Quotient of principal congruence subgroups This is a direct follow-up to this question. What is the quotient $\Gamma(2)/\Gamma(2^n)?$ (the principal congruence subgroups are in $SL(2, \mathbb{Z}).$ It is a 2-group, but what else?
 A: This is far from a full answer, but here a few things I've figured out, hope it helps.
First of all, $\Gamma(2)$ is the direct product of $\{\pm 1\}$ and the subgroup $\Gamma(2)'$ consisting of all matrices in $\Gamma(2)$ whose diagonal elements are $1$ modulo $4$.  So it suffices to determine each $\Gamma(2)' / \Gamma(2^{n})$.  Clearly $\Gamma(2)' / \Gamma(4) \cong (\mathbb{Z} / 2\mathbb{Z})^{2}$.  Also, 
$$\Gamma(2)' / \Gamma(8) \cong \langle \sigma, \tau \ | \ \sigma^{4} = \tau^{4} = [\sigma, \tau]^{2} = [\sigma^{2}, \tau] = [\sigma, \tau^{2}] = [[\sigma, \tau], \sigma] = [[\sigma, \tau], \tau] = 1 \rangle.$$
More generally, each $\Gamma(2)' / \Gamma(2^{n})$ is generated by two elements $\sigma$ and $\tau$ each of order $2^{n - 1}$, whose commutator $[\sigma, \tau]$ has order $2^{n - 2}$ and generates the whole commutator subgroup.  I think one should be able to derive a full set of relations in the form of embedded commutators as above, and there should be some recursive pattern to this set of relations as $n$ increases, but I haven't figured it out.
