TITLE: Variety having infinite and perfect $\pi_1$
QUESTION [8 upvotes]: Question. Does there exist a smooth complex projective variety with infinite and perfect fundamental group?

A group $G$ is perfect if its Abelianisation $G/[G,\, G]$ is the trivial group.

REPLY [8 votes]: The simplest example I know of such a group is given by the presentation
$$
\langle a, b, c| a^2, b^3, c^7, abc\rangle.
$$
This group $G$ is obviously perfect and (less obviously) is isomorphic to the fundamental group of a smooth complex-projective variety. The reason for the latter is that $G$ acts holomorphically, properly and cocompactly on the unit disk and contains a finite index torsion-free subgroup.
With this in mind, one uses the following lemma a form of which one can find in the book "Kahler groups":
Lemma. Suppose that $G$ is a group containing a torsion-free normal subgroup $H$ of finite index. Suppose that $G$ is acting holomorphically, properly on a simply-connected complex manifold $X$ such that $X/H$ is biholomorphic to a smooth complex-projective variety. Then $G$ is isomorphic to the fundamental group of a smooth complex projective variety.
Proof. There exists a simply-connected projective variety $Z$ with a free holomorphic action of $G/H$. Now, form the fiber product
$$
Y=(X\times Z)/G
$$
where the action of $G$ on $X$ is as above and $G$ acts on $Z$ via the action of $G/H$. Then $\pi_1(Y)\cong G$ and $Y$ is biholomorphic to a smooth projective variety, since it is the quotient of the projective variety $W=(X/H) \times Z$ by a free holomorphic action of a finite group, namely $G/H$.