So far, I know the matrices $$S=\left( \begin{matrix} 0&-1\\ 1&0 \end{matrix} \right), \quad T=\left( \begin{matrix} 1&1\\ 0&1 \end{matrix} \right) $$ generate SL$(2,\mathbb{Z})$. $S^2=(ST)^3=-I$.
I also know the matrices $$X=\left( \begin{matrix} 1&1\\ 1&2 \end{matrix} \right), \quad Y=\left( \begin{matrix} 2&1\\ 1&1 \end{matrix} \right) $$ with their inverses $X^{-1},Y^{-1}$ generate SL$(2,\mathbb{Z})'$, the commutator subgroup.
I'm interested in what the quotient $SL(2,\mathbb{Z})/SL(2,\mathbb{Z})'$ is.
In general, how would I go about computing the quotient of two matrix groups? I'm trying to use GAP but I'm having a lot of trouble!
You have the presentation $S^2=(ST)^3=-I$. You get the presentation of $G/G'$ by including all relations $ST=TS$ etc. So the presentation becomes $S^4=I$, $S^3T^3=S^2$ and $ST=TS$. This implies $S=T^{-3}$ and $T^{12}=I$. Therefore $G/G'$ is cyclic of order $12$ generated by the image of $T$.