Abelianisation of certain congruence subgroups in GL_2(Z)

Question

$\DeclareMathOperator\SL{SL}\newcommand{\ab}{\mathrm{ab}}$Denote by $$\Gamma(m) = \left\{ \left( \begin{array}{cc} 1 +ma_{11} & m a_{12} \\ ma_{21} & 1 +ma_{22} \end{array} \right) \in \SL_2(\mathbb{Z}) \mid a_{ij} \in \mathbb{Z} \right\}$$ the principal congruence subgroup of level $m$ of $\SL_2(\mathbb{Z})$. Also consider $h = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)$ and $g = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$. I was wondering about the following:

(1) Is the abelianisation of $\langle \Gamma(m), h,g \rangle$ finite?

(2) Suppose $m= 2^{n}$ is a $2$-power. Is then $\langle \Gamma(m), h,g \rangle^{\ab}$ finite and even an elementary abelian $2$-group? If yes, what about non $2$-powers?

In fact I am wondering about the above questions for the subgroup $$V_m =\left\{ \left( \begin{array}{cc} 1 +ma_{11} & m a_{12} \\ ma_{21} & 1 +ma_{22} \end{array} \right) \in \SL_2(\mathbb{Z}) \mid a_{11} \equiv a_{22}, a_{12}\equiv a_{21} \mod 2 \in \mathbb{Z} \right\}$$ of $\Gamma(m)$. Question (1) for $\langle \Gamma(m), h,g \rangle$ or $\langle V_m, h,g \rangle$ is however equivalent, but (2) a priori not.

I know that $\Gamma(m)^{\ab}$ is a free abelian group (because when $m \geq 3$ the group is free). However the abelianisation of $\mathrm{GL}_2(\mathbb{Z}) \cong D_{12} \star_{C_2\times C_2} D_8$ is finite. For $m=2,4$ one can show that (2) holds (via an explicit set of generators).

Thanks for any input and help!

Answer 1

$\newcommand{\H}{\mathcal{H}}\DeclareMathOperator{\GL}{GL}\DeclareMathOperator{\SL}{SL}\DeclareMathOperator{\Hom}{Hom}\newcommand{\Z}{\mathbb{Z}}\newcommand{\Q}{\mathbb{Q}}$No, the rank of the abelianisation goes to infinity with $m$. This can probably be obtained in many ways : Riemann-Hurwitz, trace formula, Shapiro's lemma and amalgamated product decompositions, etc.

Let me sketch one proof. Assume $m\ge 3$ and let $\H$ be the upper half-space with its usual action of $\GL_2(\Z)$. Let $\Gamma'(m) = \langle \Gamma(m),g,h\rangle$ and $\Gamma''(m) = \Gamma'(m) \cap \SL_2(\Z) = \langle \Gamma(m),S\rangle$ where $S = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$. The rank of the abelianisation of $\Gamma'(m)$ is the first Betti number of $\H/\Gamma'(m)$, which is half that of its orientation covering $\H/\Gamma''(m)$. We can compute the last Betti number as $\dim H^1(\Gamma''(m),\Q) = \dim H^1(\Gamma(m),\Q)^{\langle S\rangle} = \dim H^1(\Gamma(m),\Q)^{\langle -1, S\rangle} $ $= \dim \Hom_{G_m}(\sigma,H^1(\Gamma(m),\Q))$, where $G_m = \SL_2(\Z/m\Z)$ and $\sigma$ is the permutation representation $\Q[G_m/\langle -1,S\rangle]$ of $G_m$. You can get a formula for the last dimension by using for instance Proposition 4.18 (with $N=m$, $k=2$) in The cohomology of lattices in $\SL_2(\mathbb{C})$ by Finis, Grunewald and Tirao, and the order of magnitude is a constant times $\# G_m$, which tends to infinity with $m$.