MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\newcommand{\Z}{\mathbf{Z}}$ Given a nice infinite collection of groups, for example the symmetric groups, one can ask whether any finite group is a subgroup of one of them. Of course any finite group acts on itself, so any finite group is a subgroup of a symmetric group. Similarly any finite group acts linearly on its group ring over a finite field, so given a field $k$, any finite group embeds into $GL(n,k)$ for some sufficiently large $n$ (as permutation matrices).

Question 1: What if we do what I ask in the title, and consider the groups $SL(2,R)$ as $R$ ranges over all commutative rings. Given an arbitrary finite group, can I find a commutative ring $R$ (with a 1) such that this group is a subgroup of $SL(2,R)$?

Of course this is inspired by this pesky question which, at the time of typing, seems to remain unsolved.

Here is a more specific question:

Question 2: Is there a commutative ring $R$ (with a 1) such that the symmetric group $S_4$ injects into $SL(2,R)$?

I haven't thought much about question 1 at all. I'll tell you what I know about question 2. Let's consider first the case where $R$ is an algebraically closed field. If the characteristic is zero, or greater than 3, then by character theory any map from $S_4$ into $SL(2,R)$ must contain $A_4$ in its kernel (the map must give a semisimple representation and the irreducible 2-dimensional one has non-trivial determinant).

If the characteristic is 3 then considering the restriction of a map $S_4\to SL(2,R)$ to a Sylow 2-subgroup we see again by character theory that the kernel must contain the central element. But the kernel is a normal subgroup of $S_4$ so it must contain $V_4$ and hence factors through a map $S_3\to SL(2,R)$. Now the image of an element of order 2 must be central and it's not hard to deduce that the 3-cycles must again be in the kernel.

In characteristic 2 there are more possibilities. If I got it right, the kernel of a map $S_4\to SL(2,R)$ ($R$ alg closed char 2) is either $S_4$, $A_4$ or $V_4$ and of course the representation can be non-semisimple this time.

We conclude from this case-by-case analysis that if $R$ is any ring and $S_4\to SL(2,R)$ is any map then the image of $V_4$ is in $1+M_2(J)$, where $J$ is the intersection of all the prime ideals, that is, the nilpotent elements of $R$.

I now wanted to consider the case $J^2=0$ and check that $V_4$ must be killed mod $J^2$ and then go by induction, but I couldn't bash it out and wonder whether it's true.

It's clear that one could brute-force the argument if one could do a Groebner basis calculation over the integers. I have tried one of these in my life---when trying to solve the open problem of whether every finite flat group scheme of order 4 was killed by 4. That latter question seems to be beyond current computers, but perhaps the one I'm raising here might not be. The problem would be that one has to work over $\Z$ and this slows things down greatly.

I then looked for counterexamples, but convinced myself that $S_4$ was not a subgroup of either $SL(2,\Z/4\Z)$ or $SL(2,\Z/2\Z[\epsilon])$ with $\epsilon^2=0$ [edit: I am wrong; $SL(2,\Z/2\Z[\epsilon])$ does work, as pointed out by Tim Dokchitser]. I don't know how to get a computer algebra system to check $SL(2,\Z/2\Z[\epsilon,\delta])$ so I gave up and asked here.

I suspect I am missing some standard fact :-/

share|cite|improve this question
Kevin, if you want to check more examples with computer algebra I would really like it if you checked SL_2(Z/6Z). – Qiaochu Yuan Nov 29 '10 at 19:39
By CRT this is just $SL(2,2)\times SL(2,3)$ which is $S_3\times\tilde{S}_4$. Now $\tilde{S}_4$ is (by definition) the non-split central extension of $S_4$ by $C_2$ and contains no $S_4$, so $SL(2,6)$ can't contain an $S_4$ either (indeed any map from $S_4$ to $SL(2,6)$ will contain $V_4$ in the kernel, this being the minimal normal subgroup). – Kevin Buzzard Nov 29 '10 at 19:53
This book might be useful:… In particular, try to compute H[S_4] (in the book's notation). Any SL(2) rep. will have to factor through this quotient algebra. If S_4 doesn't inject in H[S_4], then the answer to question 2 is no. – Ian Agol Nov 29 '10 at 22:44
"the open problem of whether every finite flat group scheme of order 4 was killed by 4" Holy crow, this isn't known? – JSE Nov 30 '10 at 1:25
@JSE: any commutative group scheme of order $n$ is killed by $n$: the "old-fashioned" proof that any commutative group of order $n$ is killed by $n$ (multiply all the elements together, call the result $x$, and note that $gx=x$) generalises very nicely. But there are non-commutative group schemes of order 4. The issue of course is a base with 2 locally nilpotent. If no conceptual proof is known one can try writing down the universal group scheme of order 4 and checking it on this, but the bottom line is that this "non-conceptual" approach involves writing down about 30 generators for an... – Kevin Buzzard Nov 30 '10 at 7:10
up vote 16 down vote accepted

For Question 2, The central extension $\tilde{S}_4$ is certainly a subgroup of $\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]) \subset \mathrm{GL}_2(\mathbf{C})$. The image of the determinant is $\pm 1$. The image of $\tilde{S}_4$ in $$\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]/2) = \mathrm{GL}_2(\mathbf{F}_2[x]/x^2)$$ is $S_4$, and all the elements have determinant one. It's easy to see that the central element $$\left( \begin{matrix} -1 & 0 \\\ 0 & -1 \end{matrix} \right)$$ lies in the kernel, so it suffices to note that nothing else does. Yet it's obvious that the map surjects onto $\mathrm{GL}_2(\mathbf{F}_2) = S_3$, and (from the character table) the image is larger than $S_3$, so the image is $S_4$.

For Question 1, if $G$ injects into $\mathrm{SL}_2(R)$ for some $R$ then it injects into such a ring where $R$ is Artinian. Here is the proof.

EDIT: Step 0. (This was in my head, but I forgot to mention it, as Kevin reminds me in the comments). One may replace $R$ by the subring generated by the images of the entries of $g-1$ for all $g \in G$, and hence assume that $R$ is finitely generated over $\mathbf{Z}$ and hence Noetherian. (The Krull intersection thm requires a Noetherian hypothesis.)

Step 1. If $x$ is a non-zero element of $R$, then there exists a maximal ideal $\mathfrak{m}$ of $R$ such that $x$ is non-zero in $R/\mathfrak{m}^k$ for some $k$. Proof: Let $\mathfrak{m}$ be some maximal ideal containing the annihilator of $x$. Then $x$ is non-zero in the localization $R_{\mathfrak{m}}$, and thus $x$ is non-zero in $R/{\mathfrak{m}^k}$ by the Krull intersection theorem.

Step 2. If $x_1, \ldots, x_n$ are non-zero elements of $R$, there exists an ideal $I$ such that each $x_i$ is non-zero in $R/I$ and $R/I$ is Artinian. Proof: Apply Step 1 to each $x_i$, and let $I = \bigcap \mathfrak{m}^{k_i}_i$.

Step 3. Suppose that $G$ has $n$ non-trivial elements. Let $x_1, \ldots, x_n$ denote a non-zero entry in the matrix $g - 1$ for each element of $g$. Apply Step 2 to deduce that $g$ is not the identity in $R/I$ for some Artinian quotient for all non-zero $g \in G$.

Remark: If $G$ is simple, then $G$ is actually a subgroup of $\mathrm{SL}(k)$ for some field $k$. Proof: Artinian rings are semi-local, so $G$ is a subgroup of $\bigoplus_{i=1}^{n} \mathrm{SL}(A_i)$ for Artinian rings $A_i$. Since $G$ is simple, it must be a subgroup of $\mathrm{SL}(A)$ for one such $A$. This latter group is filtered by the groups $\mathrm{SL}(k)$ and copies of $M_0(k)$ (trace zero matrices). Since the latter is abelian and $G$ is simple, we are done.

It's easy to find examples of groups which are not subgroups of $\mathrm{SL}_n(k)$ for all fields $k$ and some fixed integer $n$.

share|cite|improve this answer
This seems to deal with everything and in a way I understand too. Thanks! Thanks also to the other two answerers. To silence dogood: in step 1 you might want to put "wlog R is Noetherian" as the first line (before you apply Krull). – Kevin Buzzard Nov 30 '10 at 10:34
In fact Step 1 itself is false without a Noetherian hypothesis: take for example the integers of Q_p-bar and set $x=p$. [Of course this isn't a problem because in Step 0 you replace $R$ by the subring of $R$ generated by the components of the image of $G$] – Kevin Buzzard Nov 30 '10 at 11:01
So this does prove that $PSL_2({\mathbb F}_7)$ is not a subgroup of $SL_2(R)$ for any $R$, does it not? (This group has no 2-dim representations over any field.) This is what Jack Schmidt said he'd expect in his answer, and something I wondered about as well. – Tim Dokchitser Nov 30 '10 at 15:28
Yes it does. Assuming R is local Artinian (i.e. taking Ai), its proper ideals are nilpotent, and so the congruence subgroup is nilpotent ( G(I) = GL(n,R,I) satisfies [ G(I), G(J) ] ≤ G(IJ), and I^n = 0 ), and so the kernel of the map from G into SL(2,R) to SL(2,R/I) is a normal nilpotent group, contained in the Fitting subgroup. I believe this shows "R" versus "product of fields" can only fix nilpotent normal subgroups. The abelian filtration is clearer, but I think only shows you cannot fix insoluble normal subgroups. – Jack Schmidt Nov 30 '10 at 16:06

The answer to question #2 is yes, by Tim Dokchitser.

$$ R = \mathbb{Z}[e]/(2,ee), (1,2,3,4) = \begin{pmatrix} 1+e & 1+e \\ e & 1 \end{pmatrix}, (1,2) = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} $$

His answer can be verified with GAP using the matrix representation of the algebra. Since Kevin asked about using more complicated algebras, I thought it might be useful to mention how easy it is to use matrix representations.

# Define a nilpotent element
e := [[0,1],[0,0]];
o := e^0;

# Define the structure of the matrices
a := [[1,1],[0,1]];
b := [[0,1],[1,1]];
c := [[1,0],[1,1]];

# Now define the generators themselves over k[e]
f := KroneckerProduct( a, o ) + KroneckerProduct( b, e );
t := KroneckerProduct( c, o );

# Now construct and identify the group:
g := Group( [f,t] * One( GF(2) ) );
IdGroup( g ) = IdGroup( SymmetricGroup(4) );
iso := IsomorphismGroups( SymmetricGroup(4), g );
Display( Image( iso, (2,3) ) );
Display( Image( iso, (3,4) ) );

If you prefer the Coxeter generators they are:

$$ (1,2) = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, (2,3) = \begin{pmatrix} e & 1 \\ 1 & e \end{pmatrix}, (3,4) = \begin{pmatrix} 1+e & 0 \\ 1 & 1+e \end{pmatrix} $$

(There are also sparser representations:

$$ (1,2) = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, (2,3) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, (3,4) = \begin{pmatrix} 1+e & 0 \\ 1 & 1+e \end{pmatrix} $$

that might also be useful.)

I think this sort of thing should only have limited utility. S4/K4 = SL(2,2) is a pretty special situation. I would be more surprised if PSL(2,7) could be embedded in a SL(2,R).

share|cite|improve this answer
As an example of how special it is: GL(2,Z[x,y]/(2,xx,yy)) is just a downward extension of GL(2,2) by a 2-group of nilpotency class 2. This sort of thing strikes me as unable to fix insoluble (maybe even non-nilpotent) problems. In the 2-adics, the congruence subgroups are something like nilpotent (they are nilpotent in the finite quotients). It seems likely the same is true in any I-adics, where I is nilpotent. – Jack Schmidt Nov 30 '10 at 0:20
@Jack: if this is right, then I'm quite cross with myself, because I had tried this ring already. Let $k$ be the field with two elements, let $R$ be $k[e]/e^2$, and let $G$ be $SL(2,R)$. Now $G$ has non-trivial centre (because $diag(1+e)$ is in it) and order 48, if I got it right, and I had assumed that $G$ was $\tilde{S}_4$, which has no subgroup isomorphic to $S_4$. Is this latter assumption my mistake? – Kevin Buzzard Nov 30 '10 at 7:17
Jack: I now believe you. I had thought that this ring didn't work, and couldn't be bothered to do the Groebner basis calculation over Z/2Z thinking it might not work. Now you've pointed out that it works, I did the calculation and I believe you. Thanks! – Kevin Buzzard Nov 30 '10 at 10:54
@Kevin: no problem. By the way, verifying matrices form the right group can be pretty easy. First you check sanity by evaluating the defining relations (and getting the identity matrix each time); I gave the Coxeter generators since some people prefer checking those easy relations. Then you check faithfulness by testing an element of the minimal normal subgroup, like (1,2)(3,4) for the Coxeter generators or (1,3)(2,4) for the 4-cycle generator; you should not get the identity. – Jack Schmidt Nov 30 '10 at 16:37

About question 1. The answer is "no". Suppose that every finite group $G$ embeds into $SL_2(R_G)$ for a commutative $R_G$. Take the ultraproduct ${\mathcal G}$ of all $G$. It embeds into the ultraproduct of $SL_2(R_G)$ which is $SL_2(R)$ where $R$ is the ultraproduct of $R_G$. Since $R$ is a commutative ring, every finitely generated subgroups of $SL_2(R)$ is residually finite by Malcev. But not all finitely generated subgroups of ${\mathcal G}$ are residually finite. See this paper for example. In case it is not clear how "approximably finite" groups relate to ultraproducts of finite groups, here is a corollary of Theorem 2 from Section 8, Chapter 4 of Malcev's "Algebraic systems": if every finite subset of a group $G$ (with induced partial operation) is embedded into a finite group, then $G$ is embedded into an ultraproduct of finite groups. (That is actually an easy statement.) Hence "approximably finite" groups are subgroups of ultraproducts of finite groups. The article I gave a link to contains examples of "approximably finite" but non-residually finite groups.

share|cite|improve this answer
Pardon my ignorance, why are not all finitely generated subgroups residually finite here? – Jonathan Kiehlmann Nov 30 '10 at 1:30
I added a reference. – Mark Sapir Nov 30 '10 at 1:44

Over a field, the answer to the first question is no. An element of $A$ of order 4 in SL(2,R) must satisfy the equation $x^{4}-1$. But it also satisfies the characteristic polynomial $x^{2}-{\rm Tr}(A) x+1$. Taking the gcd of these two polynomials, we quickly see that $A$ also satisfies $({\rm Tr}(A)^{3}-2{\rm Tr}(A)x+{\rm Tr}(A)^{2}$. If the coefficient of $x$ is nonzero then $A$ satisfies a linear polynomial, so is diagonal, thus has 4th roots of unity along the diagonal. There are only finitely many such matrices. If the coefficient of that linear polynomial is zero then ${\rm Tr}(A)=0$ and so $A$ is a root of $x^{2}+1$, which means any two elements of order 4 have equal squares (except for the finitely many possibilities expressed in the previous case).

As for the original question: $S_{4}$ has a presentation $\langle s,t| s^{2}=t^{3}=(st)^{4}=1\rangle$. I would suggest looking at the ring $R=\mathbb{Z}[a,b,c,d,e,f,g,h]/I$ where $I$ is the set of relations forcing the 2x2 matrices give by s=((a,b),(c,d)) and t=((e,f),(g,h)) to satisfy the relations for $S_4$ and have determinant 1. Finding a Grobner-type basis for $I$ should demonstrate that the resulting structure has at least 24 distinct matrices in the group generated by s and t (or it doesn't and you cannot embed $S_{4}$ in such a ring).

share|cite|improve this answer
@Pace: I did this Groebner basis calculation but with $\mathbf{Z}$ replaced by the field with two elements, and it comes out: one can check that $(st)^2$ is not congruent to the identity in $SL(2,R/I)$. – Kevin Buzzard Nov 30 '10 at 18:00

Edit: I think the answer to Question 2 is yes. Let $R = \mathbb{Z}[x, y, z]/I$ where

$$I = (2x, 2y, 2z, z - xz^2, x - x^2 z, z - yz^2, y - y^2 z).$$

Then I think the embedding

$$(12) \mapsto \left[ \begin{array}{cc} 1 & x \\\ 0 & 1 \end{array} \right], (23) \mapsto \left[ \begin{array}{cc} 1 & 0 \\\ z & 1 \end{array} \right], (34) \mapsto \left[ \begin{array}{cc} 1 & y \\\ 0 & 1 \end{array} \right]$$

works. I have not yet verified that $I$ is as small as I think it is, though. I think would be enough to verify that $I$ does not contain $xz, yz$, or $x - y$.

Here's what I know about Question 1 from tinkering with the linked question for awhile.

  • If $R$ is an algebraically closed field of characteristic zero then the classification is the same as the classification of the finite subgroups of $\text{SL}_2(\mathbb{C})$: the cyclic groups, the dicyclic groups, and the binary polyhedral groups. Hence if $R$ is an integral domain of characteristic zero then any finite subgroup must be on this list.

  • If $R$ is an integral domain of characteristic $2$ then $\text{SL}_2(R)$ has no nontrivial elements of order $2$. This rules out any finite group of even order. Whoops: what I meant to say is that $\text{SL}_2(R)$ has no nontrivial elements of order $4$. Actually it's enough that $R$ is reduced. This rules out any finite group with such an element. Slightly more generally, if $2 = 0$ and $I$ is the ideal of $R$ consisting of elements squaring to zero, then an element of $\text{SL}_2(R)$ has order $4$ if and only if its trace lands in $I$, so there are no elements of order $4$ in $\text{SL}_2(R/I)$. If $G$ is a non-abelian simple group in $\text{SL}_2(R)$ with an element of order $4$, then its image in $\text{SL}_2(R/I)$ is not faithful, so must be trivial, but that means every element of $G$ has trace landing in $I$, hence order $4$; contradiction.

  • If $2$ is not a zero divisor in $R$ (in particular, if $2 \neq 0$ and $R$ is an integral domain) then any element of order $2$ in $\text{SL}_2(R)$ is a scalar multiple of the identity, in particular central. This rules out many finite groups, including non-abelian simple groups by Feit-Thompson.

  • In general, let $I$ be the ideal of $R$ consisting of the elements annihilated by $2$. The above bullet point shows that any element of order $2$ in the quotient $\text{SL}_2(R/I)$ must be a scalar multiple of the identity, in particular central, in this quotient. Hence if $G$ is a non-abelian simple subgroup of $\text{SL}_2(R)$, it must land in the kernel of the above map: the subgroup congruent to the identity $\bmod I$.

I strongly suspect that $\text{SL}_2(R)$ can't contain a non-abelian simple group in general but have not yet been able to prove it. It just seems as if there is not enough room to be "too noncommutative." For example, the first three cases above all rule out $\text{SL}_3(\mathbb{F}_2) \sim \text{PSL}_2(\mathbb{F}_7)$.

share|cite|improve this answer
Qiaochu: if $R$ is an integral domain of char 2 then $SL(2,R)$ has a gazillion elements of order 2! And $SL(2,k)$ is a non-abelian simple group for all finite fields of characteristic 2 other than the one with two elements. – Kevin Buzzard Nov 29 '10 at 19:44
Whoops. Let me figure out what I meant to say... – Qiaochu Yuan Nov 29 '10 at 19:51
Qiaochu: in fact $x-y$ is in your ideal, according to magma (even without going mod 2) so your map, if well-defined, is not injective. You have somehow hit the nail on the head: if you want to argue like this then you can try to prove the result via a big computer algebra calculation. You simply write down some generators for $S_4$, send them to some matrices with polynomial variables as coefficients, and then interpret the relations as relations amongst the poly vars and look at the ideal they generate. If your computer can handle the algebra you can see if $V_4$ is in the kernel. – Kevin Buzzard Nov 29 '10 at 22:22
Hmm, google doesn't give me anything for Wittgenstein's Newspapers so I wonder if I've misremembered it. It was something your (Tim) boss Martin Hyland once told me---Wittgenstein observed that if you see a startling piece of news on the front page of the Times, you would be unwise to attempt to verify it by buying another copy of the Times. – Kevin Buzzard Nov 29 '10 at 22:49
@Kevin: Are you sure that $R={\mathbb F}_2[e]/e^2$ does not work? I am getting that $(1234)\mapsto\begin{pmatrix}1&1+e\cr e&1+e\end{pmatrix}, (12)\mapsto \begin{pmatrix}1&0\cr 1&1\end{pmatrix}$ is an injection. – Tim Dokchitser Nov 29 '10 at 23:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.