Could $nT$ be covered by $mT$?

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I understand that the torus of genus one, $T$, can be covered by three different spaces, each corresponding to a different subgroup of $\pi_1 (T) = \mathbb{Z} \times \mathbb{Z}$: the plane $\mathbb{R}^2$, $S^1 \times \mathbb{R}$ and the torus $T$.

Obviously, the torus of genus $n$, $nT$, can be covered by one copy of itself. We could also argue that we could use multiple copies as a covering space, with the appropriate identifications over $n$ paths. However, using multiple copies doesn't really give us anything new, since the covering space would be $nT$ again.

I'm trying to find other covering spaces for $nT$ and, specifically, if some of those covering spaces could be $mT$ for some $m\neq n, \; n>1$.

The first thing that comes to mind is to exploit the bijection between covering spaces of $nT$ and subgroups of its fundamental group. However, since

$$ \pi_1 (nT) = \langle g_1,g_2, \dots, g_{2n} | g_1 g_2 g_1^{-1} g_2^{-1} \dots g_{2n-1} g_{2n} g_{2n-1}^{-1} g_{2n}^{-1} \rangle, $$

I'm struggling to work with the subgroups of this group and any help would be appreciated.