TITLE: There is no 3-dimensional manifold with fundamental group isomorphic to direct sum of four copies of $\Bbb Z$
QUESTION [6 upvotes]: I want to show that there doesn't exist any $3$-dimensional manifold for which the fundamental group is isomorphic to the direct sum $\Bbb Z\oplus \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$
I have calculated the fundamental group of $K_{3,3}$ graph and $K_5$ graph and if i have done right it was the direct sum of four copies of $\Bbb Z$  for both of them...

REPLY [4 votes]: Here is a proof for closed manifolds avoiding the sphere theorem. It is unclear to me if something like this can be made workable for manifolds with boundary.
Given a space $X$ with fundamental group $\pi$, the natural map $f: H_2(X) \to H_2(\pi)$, induced by the map $X \to B\pi$ classifying the universal cover, is surjective. This follows because one may add cells of rank 3 and higher to $X$ to make a space $X'$ with no higher homotopy groups, and thus factor the map up to homotopy as $X \to X' \xrightarrow{\simeq} B\pi$. In particular, adding cells of degree 3 and higher can only add relations to $H_2(X)$.
Now $H_2(\Bbb Z^4) = \Lambda^2 \Bbb Z^4 = \Bbb Z^6$, but of course for an oriented closed 3-manifold, $H_2(Y) = H^1(Y)$, and $H^1(Y) = \text{Hom}(\pi_1 Y, \Bbb Z)$. In your case, this is of course $\Bbb Z^4$. This is a contradiction.
If your 3-manifold wasn't oriented, pass to the double cover just as in Camilo's answer, which in this case has the same fundamental group.