Let $M$ be a closed $3$-manifold with $\pi_2(M) = 0$.
- Can the fundamental group of $M$ contain a subgroup isomorphic to $\mathbb{Z}^4$?
- If the fundamental group of $M$ contains a subgroup isomorphic to $\mathbb{Z}^3$, must that subgroup be of finite index?
Observe that to satisfy those conditions, the universal cover $N$ of $M$ has to be non- compact. Now $\pi_2(N)=\pi_2(M)=0$. So by Poincaré duality for non-compact space $H_k(N)=0$ for all $k$ and thus Hurewich theorem implies $N$ is contractible. So $M$ is a $K(G,1)$ space. Now any cover of this is also of same type. So $\mathbb Z^4$ cannot be its cover since if it is a cover then it has to be homotopically equivalent to a closed 4 manifold $S^1\times S^1\times S^1\times S^1$ which is not possible since it's 4th homology group is trivial. And if $ \mathbb Z^3$ is not a finite index subgroup, then $M$ has a non-compact 3 manifold cover with $\mathbb Z^3$ fundamental group. But then it is homotopically equivalent to $S^1\times S^1\times S^1$ , which is not possible, since it's 3rd homology is zero.
EDIT: One can also see that the only free abelian group $\mathbb Z^n$ that can occur as $\pi_1(M)$ is $\mathbb Z^3$, the fundamental group of $3-torus$, since $n-torus$ is a $K(\mathbb Z^n,1)$ and this has $H_n$ non-zero and $H_i$ zero for $i>n$.