When does a homomorphism factor through a free group? - MathOverflow

When does a homomorphism factor through a free group?

2012-07-29T19:11:09Z

Let $f\colon\thinspace G\to H$ be a surjective homomorphism of finitely generated groups. Are there any methods to decide whether $f$ factors through a free group? That is, does there exist a free group $F$ and homomorphisms $g\colon\thinspace G\to F$ and $h\colon\thinspace F\to H$ such that $f=h\circ g$?

I know of one necessary condition, given by cohomology. If $f$ factors through a free group then $f^*\colon\thinspace H^i(H;M)\to H^i(G;M)$ is zero for all $i>1$ and all $H$-modules $M$ (since free groups have cohomological dimension one).

This question was inspired by Tom Goodwillie's answer to my earlier question on cohomological dimension of group homomorphisms.

Answer by Roberto Frigerio for When does a homomorphism factor through a free group?

2012-07-29T19:56:41Z

Even in special cases your question seems to be very hard. Let me mention just a very special case, i.e. the case when $H=G/[G,G]$ is the abelianization of $G$ and $f\colon G\to H$ is the natural projection. In this case, your problem reduces to the question whether $corank(G)=b_1(G)$, where $corank(G)$ is, by definition, the maximal rank of a free group which is a quotient of $G$, and $b_1(G)$ is the first Betti number of $G$, i.e. the rank of $H$.

It is not difficult to show that, if $H$ has torsion, then $f$ cannot factor through a free group. However, if $H$ is free, then the situation is quite complicated. For example, if $G$ is the fundamental group of a link complement, then $corank(G)=b_1(G)$ if and only if the link is a homology boundary link.

In principle, results by Makanin about equations in free groups show that, if a finite presentation of $G$ is given, then there exists an algorithm deciding whether $corank(G)=b_1(G)$ or not. However, this algorithms cannot be exploited in practice even when dealing with very short presentations. On the other hand, computable (noncomplete) obstructions to the equality $corank(G)=b_1(G)$ may be obtained via the analysis of Alexander module invariants of $G$. Several results in this spirit may be found in the book "Algebraic invariants of links" by J.A. Hillman.

Answer by Ralph for When does a homomorphism factor through a free group?

2012-07-29T20:28:30Z

$f: G \to H$ factors through a free group iff there is a subgroup $N \le \operatorname{ker}(f)$ and a free group $F$ such that $G = N \rtimes F\;\;$ ($N$ normal ).

Proof: $(\Rightarrow)$ Let $G \xrightarrow{g} E \xrightarrow{} H$ be a factorization of $f$ with $E$ free and let $N$ be the kernel of $g$. Clearly $N \le \operatorname{ker}(f)$ and as a subgroup of a free group, $F := G/N$ is itself free. Thus we have an extensions $1 \to N \to G \to F \to 1$ that splits since $F$ is free.

$(\Leftarrow)$ Suppose $N\le \operatorname{ker}(f)$ and $G = N \rtimes F$ with $F$ free. Hence $N$ is normal in $G$ and because $N \le \operatorname{ker}(f)$, $f$ has a factorization $G \to G/N=F \to H$ through a free group.

Remark: That $f$ is surjective wasn't used. But if we know that $f$ is surjective, we can conclude that the rank of $F$ is greater or equal than the minimal number of generators of $H$.