Same homology but different fundamental group

Question

Let $M$ and $N$ be connected, closed, orientable $3$-manifolds. Then I'm able to prove the following statement :

If $\pi_1(M)\simeq\pi_1(N)$ then $H_i(M)\simeq H_i(N)$ for all $i$.

But I wonder if the converse is also true.

If $H_i(M)\simeq H_i(N)$ for all $i$ then $\pi_1(M)\simeq\pi_1(N)$.

I believe the statement is false and tried to find counterexample. Since the abelianization of fundamental group is the first homology group, I think the possible counterexample would contain $\Bbb CP^n$ or $\Bbb RP^n$ with some restriction on $n$ ($M$ and $N$ should be orientable). But it seems such counterexamples is not very easy to find so now I'm guessing the statement is in fact true. Could you help?

Answer 1

The Poincare homology sphere has the same homology as the $3$ sphere but a nontrivial fundamental group.