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Is there a path connected topological space such that its fundamental group is non-trivial, but its first homology group is trivial?

Since the first homology group of a space is the abelianization of the fundamental group, we are looking for a non-trivial group whose abelianization is trivial. Is there such a group?

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A group whose abelianization is trivial is called perfect, and there are many such groups. In particular, any nonabelian finite simple group is perfect, so $A_5$, for example, is perfect. $A_5$ is in fact the smallest nontrivial perfect group. So any space with fundamental group $A_5$ is an example.

A famous example which almost has fundamental group $A_5$ is Poincare dodecahedral space. This is a closed $3$-manifold which is the quotient of $S^3$ by an action of the binary icosahedral group, which is an extension of $A_5$, and like $A_5$ is perfect. Any closed $3$-manifold with perfect fundamental group necessarily has the same homology as a sphere, and this showed Poincare that for a $3$-manifold to be a sphere it did not suffice that its homology was the same as a sphere, motivating the fundamental group condition in the statement of the Poincare conjecture.

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Yes, for any group $G$ there exists a (path connected) $K(G,1)$, and there are non-trivial groups with trivial abelianization.

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This is exactly the type of space that Poincare constructed to show that homology was not enough to distinguish three-manifolds from the three-sphere. He took a dodecahedron and glued opposite faces with a minimal clockwise twist. The resulting space is a homology sphere --- it has the homology groups of $\Bbb S^3$, but has nontrivial $\pi_1$.

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