$\DeclareMathOperator\SL{SL}\newcommand{\ab}{\mathrm{ab}}$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!