Is there any non-trivial ring such that $SL_{3}(R)$ is isomorphic to a subgroup of $SL_{2}(R)$?

$SL_{3}(\mathbb{Z})$ is not an amalgam, and has the wrong number of order $2$ elements to be a subgroup of $SL_{2}(\mathbb{Z})$.

Is there any non-trivial ring where this occurs? When can this definitely not occur? I am trying to understand if there is any sort of group-theoretically apparent notion of dimension here.

  • 4
    $\begingroup$ Your "order 2" remark I guess generalises to something like: if $R$ is a commutative ring with $2$ not a zero divisor, and if $x^2=1$ only has a finite number of solutions in $R$ (e.g. if $R$ is an integral domain then both these conditions are satisfied), then $SL(3,R)$ has more elements of order 2 then $SL(2,R)$ so there's no injection. [Hint: use Cayley-Hamilton on a matrix of order 2 to convince yourself it must be scalar, if I got it right; conversely use the fact that $-1$ isn't 1 to build more elements in $SL(3,R)$ than this] $\endgroup$ Nov 26, 2010 at 9:44
  • 3
    $\begingroup$ Alex: If you have any doubts about Cayley-Hamilton for a 2x2 matrix over an arbitrary commutative ring, why not just bash out the algebra! $\endgroup$ Nov 26, 2010 at 10:01
  • 1
    $\begingroup$ Alex: for the general case, consider the fact that you believe it for the polynomial ring $R=Z[A_{11},A_{12},...,A_{nn}]$ (which is an ID) and now apply it to the matrix whose $(i,j)$th entry is $A_{ij}$ and see what this tells you. $\endgroup$ Nov 26, 2010 at 10:08
  • 4
    $\begingroup$ Actually, these comments make a good answer to another question currently kicking around MO: "what is the point of a universal object?". An $n\times n$ matrix over a general ring $S$ is a map from that ring $R$ above to $S$, reducing questions like CH for any ring, to CH for that one universal ring. One notices this when doing the first exercise I suggested: checking CH in the 2x2 case by bashing out the algebra. One has a bunch of equations in $A,B,C,D$ all of which magically work out, but instead of thinking of $A,B,C,D$ as varying through the ring one can think of them as poly variables. $\endgroup$ Nov 26, 2010 at 10:27
  • 1
    $\begingroup$ @BS: One might also need something like: "the map $SL(3,R)\to SL(2,R)$ extends to a map of rings $M(3,R)\to M(2,R)$" for this strategy to work, right? That doesn't sound so good (e.g. I don't think it's true for maps $SL(3,R)\to SL(3,R)$---things like "inverse-transpose" give trouble...)... $\endgroup$ Nov 26, 2010 at 12:04

7 Answers 7


(Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1) Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.

Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:

2) Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. Let $\mathfrak{n}<S$ denote the nilpotent radical. It is easy to see that every finitely generated subgroup of the kernel of $\text{SL}_2(S)\to \text{SL}_2(S/\mathfrak{n})$ is nilpotent. By the facts that $\text{SL}_3(\mathbb{Z})$ is finitely generated without nilpotent quotients we deduce that it is mapped non-trivially to $\text{SL}_2(S/\mathfrak{n})$. Since $\mathfrak{n}$ is the intersection of all prime ideal we deduce that $\text{SL}_3(\mathbb{Z})$ is mapped non-trivially to $\text{SL}_2(S/\mathfrak{p})$ for some prime ideal $\mathfrak{p}<S$. By letting $k$ be the field of fractions of $S/\mathfrak{p}$ we see that it is enough to prove the following statement:

3) Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.

(here I had before an argument I liked, but I had to replace it by a simpler one.)

Here is a nice exercise:

4) Let $k$ be a field. Then for any group homomorphism $\text{H}(\mathbb{Z})\to\text{SL}_2(k)$, where $\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $\text{H}(\mathbb{Z})$ consists of scalar matrices.

Hint: Assume the image of a generator of the center is not a scalar matrix and show that $\text{H}(\mathbb{Z})$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).

Remark: Actually, the image of the center of $\text{H}(\mathbb{Z})$ will be trivial unless $\text{char}(k)=2$.

To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.

  • $\begingroup$ Great answer! I would just note that $\mathrm{SK}_1(n, R)=\mathrm{SL}(n, R)/\mathrm{E}(n, R)$ is not always abelian. In fact, it can have arbitrary large nilpotency degree for a ring of finite Bass—Serre dimension. $\endgroup$ Apr 6, 2016 at 22:43
  • $\begingroup$ Thanks, @Andrei Smolensky, I was careless here and in some other places. I will edit my answer soon. $\endgroup$
    – Uri Bader
    Apr 7, 2016 at 5:43
  • $\begingroup$ Your answer seems to implicitly assume that $R$ is infinite, else your starting claim that $SL_3(R)$ contains $SL_3(\mathbb{Z})$ is false (e.g. if $R$ is finite). $\endgroup$
    – Max Horn
    Apr 7, 2016 at 12:16
  • 1
    $\begingroup$ Oops, I meant to write: "... assume that R has characteristic 0, ..." $\endgroup$
    – Max Horn
    Apr 7, 2016 at 13:16
  • $\begingroup$ $SL(3, R)$ contains an epimorphic image of $SL(3, \mathbb{Z})$, which is still perfect. Concerning the edits: $SK_1(R)$ is not known to be always nilpotent (it most likely is not). $\endgroup$ Apr 7, 2016 at 13:27

(Edit: the arguments below build on observations by several other people in this thread.)

Case: $2 = 0$ in $R$. As has been observed elsewhere, in this case $G = \text{SL}_3(\mathbb{F}_2)$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $4$ and trace $r$. Then $g^4 = 1$ and $g^2 = rg + 1$, hence $g^4 = r^2 g^2 + 1$ and $r^2 g^2 = 0$. This gives $r^4 g^4 = 0$, hence $r^4 = 0$. But $r^2 g^2 = r^3 g + r^2 = 0$, hence $r^4 g = r^3 = 0$, hence $r^3 g = r^2 = 0$, and this is both necessary and sufficient.

The elements of $R$ which square to zero form a nilpotent ideal $I$. It follows that if $G$ embeds into $\text{SL}_2(R)$, the image of any elements of order $4$ in $\text{SL}_2(R/I)$ must, by the above computation, have order dividing $2$. In particular the image of $G$ is not isomorphic to $G$, so it must be trivial since $G$ is simple. Hence the image of $G$ in $\text{SL}_2(R)$ consists only of matrices congruent to the identity $\bmod I$. But any such matrix has trace squaring to zero, hence order dividing $4$, which contradicts the existence of elements of order $7$ in $G$. So no such embedding exists.

Case: $2$ is nilpotent in $R$. It is still true in this case that $SL_3({\mathbb F}_2)$ embeds into $\text{SL}_3(R)$, because the two complex 3-dimensional irreducible representations of $SL_3({\mathbb F}_2)=\text{PSL}_2({\mathbb F}_7)$ are realisable over ${\mathbb Z}_2$ (they need $(1\pm\sqrt{-7})/2$ which are in ${\mathbb Z}_2$), and $R$ is a ${\mathbb Z}_2$-algebra. However, as explained in the follow-up question, $SL_3({\mathbb F}_2)$ is not a subgroup of $\text{SL}_2(R)$ for any $R$.

Case: $2$ is not a zero divisor in $R$. As has been observed elsewhere, in this case $G = S_4$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $2$ and trace $r$. Then $g^2 = 1$ and $g^2 - rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so $r$ is also not a zero divisor. It follows that $g$ must be a scalar multiple of the identity, hence central. But $S_4$ contains elements of order $2$ which are not central. (As does $S_3$. Hence this argument also shows, as in Tim Dokchitser's answer, that $S_3$ does not embed into $\text{SL}_2(R)$ in this case.)

In particular the above arguments cover the case that $2$ is invertible, as well as the case that $R$ is an integral domain. The remaining case is that $2 \neq 0$, and it is a zero divisor but is not nilpotent.

One may hope that the reduction to Artinian rings as in the follow-up question and the fact that we know the answer both when $2$ is invertible and for ${\mathbb Z}_2$-algebras can actually finish this off.

  • $\begingroup$ $SL_2(R)$ always has an element of order 4, rotation by 90 degrees. $\endgroup$ Nov 26, 2010 at 23:43
  • $\begingroup$ @Peter: not in characteristic 2! $\endgroup$ Nov 27, 2010 at 0:04
  • $\begingroup$ Oh, haha, sorry about that... $\endgroup$ Nov 27, 2010 at 0:09
  • $\begingroup$ @Qiaochu: Neat! I changed "order 4" to "order dividing 4", hope it's ok (to make sure that "necessary and sufficient" is clear). Feel free to roll back if you don't like this. $\endgroup$ Nov 28, 2010 at 16:27
  • 1
    $\begingroup$ Perhaps this is useful: the complex 3-dimensional representation of $SL_3({\mathbb Z}/2)=PSL_2(7)$ is realisable over ${\mathbb Z}_2$ (it needs $(1+\sqrt{-7})/2$ which is in ${\mathbb Z}_2$). So this group embeds into $SL_3(R)$ whenever $2$ is nilpotent in $R$, not just when $2=0$. $\endgroup$ Nov 28, 2010 at 17:35

Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1 where 2 is invertible, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.

The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is $\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is

$\gt$ M = {{a, b}, {c, d}}; T = {{e, f}, {g, h}}; id = {{1, 0}, {0, 1}};

$\gt$ GroebnerBasis[{M.M.M - id, T.T - id, M.M.T - T.M, Det[T] - 1, Det[M]-1 }]


In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.

On the other hand, using $$ M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix}, $$ we can embed $S_3$ into $SL(3,R)$ for any ring $R$.

P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!

Edit: The Groebner basis computation works over $Z[1/2]$, so this only works for rings $R$ with 2 invertible. Kevin & John: thank you for pointing this out!

  • $\begingroup$ Tim: I've not had time to think about it properly, but here's an idea which might show $S_3$ can't map into $SL(2,R)$: (1) check true for $R$ a field (char not 2,3 is fine; char 2 and char 3 are special cases but this theory is well-understood---just not by me). (2) Deduce for integral domains. (3) By considering maps induced by $R\to R/P$ deduce some statement of the form "map to SL(2,R) must degenerate mod P for all P and hence (because intersection of all primes is nilradical) must land in nilradical in some way and then (4) reduce to case where nilradical is itself nilpotent and then hope! $\endgroup$ Nov 26, 2010 at 22:00
  • $\begingroup$ Stop! In trying to push through the above strategy I notice that $S_3$ does embed into $SL(2,Z/2Z)$ so something is wrong. Did you do the Groebner basis calc over Q?? $\endgroup$ Nov 26, 2010 at 22:03
  • 1
    $\begingroup$ Either I am not thinking right at all today or something is wrong here - isn't $SL(2,Z/2Z)$ isomorphic with $S_3$? $\endgroup$ Nov 26, 2010 at 22:08
  • $\begingroup$ @John: right! My (admittedly limited) experience with Groebner bases calculations on computers is that they are typically only implemented over fields, which makes me think that Tim's calculation was done over a field, probably the rationals. So probably he has proved that there is no $\mathbf{Q}$-algebra $R$ with the property we seek. $\endgroup$ Nov 26, 2010 at 22:39
  • $\begingroup$ @Kevin&John: Oops, you're absolutely right, it must be that Mathematica does it over $Q$. I'll try to check that. To be continued... $\endgroup$ Nov 26, 2010 at 22:49

Define a sequence of groups $G_i$ and associated group rings $R_i=\mathbb{Q}[G_i]$. To start put $G_0=\mathbb{Q}$. Then define $G_{i+1}=SL_3(R_i)$. The group $SL_3(R_i)$ is a subgroup of $SL_2(R_{i+1})$ because $G_{i+1}$ is a subgroup of $SL_2(R_{i+1})$ (as the group of certain diagonal matrices). Similarly, $G_i$ is a subgroup of $G_{i+1}$, and hence $R_i$ is a subring of $R_{i+1}$. Then $R=\bigcup_i R_i$ is the ring you want.

  • $\begingroup$ I think the OP assumes $R$ to be commutative, but I might be wrong. Any way what do you mean by $SL_n$ of a non-commutative ring ? The only thing I can think of is the subgroup $E_n(R)$ of $GL_n$ generated by "elementary matrices". $\endgroup$
    – BS.
    Nov 26, 2010 at 11:29
  • $\begingroup$ @BS: You are right. There is no good definition of SL_2 for noncommutative rings, and my answer is rubbish. $\endgroup$
    – Boris Bukh
    Nov 26, 2010 at 11:44
  • 1
    $\begingroup$ Presumably there is a good notion of $GL_2$ though? Something like: $R$-linear automorphisms of $R^2$? So perhaps you can give a $GL_2$ example? $\endgroup$ Nov 26, 2010 at 11:49
  • 1
    $\begingroup$ No Kevin $GL_2(R)\cong GL_3 (R)$ for a noncomm ring without IBN. It is a no brainer... $\endgroup$
    – Bugs Bunny
    Nov 26, 2010 at 12:34
  • 1
    $\begingroup$ Yes, IBN may fail: take a ring $R$ generated by coefficients $x_{ij}$ and $y_{ji}$ of generic $mxn$ and $nxm$ matrices $X$ and $Y$ with relations $XY=1$ and $YX=1$. All you need to show is that $R$ is not zero. $\endgroup$
    – Bugs Bunny
    Nov 29, 2010 at 15:46

My guess that it should not be possible because $SL_2(R)$ would not have a 3-soluble nonnilpotent subgroup. I am not sure whether it is true that any soluble non-nilpotent subgroup would lie in Borel subgroup but I imagine that this is right...

  • $\begingroup$ I agree that it should be possible to turn this into a proof. However, even when $R$ is an algebraically closed field, a soluble group is contained in a Borel only up to finite index (e.g. consider $S=N_G(T)$, where $G$ has a solvable Weyl group). $\endgroup$
    – Guntram
    Nov 26, 2010 at 13:06
  • 1
    $\begingroup$ I don't know what 3-soluble means either :-/ but if $R$ is the $p$-adic integers then the congruence subgroup $\Gamma(p)$---doesn't it have the property that if you keep taking commutator subgroups then you get smaller and smaller without ever reaching 1? So I am a bit concerned about the claims in the answer if $R=Z/p^nZ$ for some big $n$... $\endgroup$ Nov 26, 2010 at 13:41
  • $\begingroup$ More stupid comment: $SL(2,R)$ is soluble if $R$ is quite small (e.g. Artin local with residue field of order 2 should do it, right?) $\endgroup$ Nov 26, 2010 at 14:05
  • $\begingroup$ I see where things have gone wrong but they fixable: all your examples are nilpotent... I still believe that $SL_2 (R)$ would contain no 3-soluble non-nilpotent subgroup... $\endgroup$
    – Bugs Bunny
    Nov 29, 2010 at 15:49

EDIT: I misquoted the book, and the answers here show it isn't fixable.

Let $A,B \in SL_2(R)$. This book by Brumfiel and Hilden has the following two facts in it:

  1. $Tr(A)Tr(B) = Tr(AB) + Tr(AB^{-1})$.
  2. The pair $A,B$ is uniquely determined, up to simultaneous conjugation, by the elements $Tr(A)$, $Tr(B)$, and $Tr(AB)$.

Taking $A^2 = 1$ and $A = B$, the first fact implies that $Tr(A) = \pm 2$, and then the second fact implies that $A$ is conjugate (and hence equal) to $\pm Id$. Since there is an embedding $S_4 \hookrightarrow SL_3(R)$ (which is described in the other answers), we can finish by noting that the images of the transpositions are not central in $SL_3(R)$.

I would guess that there's a more elementary way to prove that involutions in $SL_2(R)$ are central, but I don't know one at the moment.

  • 1
    $\begingroup$ #2 is false. Check A=B=[0,1;1,0] versus A=B=[1,0;0,1] over the field with two elements. The conclusion is also obviously false, as already mentioned: SL(2,2) ≅ Sym(3). $\endgroup$ Nov 27, 2010 at 2:53
  • $\begingroup$ There is a similar statement to #2 in the book, but I haven't checked the hypothesis for $R$. I'll change the answer once I do. $\endgroup$ Nov 27, 2010 at 3:20

$\newcommand{\SL}{\mathrm{SL}}$ $\newcommand{\m}{\mathfrak{m}}$ $\newcommand{\F}{\mathbf{F}}$ $\newcommand{\Z}{\mathbf{Z}}$

Assume that $\SL_3(R)$ is a subgroup of $\SL_2(R)$. We wish to obtain a contradiction.

Here is the strategy. Suppose that $R$ contains a subring of the form $A \oplus B$ where $2A = 0$. Then $SL_3(\F_2)$ is a subgroup of $\SL_3(A)$, which is a subgroup of $\SL_3(A \oplus B)$, which is a subgroup of $\SL_3(R)$. Hence, under our assumption on $R$, $\SL_3(\F_2)$ is a subgroup of $\SL_2(R)$, and this is ruled out by Silence Dogood's answer. $R$ trivially admits such a decomposition when $2 = 0$. Hence we may assume that $2 \ne 0$, and thus that $S_4$ is a subgroup of $\SL_3(R)$, and hence of $\SL_2(R)$.

If $S \subset R$ contains a subring of the form $A \oplus B$ with $2A = 0$, then so does $R$. Thus, WLOG, assume that $R$ is generated by the entries of $g-1$ where $g \in S_4 \subset \SL_2(R)$. Let $K \subset S_4$ denote the Klein $4$-subgroup. Then any map $S_4 \rightarrow G$ is injective if and only if the restriction $K \rightarrow G$ is non-zero (Obvious). $K$ is the only non-trivial normal subgroup of $S_4$ which is a $p$-group (Obvious). By construction, $R$ is Noetherian. If $x \in R$ is any element, and $\m$ is a maximal ideal containing the annihilator of $x$, then $x$ is non-zero in the localization $R_{\m}$. Hence there exists an $\m$ such that $K \rightarrow \SL_2(R_{\m})$ is non-zero, so $S_4 \rightarrow \SL_2(R_{\m})$ is injective. (Choose $x$ to be a non-zero matrix entry of $g-1$ for $g \in K$.) Let $A = R_{\m}$, and let $k = A/\m$. Consider the projection map $S_4 \rightarrow \SL_2(k)$, and let $H$ denote the kernel. Let $g$ be an element of $H$ which is not the identity (if such an element exists). By the Krull intersection theorem (as in SD's answer), there exists a minimal integer $n$ such that $$g - 1 \equiv 0 \mod \m^n, \qquad (g - 1) \not\equiv 0 \mod \m^{n+1}.$$ If $i$ is co-prime to the characteristic of $k$, then it is a unit in $A$, and $$g^i - 1 = (1 + (g-1))^i - 1 \equiv i (g-1) \mod \m^{n+1} \not\equiv 0 \mod \m^{n+1}.$$ It follows that the order of $g$ is some power of the characteristic (or is trivial if $\mathrm{char}(k) = 0$), and hence $H$ is a $p$-group. Hence either $S_4$ injects into $\SL_2(k)$, or $k$ has characteristic $2$ and $H = K$. The former does not occur. We shall prove that $2 = 0$ in $A$. The image of $S_4$ in $\SL_2(k)$ is $S_3$. $S_4$ contains an element $M$ of order $2$ which maps to an element of order $2$ in $S_3$ (for example, any $2$-cycle). The matrix $M$ has order two, and hence satisfies the polynomial $M^2 - 1 = 0$. Yet $M$ also has determinant one, and thus also satisfies the polynomial $M^2 - \mathrm{trace}(M) M + 1 = 0$, by Cayley--Hamilton. It follows that $\mathrm{trace}(M) M = 2 \ne 0$ (by assumption). Yet $M$ has at least one entry that is a unit, and thus $(\mathrm{trace}(M)) = (2)$ in $A$, and it follows easily that the image of $M$ is a scalar matrix in $\SL_2(k)$. Since $k$ has characteristic $2$, this implies that the image of $M$ in $\SL_2(k)$ is trivial ($v^2 = 1$ implies that $v = 1$), a contradiction. Hence $2 = 0$ in $A$.

We have now shown that $2 = 0$ in $A = R_{\m}$. Suppose we can show in addition that $A$ has finite length, that is $A/\m^k = A$ for some $k$. Assume this is so. Let $x_1, \ldots, x_n$ be generators of $\m^k \subset R$. By definition, $x_i$ maps to zero in the localization map $R \rightarrow R_{\m} = A$. Thus there exists an element $y_i \notin \m$ such that $y_i x_i = 0$. Let $y = y_1 \times \ldots \times y_n$. Since $y_i \notin \m$, the product $y \notin \m$. It follows that $$y + \m^k = R,$$ as the ideal on the LHS is not contained in any maximal ideal. On the other hand, $y$ annihilates $\m^k$ by construction. Thus, by the Chinese remainder theorem, $$R = R/y \m^k = R/y \oplus R/\m^k = R/y \oplus A.$$

Since $2 = 0$ in $A$, this shows that $R$ has the required decomposition. Thus we will be done if we can show that $A$ has finite length. Equivalently, we are done if we can show that the non-unit elements of $A$ are nilpotent.

It seems according to Tim that this won't work, since $S_4$ injects into $\SL_2(\F[[x]])$ via the map $$(12) \mapsto \left( \begin{matrix} 0 & 1 \\\ 1 & 0 \end{matrix} \right)$$ and $$(1234) \mapsto \left( \begin{matrix} 1+x+x^2 & 1+x^2 \\\ x^2 & 1+x+x^2 \end{matrix} \right)$$

Hence this answer, for the time being, is a complete fail.

  • $\begingroup$ @Jack: I think your group lands in $GL$ rather than $SL$, the generators have determinant $-1=3$. $\endgroup$ Dec 4, 2010 at 19:06
  • 1
    $\begingroup$ @frogger: my girlfriend is on call this weekend and I've got all the kids. I noticed the post but didn't yet read it. I'm really pleased the question is answered but didn't just want to mindlessly upvote until I'd read what you had done. [which I will, soon...] $\endgroup$ Dec 5, 2010 at 18:37
  • 2
    $\begingroup$ There is no need to be rude. The claim is still false, and several of the steps are quite vague. S4 embeds in SL(2,Z[x]/(6,xx)) as (1,2) = [ 1, 3 ; 0, 1 ], (2,3) = [ 4, 3 ; 3, 4 ], (3,4) = [ 1 + 3x, 3 ; 0, 1 + 3x ]. Check that AA = BB = CC = 1 mod (6,xx), ABA-BAB = BCB-CBC = AC-CA = 0 mod (6,xx), and that AC ≠ 1 mod (6,xx) (and check that the determinants are all 1 mod 6 :-). This does not seem like a serious problem, but it requires the proof to be reorganized a bit. $\endgroup$ Dec 5, 2010 at 19:21
  • 1
    $\begingroup$ This argument does point out some strange (to me) things in the Artinian local case: Let G(n) be SL(2,ZZ[x]/(n,xx)). I would have expected G(4) to be more flexible in all ways than G(2), but in fact they each appear to have strengths. G(2) contains dihedral groups of order 6, 8, and 12, and the symmetric group S4, while G(4) contains none of those. On the other hand, G(4) contains elements of 8 and elementary abelian groups of rank 6, while G(2) only has elements of order 4 and subgroups of rank 3. Is there some sense in which G(4) is more abelian? $\endgroup$ Dec 5, 2010 at 20:03
  • 1
    $\begingroup$ @Jack: good question, but one that, slightly more out of desire to keep these comments from growing too convoluted and get it better attention than to create as many spin-offs from my first question as possible, I think you should ask as its own question :) $\endgroup$ Dec 5, 2010 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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