# Small-index subgroups of SL(3,Z)

I would like to know the smallest-index subgroups of ${\rm SL}(3,\mathbb{Z})$.

The smallest I could find has even entries $a_{3,1}$ and $a_{3,2}$, along the bottom row. I could not figure out whether there are subgroups of index $2$ or $3$.

A search found lots of information about ${\rm SL}(2,\mathbb{Z})$, but not about ${\rm SL}(3,\mathbb{Z})$.

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Since $SL(3,\mathbb{Z})$ has the congruence subgroup property, I believe that every finite-index subgroup is pulled back from some subgroup of $SL(3,\mathbb{Z}/n)$ for some $n$. So your question can be reduced to the same question over $\mathbb{Z}/n$ (for all $n$). – HJRW Feb 25 '13 at 15:03
It seems to me that $SL(3,{\mathbb Z})$ is its own commutator. Hence it cannot have a subgroup of index $2$ or $3$. – Venkataramana Feb 25 '13 at 15:59
I would guess that the smallest index is $Card({\mathbb P}^2 ({\mathbb F}_2)=7$. – Venkataramana Feb 25 '13 at 16:06
@David: Concerning $\mathrm{SL}(2,\mathbb{Z})$, or its quotient by scalars the modular group, the congruence subgroup property fails and the subgroup structure is much richer than in higher ranks. But the modular group has a lot of older literature relative to its action on the complex upper half plane, etc. – Jim Humphreys Feb 26 '13 at 14:06

To fill in the comments, there are basically two serious issues involved.

1) You want to know that every subgroup of finite index in $\mathrm{SL}(3,\mathbb{Z})$ contains some congruence kernel: the kernel of the natural reduction homomorphism induced by $\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$. In other words, the original group satisfies the congruence subgroup property (as HW comments). This is already quite nontrivial to prove and evolved from work of Bass-Lazard-Serre here followed by more definitive work by Bass-Milnor-Serre here. (There is some exposition in the last part of my 1980 Springer Lecture Notes 789.)

2) Now you have to pin down the maximal subgroups of a typical finite group $\mathrm{SL}(3,n)$ and pull these back to the big group in order to sort out which give you subgroups of minimal index there. Maximal subgroups of the finite groups in question have been extensively studied, especially over finite fields, but in this case some trial-and-error leads pretty quickly to the number 7, as indicated by Aakumadula. This results from the first possibility $n=2$, where you get the simple nonabelian group $\mathrm{SL}(3,2)$ of order $8 \cdot 3 \cdot 7$, from the general formula for a prime $p$ given by $p^3 (p^2-1)(p^3-1)$. Here there is a maximal parabolic subgroup of index 7.

The search for maximal subgroups is itself a major project, for which I don't have all relevant references at hand. But over a field some basic insight comes from algebraic groups, where you see that a maximal (closed) proper subgroup will be either parabolic or reductive. I'm not sure in the special case at hand how hard it is to pin down the maximal subgroups of the various finite groups directly.

ADDED: As pointed out in comments, it's possible here to rule out subgroups of index 2 or 3 by combining the known commutator group result with a study of actions of the big group on left cosets of a hypothetical subgroup. But a systematic study (especially for other Chevalley-type groups) probably requires something like the program I've sketched above. Given the congruence subgroup property, the study of subgroups of finite index then reduces quickly to the study of finite groups of the same type over finite rings and then prime fields. In your case, this helps to provide a conceptual answer to such questions as: Why does $\mathrm{SL}(3, \mathbb{Z})$ have a subgroup of index 7?

There is a vast literature by now on maximal subgroups of known finite simple groups, by people including Aschbacher, Kleidman, Liebeck, Scott, Seitz, .... An older brief survey by Kleidman-Liebeck, with an impressive bibliography, appeared in Geom. Dedicata 25 (1988). For your specific group there are very old results, but the general approach is more unified even though intricate.

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Given an index $k$ subgroup of $SL(3,Z)$, $k\leq 6$, one obtains a homomorphism to $A_k$ from permuting cosets. By the congruence subgroup property, the image must be congruence, and therefore contains the simple $PSL(3,p)$ as a quotient for some $p$. But we see that no such simple group divides $360$ from your formula. – Ian Agol Feb 25 '13 at 19:24
A step I am missing is why the subgroup of $SL(3,Z)$ has to be the pullback of a subgroup of $SL(3,n)$. Is maximality being used? If $G$ is the subgroup, I get that there exists $H<G$ so that $H$ is the matrices congruent to the identity mod $n$, for some integer $n$. Probably I am missing something easy for the next step. – David Farmer Feb 25 '13 at 20:55
Your $H$ is normal in $SL(3,Z)$ so the only ingredient not explicitly mentioned is the undergraduate-level fact that subgroups of $G$ containing the normal subgroup $N$ biject canonically with subgroups of $G/N$ via the pullback map. – user30035 Feb 25 '13 at 21:16
If a group has a subgroup of index $4$ then it also has either a subgroup of index $2$ or $3$. – Tom Goodwillie Feb 26 '13 at 15:45
The names of Mitchell, Bloom and Hartley come to mind when thinking about three dimensional linear groups over finite fields. – Geoff Robinson Apr 13 at 16:06

In order to answer the question we need a finite presentation of ${\rm SL}(3,\mathbb{Z})$ and a general method to find all subgroups of index $\leq n$ of a finitely presented group:

• A finite presentation for ${\rm SL}(3,\mathbb{Z})$ can be found for example in Theorem 2 in

Marston Conder, Edmund Robertson, Peter Williams: Presentations for 3-dimensional special linear groups over integer rings, Proc. Amer. Math. Soc. 115 (1992), no. 1, 19-26. http://www.ams.org/journals/proc/1992-115-01/S0002-9939-1992-1079696-5/S0002-9939-1992-1079696-5.pdf.

The finite presentation given in this paper is $${\rm SL}(3,\mathbb{Z}) \cong \left< x, y, z \ | \ x^3 = y^3 = z^2 = (xz)^3 = (yz)^3 = (x^{-1}zxy)^2 = (y^{-1}zyx)^2 = (xy)^6 = 1 \right>$$ on the generators $$x \ = \ \left( \begin{array}{rrr} 1 & 0 & 1 \\\ 0 & -1 & -1 \\\ 0 & 1 & 0 \end{array} \right), \ \ y \ = \ \left( \begin{array}{rrr} 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ 1 & 0 & 0 \end{array} \right), \ \ z \ = \ \left( \begin{array}{rrr} 0 & 1 & 0 \\\ 1 & 0 & 0 \\\ -1 & -1 & -1 \end{array} \right).$$

• A general method to find all subgroups of index $\leq n$ of a finitely presented group is the so-called low index subgroups procedure. This algorithm is described in Section 5.4 in

Derek F. Holt, Bettina Eick, and Eamonn A. O'Brien, Handbook of computational group theory, Discrete Mathematics and its Applications (Boca Raton), Chapman & Hall / CRC, Boca Raton, FL, 2005. MR 2129747 (2006f:20001).

For an online resource, see e.g.

Marston Conder: Applications and adaptations of the low index subgroups procedure, http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.107.5164.

The low index subgroups procedure is implemented in the GAP computer algebra system (cf. http://www.gap-system.org). Hence all we need to do is to enter the presentation of ${\rm SL}(3,\mathbb{Z})$ taken from the above paper into GAP ...

gap> F := FreeGroup("x","y","z");;
gap> AssignGeneratorVariables(F);
#I  Assigned the global variables [ x, y, z ]
gap> G := F/[x^3,y^3,z^2,(x*z)^3,(y*z)^3,(x^-1*z*x*y)^2,(y^-1*z*y*x)^2,(x*y)^6];
<fp group on the generators [ x, y, z ]>


... and to run the algorithm on it:

gap> sub := LowIndexSubgroupsFpGroup(G,7);;
gap> List(sub,H->Index(G,H));
[ 1, 7, 7 ]
gap> gens := List(sub,GeneratorsOfGroup);
[ [ x, y, z ], [ x, z, y*z*y^-1, (y*x)^2*y ],
[ x, y*x^-1*z^-1, y^-1*z*y, z*y^-1*x*y, y^-1*x*y*x^-1*y ] ]


This tells us that the smallest index of a proper subgroup of ${\rm SL}(3,\mathbb{Z})$ is 7, and that there are 2 conjugacy classes of subgroups of index 7. Now it is straightforward to obtain generators for our subgroups in terms of matrices:

gap> x := [ [ 1, 0, 1 ], [ 0, -1, -1 ], [ 0, 1, 0 ] ];;
gap> y := [ [ 0, 1, 0 ], [ 0, 0, 1 ], [ 1, 0, 0 ] ];;
gap> z := [ [ 0, 1, 0 ], [ 1, 0, 0 ], [ -1, -1, -1 ] ];;
gap> List(gens[2],g->MappedWord(g,GeneratorsOfGroup(G),[x,y,z]));
[ [ [ 1, 0, 1 ], [ 0, -1, -1 ], [ 0, 1, 0 ] ],
[ [ 0, 1, 0 ], [ 1, 0, 0 ], [ -1, -1, -1 ] ],
[ [ 0, 0, 1 ], [ -1, -1, -1 ], [ 1, 0, 0 ] ],
[ [ -1, -1, -1 ], [ 0, 0, 1 ], [ 0, 1, -1 ] ] ]
gap> List(gens[3],g->MappedWord(g,GeneratorsOfGroup(G),[x,y,z]));
[ [ [ 1, 0, 1 ], [ 0, -1, -1 ], [ 0, 1, 0 ] ],
[ [ -1, -1, -1 ], [ 0, 1, 1 ], [ 0, 0, -1 ] ],
[ [ -1, -1, -1 ], [ 0, 0, 1 ], [ 0, 1, 0 ] ],
[ [ 1, 1, 0 ], [ 0, 0, 1 ], [ 0, -1, 0 ] ],
[ [ -1, 0, -1 ], [ 2, 1, 1 ], [ 0, -1, 0 ] ] ]


So our representatives for the conjugacy classes of subgroups of ${\rm SL}(3,\mathbb{Z})$ of index 7 are $$G_{7,1} \ = \ \left< \left(\begin{array}{rrr} 1&0&1\\\ 0&-1&-1\\\ 0&1&0 \end{array}\right), \ \left(\begin{array}{rrr}% 0&1&0\\\ 1&0&0\\\ -1&-1&-1 \end{array}\right), \\ \left(\begin{array}{rrr}% 0&0&1\\\ -1&-1&-1\\\ 1&0&0 \end{array}\right), \ \left(\begin{array}{rrr}% -1&-1&-1\\\ 0&0&1\\\ 0&1&-1 \end{array}\right) \right>$$ and $$G_{7,2} \ = \ \left< \left(\begin{array}{rrr}% 1&0&1\\\ 0&-1&-1\\\ 0&1&0 \end{array}\right), \ \left(\begin{array}{rrr}% -1&-1&-1\\\ 0&1&1\\\ 0&0&-1 \end{array}\right), \ \left(\begin{array}{rrr}% -1&-1&-1\\\ 0&0&1\\\ 0&1&0 \end{array}\right), \\ \left(\begin{array}{rrr}% 1&1&0\\\ 0&0&1\\\ 0&-1&0 \end{array}\right), \ \left(\begin{array}{rrr}% -1&0&-1\\\ 2&1&1\\\ 0&-1&0 \end{array}\right) \right>.$$ The computations above take just a few milliseconds. If one is willing to put in a minute or so, then one can go a bit further and compute representatives for the conjugacy classes of subgroups of ${\rm SL}(3,\mathbb{Z})$ of index $\leq 30$:

gap> sub := LowIndexSubgroupsFpGroup(G,30);;
gap> List(sub,H->Index(G,H));
[ 1, 8, 7, 28, 14, 13, 7, 13, 28, 26, 14, 26, 28, 28, 28, 24, 21 ]


So we have subgroups of indices 7, 8, 13, 14, 21, 24, 26 and 28, and there are no proper subgroups of other indices $\leq 30$. Generators of the subgroups in terms of our generators $x, y, z$ of ${\rm SL}(3,\mathbb{Z})$ can be determined easily as well:

gap> List(sub,GeneratorsOfGroup);
[ [ x, y, z ], [ x, y ], [ x, z, y*z*y^-1, (y*x)^2*y ],
[ x, z, y*z*y^-1, y*x*(y*x^-1)^2*y, (y*x)^2*(y^-1*x)^2*y^-1 ],
[ x, z, (y*x)^2*y ], [ x, y*x^-1*z^-1, y^-1*z*y, z*y^-1*x^-1*y ],
[ x, y*x^-1*z^-1, y^-1*z*y, z*y^-1*x*y, y^-1*x*y*x^-1*y ],
[ x, y*x^-1*z^-1, y^-1*z*y, z*y^-1*x*y, y^-1*(x*y)^2*x^-1*y^-1*x^-1*y ],
[ x, y*x^-1*z^-1, y^-1*z*y, z*y^-1*x*y,
y^-1*(x*y)^2*x^-1*y^-1*x*y*x*y^-1*x^-1*y,
y^-1*(x*y)^2*x^-1*y*x*y^-1*x*y*x^-1*y^-1*x^-1*y ],
[ x, y*x^-1*z^-1, z*y^-1*x^-1*y ],
[ x, y*z*x^-1*y^-1, z*y*x*z^-1, (y*x)^2*y, y^-1*x*y*x^-1*y ],
[ x, y*z*x^-1*y^-1, z*y*x*z^-1, z*y^-1*x*y, y^-1*(x*y)^2*x^-1*y^-1*x^-1*y ],
[ x, y^-1*z*y, (y*x)^2*y, y^-1*x*y*x^-1*y ],
[ x, y^-1*z*y, (y*x)^2*y, z*y^-1*(x^-1*y)^2 ],
[ x, y^-1*z*y, y*x^-1*z*y^-1*x^-1*y^-1, y^-1*x*y*z^-1*x^-1*y^-1 ],
[ y*x^-1, y^-1*x ], [ z, x*z*x^-1, y*z*y^-1, (y*x)^3 ] ]

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Great answer. I wish I could "accept" two answers. – David Farmer Feb 28 '13 at 18:59

You don't need the congruence subgroup property to see that this group has no proper subgroups of index less than $7$, in other words that it has no nontrivial homomorphism to $S_6$.

Nor do you need a presentation of the group; all you need is that it is generated by six elements satisfying certain relations, and a little patience.

Let $x_{i,j}$ be the matrix with $1$ on the diagonal, $1$ in the $(i,j)$, spot, and $0$ otherwise. It is not hard to see, by row reduction, that $SL(3,\mathbb Z)$ is generated by these elements. When $i$, $j$, and $k$ are distinct then $x_{i,j}$ is the commutator of $x_{i,k}$ and $x_{k,j}$, and also $x_{i,j}$ commutes with $x_{i,k}$ and with $x_{k,j}$.

Now suppose for contradiction that $S_6$ has elements $x_{i,j}$ satisfying these same relations and not all equal to the identity. No $x_{i,j}$ can be the identity because then by taking commutators they all would be the identity. Being a commutator, $x_{i,j}$ must belong to $A_6$. It is the commutator of $x_{i,k}$ and $x_{k,j}$, two elements of $A_6$ that commute with it. Looking at all cases, you see that the only nontrivial elements of $A_6$ whose centralizers are nonabelian are those of the form $(ab)(cd)$, product of two disjoint $2$-cycles. If $x_{i,j}=(ab)(cd)$ then $x_{i,k}$ and $x_{k,j}$ must be elements of its centralizer and again of the same type. But the only such elements are $(ab)(cd)$, $(ac)(bd)$, $(ad)(bc)$, $(ab)(ef)$, and $(cd)(ef)$. There are only a few possibilities for pairs of these whose commutator is $(ab)(cd)$. You'll find that none of them will lead to a solution, a choice of $x_{i,j}$ for all $i$ and $j$.

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When I posted an erroneous version of this answer earlier today it garnered several votes. How do I give those votes back? – Tom Goodwillie Feb 28 '13 at 17:47
@Tom: For the $3 \times 3$ case here, I agree one doesn't need the congruence subgroup property (to find a subgroup of index 7, in particular). I wanted to conceptualize the approach to Chevalley groups of rank >1 over $\mathbb{Z}$. But ruling out index below 7 here gets messy. Your argument against subgroups of small index looks most efficient. Note: the commutator relations here (which give a presentation with one more relation added) go back to Nielsen-Magnus; see Steinberg's Yale lectures (p. 96) or Cor. 10.3 in Milnor's Introduction to Algebraic K-Theory. – Jim Humphreys Feb 28 '13 at 18:40
P.S. An interesting conference article by Steinberg has related results. For example, his Theorem 3 should reduce the problem here to looking at small index subgroups of $\mathrm{SL}(3,2)$: R.Steinberg, Some consequences of the elementary relations in $\mathrm{SL}_n$. Finite groups—-coming of age (Montreal, Que., 1982), 335–350, Contemp. Math., 45, Amer. Math. Soc., Providence, RI, 1985. – Jim Humphreys Mar 1 '13 at 21:46