbb{Z}$ (implicitly as a group scheme) especially when $n$ is a prime power: here and here.

It's reasonable to ask in what generality questions of the second kind can be treated uniformly. Here for example the question doesn't seem to require any knowledge of the structure of symplectic matrices, once one knows the orders of the finite symplectic groups over fields. (These have been written down in numerous papers and books.) In particular:

For group schemes $G$ over $\mathbb{Z}$, in what natural generality is there a uniform procedure for computation of the orders of finite groups $G(\mathbb{Z}/p^k\mathbb{Z})$ as $p$ ranges over the prime numbrs?

An old computation having a similar flavor occurs for the multiplicative group scheme (which is reductive but not simple) in the determination of the group of units in the ring $\mathbb{Z}_p$ of $p$-adic integers, written as an inverse limit of finite groups with successive quotients of order $p-1$ or $p$. Here the additive group scheme (viewed as the Lie algebra of the multiplicative group scheme) enters the picture, providing an iterated extension of the group of units of the residue field $\mathbb{F}_p$. See for example the book by Serre A Course in Arithmetic, II, section 3.

As George McNinch points out in his comment on Scott Carnahan's direct computation of the order for finite symplectic groups, a uniform approach is suggested by a paper of Serre, "Exemples de plongements des groupes PSL$_2(\mathbb{F}_p)$ dans des groupes de Lie simples", Invent. Math. 124 (1996), 3.1. Here G is a connected simple algebraic group over an algebraically closed field, essentially treated as a group scheme. Serre refers in turn to Demazure-Gabriel, Groupes Algebriques (1970), II, section 4, no. 3, a treatise influenced by the earlier Demazure-Grothendieck seminar SGA3.

In a Bourbaki talk, Chevalley showed how to view the simple adjoint groups, which he had constructed in a uniform way in 1955, as group schemes over $\mathbb{Z}$, a theme refined further by Kostant, and more recently by Lusztig in J. Amer. Math. Soc. 22 (2009). That seems to be a good setting for the question I've raised, though perhaps one can go further in the direction of reductive groups?

ADDED: The answers and comments are very interesting, but while I think further about them and the literature I should clarify that I'm taking for granted the standard (though nontrivial) formulas for the group orders over finite fields. And while it's natural (at least for Chevalley groups) to start over the ring of integers, there is certainly a passage to local rings implied here. The specific prime stays in the background, since unlike many questions in Lie theory this kind of computation doesn't distinguish "good" and "bad" primes: whatever is done should apply uniformly to all primes. Meanwhile a result in Appendix A.5 of the recent book by Conrad-Gabber-Prasad on pseudo-reductive groups has been pointed out to me. This suggests that groups aren't so essential to my question, but only a well-behaved class of schemes (again assuming that one can already count their points over finite fields).

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    $\begingroup$ Hi Jim, I guess an example for the reason that I mumbled about local rings in my answer is this: s'pose that $G$ is a reductive group scheme over $\mathbf{Z}$, that $x \in G(\mathbf{Z})$, and that $C=C_G(x)$ is the centralizer group scheme . For any prime $p$ for which $C$ is smooth over $\mathbf{Z}_{(p)}$, you can then hope to compute $|C(\mathbf{Z}/p^r\mathbf{Z})|$ basically by knowing the centralizer of $x$ "mod p" -- as indicated in the answers. But of course in general $C$ won't be smooth over $\mathbf{Z}$. $\endgroup$ – George McNinch Feb 12 '12 at 12:48

I think the assumptions needed on a group scheme $G$ over $\mathbf{Z}$ are that $G$ should be smooth, affine, and of finite type over $\mathbf{Z}$. Actually, I'm going to take the point of view that $p$ is fixed and so I'll suppose that $G$ is smooth, affine, and of finite type over the local ring $\mathbf{Z}_{(p)}$.

I'm going to write $\Lambda = \Lambda_r = \mathbf{Z}/p^r\mathbf{Z}$ and $F = \mathbf{F}_p = \Lambda/p\Lambda$.

To give a formula for $|G(\Lambda)|$, you should have a formula for the number of points of the special fiber -- i.e. for $|G(\mathbf{F}_p)|$ (which is OK e.g. if the special fiber is reductive by well-known order formulas as mentioned in Marty's answer).

Let $H$ be an affine group scheme over $\Lambda$ (say, arising by base-change from the group scheme $G$ over $\mathbf{Z}_{(p)}$).

The result in Demazure-Gabriel to which Serre refers in the article "Exemples de plongements..." (as discussed in the question) implies that the kernel $I$ of the natural mapping $H(\Lambda_k) \to H(\Lambda_{r-1})$ identifies naturally with the $F$-vector space


where $\omega_{H/\Lambda}$ is the module of Kahler differentials of $H$ over $\Lambda$.

In particular, $|I| = p^d$ where $d = \dim_F (\omega_{H/\Lambda} \otimes_{\Lambda} F)$.

If $H$ is smooth and of finite type over $\Lambda$ (e.g. if $H = G_{/\Lambda}$ for $G$ smooth over $\mathbf{Z}_{(p)}$) then the natural mapping $H(\Lambda_r) \to H(\Lambda_{r-1})$ is surjective (say, by [SGA1, Corr. III.5.3]) and $\omega_{H/\Lambda}$ is a free $\Lambda$-module of rank $\dim H_{/F}$. And in this case, one gets the formula for $|H(\Lambda)|$ given in Stasinski's answer.

And if $H = G_{/\Lambda}$ then of course $H(\Lambda) = G(\Lambda)$.

Remarks: (1) It seems to me (?) that the result just quoted from Demazure-Gabriel answers the conjecture from the end of Greenberg's 1963 paper mentioned in Stasinski's answer.

(2) Let $k$ be (say) an algebraically closed field, and let $W = W_r$ be the ring of Witt vectors of length $r$ over $k$. Then our smooth group scheme $G$ as before determines a group scheme $H = G_{/W}$ over $W$. By "Greenberg's functor" we can view the group of points $G(W) = H(W)$ as a linear algebraic group over $k$.

If say $r=2$, then there is a strictly exact sequence of linear algebraic groups $$ (*) \quad 0 \to V \to H(W) \to G_{/k} \to 1$$ where $V = $ Lie$(G_{/k})^{[1]}$ is the vector group on which $G_{/k}$ acts via the first Frobenius twist of the adjoint representation Lie$(G_{/k})$.

I mention this because one somehow doesn't "see" the Frobenius twist when the residue field is $\mathbf{F}_p$, but the Frobenius twist is important in general.

Indeed, if say $G_{/k}$ is a simple alegbraic group, the exact sequence $(*)$ is not split -- it corresonds to a non-zero class in $H^2(G_{/k},$Lie$(G_{/k})^{[1]})$. So the Frobenius twist is crucial since most of the time $H^2(G_{/k},$Lie$(G_{/k})) = 0$.

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    $\begingroup$ Dear George, I agree that it seems the results in Demazure-Gabriel imply the conjecture in Greenberg's paper. $\endgroup$ – A Stasinski Feb 11 '12 at 15:30
  • $\begingroup$ @George: This approach strikes me as economical, while being consistent with the older literature. One newer reference which is fairly close to your viewpoint is Proposition A.5.12 in Pseudo-reductive groups by Conrad-Gabber-Prasad (Cambridge Univ. Press, 2010); they integrate some of Oesterle's older ideas. $\endgroup$ – Jim Humphreys Feb 12 '12 at 15:28

Let $G$ be a smooth group scheme of finite type over $\mathbb{Z}/p^{r}$ (one can also do this over any Artinian local ring with perfect residue field, but then some details below change). For example, $G$ could be a reductive group scheme over $\mathbb{Z}/p^{r}$. This also includes Chevalley groups over $\mathbb{Z}$ after extension of scalars to $\mathbb{Z}/p^{r}$. In this setting the Greenberg functor $\mathcal{F}$ transforms $G$ into a smooth group scheme $\mathcal{F}G$ of finite type over $\mathbb{F}_{p}$, such that $G(\mathbb{Z}/p^{r})\cong(\mathcal{F}G)(\mathbb{F}_{p})$. The smoothness of $G$ implies that the natural map \[ G(\mathbb{Z}/p^{r})\longrightarrow G(\mathbb{F}_{p}) \] is surjective. The key is now Greenberg's structure theorem (see the end of Greenberg's paper Schemata over local rings II), which implies that for any $i\geq1$, the kernel of the natural map (which is a surjection, thanks to the smoothness of $G$) \[ G(\mathbb{Z}/p^{i+1})\longrightarrow G(\mathbb{Z}/p^{i}) \] is isomorphic to the $\mathbb{F}_{p}$-points of an algebraic group over $\mathbb{F}_{p}$ whose underlying variety is affine $d$-space where $d=\dim G_{\mathbb{F}_{p}}$. Greenberg conjectured that the kernel is always isomorphic to the vector group $\mathbb{G}_a^d$. If I've understood things correctly, this would imply that $$ |G(\mathbb{Z}/p^{r})|=p^{d(r-1)}|G(\mathbb{F}_{p})|. $$ I do not know about the status of Greenberg's conjecture or the stronger conjecture right at the end of his paper referred to above. It would be very interesting to know if any progress has been made on this since Greenberg's paper appeared.

Edit: It may not be necessary to go via Greenberg's conjecture. It should be enough to show that the $\mathbb{F}_{p}$-rational structure on the kernel (viewed as a connected unipotent group) is sufficiently simple so that the order of the finite kernel is $p^{d}$.

  • $\begingroup$ Good point.. I neglected to say that the kernel of reduction is the easy part -- just powers of $p$ show up like you wrote. $\endgroup$ – Marty Feb 10 '12 at 23:06
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    $\begingroup$ My interpretation of Jim's question was that it was about a general setting for the kernels and reduction maps and that the $\mathbb{F}_p$-points were assumed to be known. $\endgroup$ – A Stasinski Feb 10 '12 at 23:44
  • $\begingroup$ I see. My interpretation of his "uniform...as p ranges over the prime numbers" was that the question was more about variation in p, and not in the power of p. Since both $p$ and $k$ are given as variables in $p^k$, I guess both our answers are appropriate. $\endgroup$ – Marty Feb 11 '12 at 5:41

The places to look are:

  1. Steinberg, "Endomorphisms of linear algebraic groups." Memoir AMS 80, (1968), and
  2. Gross, "The motive of a reductive group" Invent. math. 130, 287 ± 313 (1997).

(I learned about the former from the latter).

To any (quasi-split) group $G$ with maximal torus $T$, there is an associated Artin-Tate motive $M$ (i.e., a Galois representation).

The point-count over finite fields is related to the action of Frobenius on the twisted dual motive: equation (3.1) of Gross's article (citing Steinberg) gives us: $$ \vert G(k) \vert / q^{dim(G)} = \prod_{d \geq 1} det(1 - Fr \cdot q^{-d} : V_d ),$$ where $Fr$ denotes the Frobenius, $q$ the order of the finite field $k$, and $V = \bigoplus V_d$ a graded ${\mathbb Q}$ vector space as follows: $V = R / R^+$ where $R$ is the graded algebra $Sym^\bullet(E)^W$, where $E$ is the character lattice of $T$ tensored with ${\mathbb Q}$ (section 1 of Gross).

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    $\begingroup$ It seems that the definition of $M$ formally (although perhaps not essentially) depends on a fixed ground field (for $G$). If $G$ comes from a group scheme over $\mathbb{Z}$, is there an analog of $M$ such that the various objects $M$ are obtained from the one over $\mathbb{Z}$ by base extension? This would make the uniformity in $q$ clearer. $\endgroup$ – A Stasinski Feb 11 '12 at 14:52

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