Theorem: If $M$ is a compact Riemannian manifold of dimension $n$ and has nonnegative Ricci curvature, then $b_1(M) \le \operatorname{dim}M = n$, where $b_1$ is the first Betti number.
Proof: There is a natural surjection $h : \pi_1(M) \to H_1(M,\mathbb Z) \cong \mathbb Z^{b_1} \times T$ that maps loops to cycles, and where $T$ is a finite abelian group. The structure of the fundamental group shows that $h(G) \subset T$ since $G$ is finite. Thus, we obtain a surjective homomorphism $\pi_1(M) / G \to \mathbb Z^{b_1}$, where $\pi_1(M)/G = \pi_1(M) \cap \operatorname{Iso}(\mathbb R^k)$. Moreover, the image of $\pi_1(M) \cap \mathbb R^k = \mathbb Z^k$ in $\mathbb Z^{b_1}$ has finite index. This shows that $b_1\le k \le n$.
My question:
How does $h(G) \subset T$ imply that the map $\pi_1(M) / G \to \mathbb Z^{b_1}$ is surjective?
I think we are restricting $h$ to the domain of $\pi_1(M) / G$, but if so I am not sure how $h(G) \subset T$ implies that the homomorphism from $G/M$ takes elements only into $\mathbb Z^{b_1}$ and not $T$.
Consider the surjective homomorphism $p : \mathbb{Z}^{b_1}\times T \to \mathbb{Z}^{b_1}$ given by $(z, t) \mapsto z$. The composite $p\circ h : \pi_1(M) \to \mathbb{Z}^{b_1}$ is a surjective homomorphism.
Note that $(p\circ h)(G) = 0$ as $h(G) \leq T$, so $G \leq \ker(p\circ h)$. Therefore, there is a surjective homomorphism $\pi_1(M)/G \to \pi_1(M)/\ker(p\circ h)$.
By the first isomorphism theorem, we have $\pi_1(M)/\ker(p\circ h) \cong \mathbb{Z}^{b_1}$. This identification gives us a surjective group homomorphism $\pi_1(M)/G \to \mathbb{Z}^{b_1}$.