Some Subgroups of SL(2,Z) of finite index
Marc Bogaerts
http://math.arizona.edu/~asp/2008/problems3.pdf
The group G := SL(2, Z) is generated by the matrices
0 1
1 1
and
1 1
0 1
Now consider the subgroup
S := {
a b
c d
| c = 0 (mod 3) }
What is [G : S]? Determine coset representatives for S in G and a
generating set for S.
I solved this case, but the general case where 3 is replaced by a prime
p is even more interesting.
I prefer the Group SL(n,Z) to be generated by
N= [[1,1],[0,1]] and P := [[0,-1],[1,0]] because these matrices, when
operating on the left, N represents the replacement of the first row by
its sum with the second row, and P represents the permutation of the two
rows (changing one of them by sign in order to preserve the determinant).
The group H, consisting of the upper triangular matrices is of finite
index p+1. I'll give a short outline of the proof here.
First any matrix M in G can be written as a product of the form
P * N^a1 * P * N^a2 * .. * N^ai * P * Q, with Q in the center of G (1 or
P^2, in the following I will neglect any factors P^2 because they
commute with all matrices, in fact, they only mean a change of sign).
Using any of the "building blocks" of the form
P * N^a * P * N^a' * P,
where a' denotes the inverse of a modulo p, and the extra block
P * N^p * P, and I suppose H being generated by these elements.
I can reduce any matrix M in G in the following way:
Let M = P * N^a1 * P * N^a2 * .. * N^ai * P then if a1 > p then I
consider the product (P * N^-p)^n * P * M, this leads to a product of
the form
P * N^(a1-n * P) * P * N^a2 * .. * N^ai * P, so I consider only the
cases where the ai < n.
Then the following product
(P * N^-ai' * P * N^-ai * P) * P * N^a1 * P * N^a2 * .. * N^ai * P, with
ai < p reduces to
P * N^a2-ai' * P * N^a2 * .. * N^ai * P, belonging to the same coset of
H and having one factor less.
If I repeat this procedure until I'm left with one of the following
N, P , P * N, P * N^2, .. P * N^(p-1)
So there are exactly p+1 cosets, and the "building blocks" are a set of
generators for H.
I'm working on the case SL(3,Z) but this seems more difficult.
Any suggestions?
Derek Holt
> I found an interesting exercise (nr. 1) inhttp://math.arizona.edu/~asp/2008/problems3.pdf
> The group G := SL(2, Z) is generated by the matrices
> 0 1
> 1 1
> and
> 1 1
> 0 1
> Now consider the subgroup
> S := {
> a b
> c d
> | c = 0 (mod 3) }
>
> What is [G : S]? Determine coset representatives for S in G and a
> generating set for S.
>
> I solved this case, but the general case where 3 is replaced by a prime
> p is even more interesting.
Let phi be the natural projection G -> SL(2,p). Then phi(S) is the
stabilizer of a 1-dimensional subspace in SL(2,p), which has index p+1
in SL(2,p) (since there are p+1 1-dimensional subspaces), and S is the
complete inverse image of phi(S) under phi, so it has index p+1 in G.
You describe coset representatives below. For any subgroup H of any
group G, there is a standard procedure for getting generators (the
Schreier generators) of H from generators of G and a set of coset
representatives of H in G. I haven't read your argument in detail, but
I think the generators that you are coming up with are more or less
the same as the Schreier generators.
>
> I prefer the Group SL(n,Z) to be generated by
> N= [[1,1],[0,1]] and P := [[0,-1],[1,0]] because these matrices, when
> operating on the left, N represents the replacement of the first row by
> its sum with the second row, and P represents the permutation of the two
> rows (changing one of them by sign in order to preserve the determinant).
> The group H, consisting of the upper triangular matrices is of finite
> index p+1. I'll give a short outline of the proof here.
You cannot mean H is just the upper triangular matrices, since that
does not involve p. Presumably H is what you called S earlier.
What is your question about SL(3,Z)?
Derek Holt.
Marc Bogaerts
> On 12 Mar, 23:33, Marc Bogaerts<mbg-dot-ni...@gmail.com> wrote:
>>
>> I solved this case, but the general case where 3 is replaced by a prime
>> p is even more interesting.
>
> Let phi be the natural projection G -> SL(2,p). Then phi(S) is the
> stabilizer of a 1-dimensional subspace in SL(2,p), which has index p+1
> in SL(2,p) (since there are p+1 1-dimensional subspaces), and S is the
(I preferred to rename S in H in my text.)
> complete inverse image of phi(S) under phi, so it has index p+1 in G.
>
> You describe coset representatives below. For any subgroup H of any
> group G, there is a standard procedure for getting generators (the
> Schreier generators) of H from generators of G and a set of coset
> representatives of H in G. I haven't read your argument in detail, but
> I think the generators that you are coming up with are more or less
> the same as the Schreier generators.
Indeed, that's how I found representatives for G/H, and, using
Schreier's lemma to find generators for H. But it took a bit of
imagination to shorten them in length and find a minimal set.
Same thing for the coset representatives.
>
>>
>> I prefer the Group SL(n,Z) to be generated by
>> N= [[1,1],[0,1]] and P := [[0,-1],[1,0]] because these matrices, when
>> operating on the left, N represents the replacement of the first row by
>> its sum with the second row, and P represents the permutation of the two
>> rows (changing one of them by sign in order to preserve the determinant).
>> The group H, consisting of the upper triangular matrices is of finite
>> index p+1. I'll give a short outline of the proof here.
>
> You cannot mean H is just the upper triangular matrices, since that
> does not involve p. Presumably H is what you called S earlier.
>
Sorry, I meant upper triangular mod p.
In the case of SL(3,Z) (and SL(n,Z) in general) the group can be
generated by matrices that reflect Gaussian elimination, and
permutations of rows (changing sign when necessary) respectively. For
n=3 there are the matrix
N = [[ 1, 1, 0], [0, 1, 0], [0, 0, 0]]
and the two permutation matrices
P = [[0, -1, 0 ],[1, 0, 0 ],[0, 0, 1]]
Q = [[0, 0, -1 ],[0, 1, 0 ],[1, 0, 0]]
The question is to find representatives of the cosets
>
> Derek Holt.
>
Derek Holt
Cosets of what? Of the subgroup H of matrices that are upper
triangular mod p? The index is
(p^2+p+1)(p+1) for n=3, or
(p^n-1)...(p^2-1)(p-1)/(p-1)^n for general n.
I expect it is possible to find a generic set of coset representatives
as word in your generators.
It might help to use a chain of subgroups
H=H_1 < H_2 < ... < H_n = G.
with |H_i:H_{i-1}| = (p^i-1)/(p-1)
Good luck!
Derek Holt.
Marc Bogaerts
>
> Cosets of what? Of the subgroup H of matrices that are upper
> triangular mod p? The index is
>
> (p^2+p+1)(p+1) for n=3, or
>
> (p^n-1)...(p^2-1)(p-1)/(p-1)^n for general n.
>
This makes me think as the product of numbers of points in projective
spaces GF(p^(n-i))^i
> I expect it is possible to find a generic set of coset representatives
> as word in your generators.
>
> It might help to use a chain of subgroups
>
> H=H_1< H_2< ...< H_n = G.
>
> with |H_i:H_{i-1}| = (p^i-1)/(p-1)
>
> Good luck!
Thanks a lot for your valuable advice.
>
> Derek Holt.
>
>