I'm working on exercise 1.3.19 of Hatcher's book, more specifically in first part of the problem
Use the preceding problem to show that a closed orientable surface $M_g$ of genus $g$ has a connected normal covering space with deck transformation group isomorphic to $\mathbb{Z}^n$ iff $n \leq 2 g$.
My attempt
So, let $p:\tilde{X}\to M_g$ be an abelian covering space such that the group of deck transformation $G(\tilde{X}) = \pi(M_g)/p_*(\pi(\tilde{X}))$ is isomorphic to $\mathbb{Z}^n$. We know that
$$ \pi(M_g) = \langle a_1,b_1,\cdots, a_g,b_g| \left[a_1,b_1\right] \cdots \left[a_g,b_g\right] \rangle $$
and $(\pi(M_g))^{ab} = \mathbb{Z}^{2 g}$ is the universal abelian covering of $M_g$. Observe that $N =p_*(\pi(\tilde{X})) \lhd \pi(M_g)$, $ A = \left[ \pi(M_g), \pi(M_g)\right] \lhd \pi(M_g)$ and $A$ is the smaller normal subgroup of $\pi(M_g)$ such that $\pi(M_g)/A$ is abelian, therefore, since $\pi(M_g)/N$ is abelian, we have $A \subset N$. By the third isomorphism theorem, there is a 1-to-1 correspondence between the normal subgroup of $\pi(M_g)$ and $\pi(M_g)/A$, futhemore $N$ is a normal contanning $A$ iff $N/A \lhd \pi(M_g)/A$, then $$ \frac{\pi(M_g)/A}{N/A} \cong \frac{\pi(M_g)}{N}\cong \mathbb{Z}^n $$
But how can I conclude that $n \leq 2g$? I think that its follows from the fact that, if $q:\tilde{A} \to X$ is the universal abelian covering of $X$, then $q:\tilde{A} \to \tilde{X}$ is a covering space of $\tilde{X}$, futhemore, $\mathbb{Z}^n \leqslant \mathbb{Z}^{2g}$ iff $n\leq 2g$ but I cannot conlude not from this. There is some relation between the abeilan universal covering, the deck transformation of $G(\tilde{A})$ and $G(\tilde{X})$? I'm confused about the reciprocal, could someone help me?