## Describing a matrix group (with integer coefficients) through conditions on the coefficients.

I'm wondering if there's always a (not too complicated?) way to characterize a matrix group by conditions on the coefficients.

I know if I'm dealing with matrix groups over a field, then it's sort of a question of seeing if the matrix group is Zariski-closed, which would correspond to polynomial conditions on the coefficients, though I'm not necessarily only interested in polynomial conditions, and I want to work with integer coefficients.

More specifically, I've got four matrices $A,B,C,D$ in $SL(2,\mathbb{Z})$, and I'm looking at the subgroup of the group generated by these four matrices (which have the sole relation $DCBA = 1$), such that the number of times $A$ or $C$ appears in any word is even.

Actually, (Formatting looks weird, these are meant to be $2\times2$ matrices and the coefficients are ordered as top left, top right, bottom left, bottom right) $$A = \left[\begin{array}{ll} 1 & 5 \\ 0 & 1\end{array}\right]$$ $$B = \left[\begin{array}{ll} 1 & 0 \\ -1 & 1\end{array}\right]$$ $$C = \left[\begin{array}{ll} 11 & 20 \\ -5 & -9\end{array}\right]$$ $$D = \left[\begin{array}{ll} 11 & 25 \\ -4 & -9\end{array}\right]$$

($A,B,C,D$ actually generate the congruence group $\Gamma^1(5)$)

In fact, if we let $$M = \left[\begin{array}{ll} -2 & -5 \\ 1 & 2\end{array}\right]$$ then $C = MAM^{-1}$, and $D = MBM^{-1}$.

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I took the liberty of fixing the formatting (see the sidebar item "Basic solution") – Yemon Choi Oct 16 at 4:46
I took the liberty of changing $B$ to a matrix with determinant $1$ rather than $0.$ – Will Jagy Oct 16 at 4:56
Note that your matrices do not seem to fit the group, see en.wikipedia.org/wiki/… – Will Jagy Oct 16 at 5:19
@Will: $\Gamma^1(5)$, not $\Gamma_1(5)$; $\Gamma^1$ has the top right entry divisible by 5. I corrected $B$ again, so it is now in $\Gamma^1(5)$ and the claimed relation $D = MBM^{-1}$ holds. But it's not totally clear to me what the question here is, and as far as I can see it has nothing whatsoever to do with modular forms. – David Loeffler Oct 16 at 9:04

The question is rather vague, but I hope that the following is helpful.

You seem to want a solution to the membership problem for your subgroup of $SL(2,\mathbb{Z})$. Note that such a solution is known not to exist in $SL(4,\mathbb{Z})$, which contains a copy of $F_2\times F_2$; I suspect its existence is open for $SL(3,\mathbb{Z})$.

However, $SL(2,\mathbb{Z})$ is group-theoretically much nicer, being as it is virtually free. The membership problem is certainly solvable here quite efficiently, though I don't know an implementation off the top of my head. The methods involved go back to Stallings's notion of folding. A google search turned up the paper 'Membership problem for the modular group' by Gurevich and Schupp.

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By the way, if you are correct and the subgroup generated by $A,B,C,D$ is free on $A,B,C$ (I haven't checked), then 'the subgroup ... such that the number of times A or C appears in any word is even' is generated by $A^2, C^2, AC, B, ABA^{-1}$. – HW Oct 16 at 10:22
Note also that the fact that the membership problem is not solvable in general shows that there are subgroups which are not cut out by a family of polynomials. – HW Oct 16 at 13:09