Let $M$ be a compact n-dimensional Riemannian manifold with non-negative Ricci curvature. Then its universal cover $\tilde{M}$ is isometric to $\mathbb{R}^p\times N$ for some $p\leqslant n$ and $N$ compact. Consider the group $G$ of isometries on $N$ that are split off by the action of $\pi_1(M)$ on $\tilde{M}$.

**We know that $\pi_1(M)$ act discretely on $\tilde{M}$, why does G act discretely on N? Hence it's finite.**

Denote the isometry group of $\mathbb{R}^p$ by $Iso(\mathbb{R}^p)$. Consider the projection map $$ \pi_1(M) \to Iso(R^p), $$ the kernel is $G$. The image $H$ is a subgroup of $Iso(\mathbb{R}^p)$ acting discretely on $\mathbb{R}^p$. The Bieberbach theorem says that $H$ contains a normal free abelian subgroup $\mathbb{Z}^p$ such that $[H:\mathbb{Z}^p]<C$.

Let us prove this. Since any element $\alpha \in Iso(\mathbb{R}^p)$ is in the form: $\alpha(x)=Ax+a$, where $A\in O(p-1)$ is a rotation, $a\in \mathbb{R}^p$ is a translation. Consider the projection $$ \alpha=(A,a) \in H \to A \in O(p-1). $$ The kernel is a subgroup acting discretely on $\mathbb{R}^p$ by translation, it's $\mathbb{Z}^p$.

**Why is the image finite? i.e. Why does the rotation part act discretely on $\mathbb{R}^p$?**