This is a direct followup 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 2group, but what else?

$\begingroup$ The derived length and nilpotence class aren't difficult to calculate. What sort of things are you looking for? $\endgroup$– Geoff RobinsonMar 15, 2015 at 18:34

$\begingroup$ @GeoffRobinson Well, for the source question, it would be good to get enough information to find the automorphism group... $\endgroup$– Igor RivinMar 15, 2015 at 19:32

1$\begingroup$ If you have a finite 2group $G$, then the set of automorphisms acting as the identity on $G/G^2$ is a normal 2subgroup, and the quotient is a subgroup of automorphisms of $G/G^2$ which should be computable with not too much effort in your case. $\endgroup$– YCorMar 15, 2015 at 20:27
1 Answer
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.