Return to Answer

added two relations I left out

Source Link

edited Mar 15, 2015 at 19:36

1.3k
6
12

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, $$\Gamma(2)' / \Gamma(8) \cong \langle \sigma, \tau \ | \ \sigma^{4} = \tau^{4} = [\sigma, \tau]^{2} = [[\sigma, \tau], \sigma] = [[\sigma, \tau], \tau] = 1 \rangle.$$$$\Gamma(2)' / \Gamma(8) \cong \langle \sigma, \tau \ | \ \sigma^{4} = \tau^{4} = [\sigma, \tau]^{2} = [\sigma^{2}, \tau] = [\sigma, \tau^{2}] = [[\sigma, \tau], \sigma] = [[\sigma, \tau], \tau] = 1 \rangle.$$ 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.

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, $$\Gamma(2)' / \Gamma(8) \cong \langle \sigma, \tau \ | \ \sigma^{4} = \tau^{4} = [\sigma, \tau]^{2} = [[\sigma, \tau], \sigma] = [[\sigma, \tau], \tau] = 1 \rangle.$$ 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.

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.

Source Link

created Mar 15, 2015 at 19:30

Jeff Yelton

1.3k
6
12

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, $$\Gamma(2)' / \Gamma(8) \cong \langle \sigma, \tau \ | \ \sigma^{4} = \tau^{4} = [\sigma, \tau]^{2} = [[\sigma, \tau], \sigma] = [[\sigma, \tau], \tau] = 1 \rangle.$$ 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.