Does $S_4$ inject into $SL(2,R)$ for some commutative ring $R$?

Question

$\newcommand{\Z}{\mathbf{Z}}$ 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 :-/

Answer 1

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 $n$.

Answer 2

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).

Answer 3

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.

Answer 4

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).

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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).

Answer 5

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 $x - y$.

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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 $\text{SL}_3(\mathbb{F}_2) \sim \text{PSL}_2(\mathbb{F}_7)$.