Is it possible to have a finite first homology group but infinite fundamental group?

Question

Question is in the title. Is it possible that $H_1(X) = \pi_1(X) / [\pi_1(X) ,\pi_1(X) ]$ is finite but $\pi_1(X) $ isn't?

How about if $X$ is a smooth manifold? How about if it's a Lie group?

Answer 1

1 + 2. Yes, every finitely presented group is realized as the fundamental group of a smooth, compact manifold (Showing that every finitely presented group has a $4$-manifold with it as its fundamental group), so pick an infinite perfect group.

3. No, a Lie group has an abelian fundamental group so it agrees with its homology.

Answer 2

A Hantzsche-Wendt manifold $M$ is a $3$-dimensional flat manifold with finite $H^1(M,\Bbb Z)$. However, its fundamental group is a Bieberbach group which is an extension of $\Bbb Z^3$ by the finite group $ℤ/2ℤ×ℤ/2ℤ$, and hence it is infinite.

Answer 3

A quick example of such a space is $\mathbb RP^2 \vee \mathbb RP^2$. The fundamental group is the infinite dihedral group by Van Kampen, but Mayer Vietoris shows that its first homology is the klein 4 group.

As previously mentioned, an Eckmann-Hilton argument shows that the fundamental group of a Lie group is Abelian.

Answer 4

Yes, the Hantze wendt manifold is an example, it is finitely covered by a torus, see page 5 of this article.

http://www.personal.soton.ac.uk/np3u12/Cohomology%20of%20Hantzsche-Wendt%20Groups.pdf