Converse to Milnor's theorem on manifolds with nonnegative Ricci curvature

A better question is:

Given a group $\Pi$, is there a compact manifold $M$ with non-negative Ricci curvature such that $\pi_1(M)=\Pi$?

The answer is given in "On fundamental groups of manifolds of nonnegative curvature" by Wilking. Here is the main result:

_{(source: psu.edu)}

For a compact counterexample, take any nilmanifold $N/H$ modulo the action of a freely acting cocompact lattice $\Lambda$, assuming $N/H$ is not just Euclidean space and $\Lambda$ is not just virtually abelian. For instance, $N=N/H$ is the $3 \times 3$ real Heisenberg group and $\Lambda$ is the $3 \times 3$ integer Heisenberg group. The proof that this is a counterexample is to apply Wilking's theorem in Anton's answer, and the theorem that a finitely generated nilpotent group has polynomial growth.