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The subgroup of $SL(2,\mathbb{Z})$ generated by

$\begin{pmatrix}1&0\\1&1\end{pmatrix}$ and $\begin{pmatrix}1&5\\0&1\end{pmatrix}$

has come up in a research question in string theory, and I am interested in determining whether or not its index is infinite.

I found the article "Manipulating Subgroups of the Modular Group" by Daniel Schultz in the Mathematica Journal, and it seems that the package referenced therein (ModularSubgroups) could answer this question for me. However, I have not been able to find where to download this package.

Does anyone know if this particular subgroup has infinite index?

More Details: I actually have the subgroup of $Sp(4,\mathbb{Z})$ generated by

$\begin{pmatrix}1&0&0&0\\1&1&0&0\\0&-5&1&0\\0&0&-1&1\end{pmatrix}$ and $\begin{pmatrix}1&0&0&1\\0&1&0&1\\0&0&1&-5\\0&0&0&1\end{pmatrix}$

And I want to know if the action of this subgroup on the lattice $\mathbb{Z}^4$ has a finite or an infinite number of orbits. Since the third row of matrices in this subgroup is always $(0,0,1,0)$ mod 5, the answer to the modular group question will help me to guide my efforts toward solving the actual problem that I have.

Daniel Longenecker
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    Yes the index of this subgroup (which is free on its two generators) is infinite. The modular group is virtually free, so it is possible to do calculations like these. If no-one else answers I will try and give more details of my own computer calcuations, which I did in my KBMAG Knuth-Bendix package, later. – Derek Holt Nov 27 '19 at 08:35
  • That’s great news, thanks. I looked at the package documentation, and it seems like exactly what I need. Thanks for writing the package and for letting me know about it. Also, can that computer package also answer questions about the group action on the integer lattice (or better yet the ordered sets of coprime integers, which can be thought of as the group acting on itself in the defining representation, modulo the second column of the matrix)? Lastly, is there an algorithm for generating a representative for each coset? – Daniel Longenecker Nov 29 '19 at 04:09

1 Answers1

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It is well known that ${\rm SL}(2,{\mathbb Z})$ is generated by the matrices $$a = \left(\begin{array}{rr}0&1\\-1&0\end{array}\right)\ \ \ {\rm and}\ \ \ b = \left(\begin{array}{rr}0&1\\-1&1\end{array}\right),$$ and that $\langle a,b \mid a^2=b^3, a^4=1 \rangle$ is a presentation on these generators. Furthermore, we have $$(b^{-1}a)^5 = \left(\begin{array}{rr}1&5\\0&1\end{array}\right)\ \ \ {\rm and}\ \ \ ba^{-1} = \left(\begin{array}{rr}1&0\\1&1\end{array}\right),$$ which generate your subgroup.

My KBMAG package can often compute presentations of subgroups of automatic groups (when thyey are quasi-isometrically embedded). The group ${\rm SL}(2,{\mathbb Z})$ is virtually free and is an easy calculation.

I will demonstrate how to do it using the GAP interface to KBMAG.

gap> LoadPackage("kbmag");
true
gap> F := FreeGroup(2);; a:=F.1;; b:=F.2;;
gap> G := F/[a^2*b^-3, a^4];;
gap> R := KBMAGRewritingSystem(G);
gap> S := SubgroupOfKBMAGRewritingSystem(R, [(b^-1*a)^5, b*a^-1] );;
gap> A := AutomaticStructureOnCosetsWithSubgroupPresentation(R, S);
true
gap> Index(R, S);
infinity
gap> presH := PresentationOfSubgroupOfKBMAGRewritingSystem(R, S);
<fp group of size infinity on the generators [ f1, f2 ]>
gap> RelatorsOfFpGroup(presH);
[  ]

So we see that the index of the subgroup is infinite, and it has a presentation with two generators and no relators, so it is free.

Derek Holt
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