Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
4  
So the question is whether this necessary condition is sufficient? Nice question. –  Lee Mosher Jul 29 '12 at 20:51
3  
@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
5  
@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
4  
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
2  
@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
show 7 more comments

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 $H$.

share|improve this answer
    
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
show 4 more comments

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.