How to prove that $\Gamma$ is finitely generated?

Let $\pi \colon \tilde{M} \to M$ be the abelian cover, where $M$ is a compact, oriented smooth and finite-dimensional manifold. Let $\alpha \in \Omega^1(M)$ be a closed 1-form such that the kernel of its period homomorphism $$\phi_{[\alpha]} \colon \pi_1(M) \to \mathbb{R}$$ is equal to the group $\pi_\#(\pi_1(\tilde{M}))$.

Now we can define the quotient $\Gamma=\pi_1(M)/\ker\phi_{[\alpha]}$ which is isomorphic to the group of deck transformations $\text{Deck}(\tilde{M}).$

Apparently $\Gamma$ turns out to be isomorphic to $\mathbb{Z}^m$ which I tried to prove as follows:

The period homomorphism descends to an injective homomorphism $\tilde{\phi}_{[\alpha]} \colon \Gamma \to \mathbb{R}$. With this it is easy to verify that $\Gamma$ is abelian and torsion-free.

By showing that $\Gamma$ is finitely generated one could conclude that $\Gamma \cong \mathbb{Z}^m$ using the Classification of Finitely Generated Abelian Groups, but I was not able to do so:

One idea was to somehow use the fact $M$ is a compact manifold and then apply Corollary A.8./A.9. in Hatcher, but my algebra is rusty so I'm not sure if I can recover the needed information on the quotient $\pi_1(M)/\ker\phi_{[\alpha]}$.

Or maybe this follows from some "covering space" theory.

In any case, any help is greatly appreciated!

$\textbf{Edit:}$ Thanks to anomaly for the answer. Maybe someone can give an alternate proof that does not invoke seriously involved results like the ones I quoted from Hatcher.

• Quotients of f.g. groups are also f.g. (The projections of the generators also generate the quotient.) – anomaly Dec 12 '17 at 17:20
• @anomaly lol, I had the feeling this might be a dumb question. Thanks! – noctusraid Dec 12 '17 at 17:23