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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.

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So the question is whether this necessary condition is sufficient? Nice question. –  Lee Mosher Jul 29 '12 at 20:51
@Ralph: The Stallings-Swan theorem says that if $H^i(G;\mathbb{Z}G)$ is zero for all $i > 1$ then $G$ is free. So Higman's group, despite being acyclic, should have nontrivial $H^i(G;\mathbb{Z}G)$ for some $i>1$, although I do not know for what value of $i$ this would be true. –  Lee Mosher Jul 29 '12 at 22:41
@Ralph: Higman's group is not a counter-example since it has nonzero $H^2$ with $ZG$ coefficients (Higman's group has cohomological dimension 2). –  Misha Jul 30 '12 at 4:26
Mark - it can't factor through a free group because free groups are Hopfian. Specifically, let $f:G\to H$ be Tom's example and suppose it factors as $f_1:G\to F_2$ composed with $f_2:F_2\to H$. (Note that the intermediate free group must be of rank two.) Now let $g: F_2\to G$ be any epimorphism. Then $f_1\circ g:F_2\to F_2$ is an epimorphism with non-trivial kernel, which contradicts the fact that free groups are Hopfian. –  HJRW Aug 1 '12 at 10:17
@HW: That's great, thanks! And doesn't your argument generalize? Can't we say that if $f\colon G\to H$ is an epimorphism between f.g. groups of the same rank, then $f$ does not factor through a free group? –  Mark Grant Aug 1 '12 at 10:52
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2 Answers

$f: G \to H$ 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 $H$.

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AS N is typically infinitely generated, this criterion is not very useful in practice. –  HJRW Jul 31 '12 at 16:41
@HW: Can you give an example of a hom. $f$ where you find the criterion difficult to apply ? –  Ralph Jul 31 '12 at 22:12
I would be very interested in a "simple" proof of the fact that the abelianization map $G\to G/[G,G]$ factors through a free group (which needs to be of rank 2) when $G$ has has a presentation with three generators $x_1,x_2,x_3$ and one relator $x_1x_2x_1^{-1}x_3x_1x_3^{-1}x_2x_3x_2^{-1}$ –  Roberto Frigerio Jul 31 '12 at 23:26
Sorry, in the example above I meant "does not factor". –  Roberto Frigerio Jul 31 '12 at 23:27
Roberto, thanks for the example. To have the right understanding: With the change "does not factor" the bracket "(which needs to be of rank 2)" is obsolete ? –  Ralph Aug 1 '12 at 7:43
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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.

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You say, if $H$ has torsion, $f$ can't factor through a free group. If I'm not missing something $f: \mathbb{Z} \to \mathbb{Z}/2$ is an example that has (trivially) such a factorization. –  Ralph Jul 29 '12 at 21:14
I think he is restricting to the case that $H$ is the abelianization of $G$. –  Lee Mosher Jul 29 '12 at 21:16
@Lee: Ok, I see. Thanks for the clarification. –  Ralph Jul 29 '12 at 21:24
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