there is no 3-manifold such its fundamental group is isomorphic to direct product Z⊕Z⊕Z⊕Z.

Question

Prove that there is no 3-manifold such its fundamental group is isomorphic to Z⊕Z⊕Z⊕Z.I tried to use Betti number or use the theorem that says if it is 3-manifold then if it is orientable any Alexander module produced in G is self-reciprocal and if it is non-orientable there exists an index 2 subgroup H of G such that any Alexander module produced in H is self-reciprocal.

Answer 1

Here is an elementary argument. Suppose $M$ is a closed 3-manifold such that $\pi_1 M \cong \mathbb{Z}^4$. Without loss of generality, we may suppose $M$ is orientable by passing to the orientation double cover $\tilde{M}$ and noting that $\pi_1 \tilde{M}$ is also free abelian of rank 4 because it is an index two subgroup of $\pi_1 M \cong \mathbb{Z}^4$.

Now by the Hurewicz isomorphism and then the universal coefficients theorem and Poincare duality, we deduce that $H_2(M; \mathbb{Z}) \cong \mathbb{Z}^4$.

On the other hand, we can attach cells of dimension $\geq 3$ to $M$ to obtain an Eilenberg-Mac Lane space $K(\mathbb{Z}^4,1)$. Since we are only attaching cells of dimension $\geq 3$, the map $M \hookrightarrow K(\mathbb{Z}^4,1)$ induces a surjection in $H_2$.

But $K(\mathbb{Z}^4, 1) \simeq (S^1)^4$, and $H_2((S^1)^4; \mathbb{Z}) \cong \mathbb{Z}^6$. So we have a surjection $$H_2(M) \cong \mathbb{Z}^4 \to \mathbb{Z}^6 \cong H_2(K(\mathbb{Z}^4, 1)).$$ This gives a contradiction.