Is there a flat manifold with trivial first homology?

Question

Is there a closed flat manifold whose fundamental group has trivial abelianization? The famous Hantzsche–Wendt flat manifold has fundamental group with finite abelianization.

Answer 1

Andrzej Szczepański pointed me to Proposition 2.3.13 in the book [Perfect Groups, Derek F. Holt and Wilhelm Plesken, 1989], which gives an answer to my question.

Namely, in a slightly different terminology Proposition 2.3.13 shows that if $G$ is the fundamental group of a closed flat manifold with holonomy group $Q$, then its commutator subgroup $G^\prime = [G,G]$ is the fundamental group of a closed flat manifold with holonomy group isomorphic to $Q^\prime=[Q,Q]$, and moreover, if $Q$ is perfect, then so is $G^\prime$. Recall that a group is perfect if it has no nontrivial abelian quotients, or equivalently, the group equals its commutator subgroup.

It is known that every finite group occurs as holonomy group of a closed flat manifold, see [On the Holonomy Group of Locally Euclidean Spaces, L. Auslander and M. Kuranishi, Annals of Mathematics, 1957].

Thus if we start from any finite perfect group $Q$, and let $G$ be the fundamental group of a closed flat manifold with holonomy $Q$, then $G^\prime$ is perfect and also the fundamental group of a closed flat manifold with holonomy isomorphic to $Q$.

To see why we replace $G$ by $G^\prime$ note that $G^\prime$ and $G^\prime\times \mathbb Z$ have the same holonomy, and the latter group is not perfect.

Answer 2

Here is an idea for making examples. Let $F$ be a free group and let $N$ be a normal subgroup of $F$. Then $F/[N,N]$ is torsion-free. To see this, suppose $w\in F$ has finite order modulo $[N,N]$, and let $E$ be the subgroup of $F$ generated by $w$ and $N$. Note that $E$ and $N$ are free groups, $E/[E,E]$ and $N/[N,N]$ are free abelian groups, and we have $[N,N]\subseteq[E,E]\subseteq N\subseteq E$. Therefore $E/[N,N]$ is torsion-free, being an extension of $[E,E]/[N,N]$ (which is a subgroup of $N/[N,N]$) by $E/[E,E]$. This being true for any choice of $w$ it follows that $F/[N,N]$ is torsion-free.

Now let $P$ be a finite perfect group. Choose a surjective homomorphism $F\twoheadrightarrow P$ with $F$ being a free group of finite rank. Let $N$ be the kernel of this homomorphism and let $G$ denote the quotient $F/[N,N]$. Let $A$ denote the subgroup $N/[N,N]$ of $G$. Then $A$ is a normal abelian subgroup of $G$ and $G/A$ is isomorphic to the finite perfect group $P$. Let $K$ be a normal subgroup of $G$ such that $G/K$ is solvable. Then $G/AK$ is both solvable and perfect and therefore $AK=G$. It follows that $G/K = AK/K \cong A/A\cap K$ is abelian. Thus the solvable quotients of $G$ are abelian and the derived subgroup $H:=[G,G]$ is perfect. Since $G$ is finitely generated and abelian by finite, all its subgroups and quotients are finitely generated and abelian by finite. Thus $H$ is abelian by finite and also normal in $G$. Since $HA=G$ we have $H/H\cap A$ is isomorphic to $P$. Now $H$ is a perfect closed flat manifold group mapping onto the original finite perfect group $P$.