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?

  • $\begingroup$ The derived length and nilpotence class aren't difficult to calculate. What sort of things are you looking for? $\endgroup$ Mar 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 Rivin
    Mar 15, 2015 at 19:32
  • 1
    $\begingroup$ If you have a finite 2-group $G$, then the set of automorphisms acting as the identity on $G/G^2$ is a normal 2-subgroup, and the quotient is a subgroup of automorphisms of $G/G^2$ which should be computable with not too much effort in your case. $\endgroup$
    – YCor
    Mar 15, 2015 at 20:27

1 Answer 1


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.


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