There is a well known theorem which states that:

Theorem(Stallings): For any immersion $f$ from a finite graph $D$ to $G$ there is a finite-sheeted covering space $D '$ of $G$ that extends $f$. More precisely, there is an embedding of $D$ into $D'$ and the restriction of the covering map to $D$ coincides with $f$.

This implies that every f.g. subgroup $H$ of a free group $F$, then $H$ is a free factor in a finite-index subgroup of $F$.

My question is if we have a group that is a free product of groups, which has finite Kurosh rank (it can be written as a free product of finitely many freely indecomposable groups), then if there is in the literature something similar to the Stallings' theorem, using graph of groups or Bass- Serre trees instead of finite graphs.

To be more precise, if we start with a graph of groups $\Gamma$ which corresponds to a free product decomposition of $G$, so in particular it has trivial edges stabilisers, then is there something like 'finite covering' of $\Gamma$ which corresponds to finite index subgroups of $G$, and starting with any immersion of a graph of groups to $\Gamma$ is it possible to be completed to a 'finite cover'?

If anyone knows something similar or at least some special cases, it would be very helpful. Thanks a lot in advance.

• Maybe Theorem 14, p. 56, in Serre's book on trees is what you are looking for? – Matthias Wendt Jul 9 '14 at 15:09
• Τhanks a lot for your answer, but if we have the same edition this is Kurosh Theorem. I will edit my question to be more precise. – user75691 Jul 9 '14 at 15:46

This is one of those 'well known' things that every one does in different ways. I believe the notion of a covering space of a graph of groups was worked out by Bass. The details are rather technical - I prefer to think about coverings of graphs of spaces, which comes to the same thing. (I believe Scott and Wall were the first to adopt this point of view.)

In your case, every edge space can be taken to be a point. I proved some generalizations of the theorem of Stallings you mention in the following papers:

• Elementarily free groups are subgroup separable. Proc. Lond. Math. Soc. (3) 95 (2007), no. 2, 473–496.
• Hall's theorem for limit groups. Geom. Funct. Anal. 18 (2008), no. 1, 271–303.

The ideas in these papers are similar to some of Wise's work (prior to the theory of special cube complexes).

In the language of those papers, the theorem you want can be stated as follows:

Theorem: Let $X$ be a graph of spaces in which every edge space is a point and $\pi_1X$ is finitely generated. If $Y\to X$ is a finite-sheeted precovering and $\pi_1Y$ is finitely generated then $Y\to X$ can be completed to a finite-sheeted covering map $\widehat{X}\to X$.

You should look at the linked papers for the formal definition of a precovering, but it's essentially an immersion that restricts to a covering map of the vertex spaces.

It's not at all hard - the proof is a direct generalization of the theorem of Stallings that you quote.

• @HRJW, you need some separability hypotheses here don't you? – Benjamin Steinberg Jul 9 '14 at 19:53
• @BenjaminSteinberg - no, it's all bound up in the definition of a precovering! – HJRW Jul 9 '14 at 20:46
• @HRJW, perhaps you should define a precovering? – Benjamin Steinberg Jul 9 '14 at 20:58
• @BenjaminSteinberg, well, I've added a rough definition. I can't define it formally without recalling the definition of graphs of spaces etc. – HJRW Jul 9 '14 at 21:09
• Thanks a lot both of for your answers. This theorem is very similar to that I want. Can I ask you something? I would like to ask if we need the hypothesis of finitely generated, because we want that every vertex space to be finite? – user75691 Jul 10 '14 at 9:57

I believe the algebraic variant of what you are asking is the property of subgroup separability in groups (which is very well studied). Recall that a group, $G$, is subgroup separable if for any finitely generated group, $\Delta \leq G$ and any $g \notin \Delta$, there exists a finite index subgroup $H \leq G$ such that $g \notin \Delta H$.

You cannot expect to generalize subgroup separability of free groups to arbitrary free products of freely indecomposable groups, as subgroup separability implies residual finiteness (and any infinite simple group is not residually finite). However, free groups and, more generally, surface groups are known to be subgroup separable. In fact, any group that contains a subgroup of finite index that is subgroup separable is subgroup separable. For the proofs of the aforementioned facts see Peter Scott's Subgroups of surface groups are almost geometric from JLMS, 1978. The main ideas in this paper were greatly generalized: the modern form of this is now called the canonical completion of Haglund and Wise. See Section 6 in Special Cube Complexes from GAFA, 2008.

• Thanks a lot for your answer. Yes I understand that it is not possible to expect that the algebraic property can be generalised so much, but this is that I am actually asking is some notion of 'covering' using some space that involves the graph of groups decomposition of a free product(or the Bass-Serre tree) and somehow related to finite index subgroups. I am not sure of course, if this possible. – user75691 Jul 9 '14 at 17:25
• Ok :) A remark: the canonical completion works in the context of virtually special groups and it is not hard to see that free products of virtually special groups are virtually special. – Khalid Bou-Rabee Jul 9 '14 at 17:42
• Thanks a lot again for your answer. I will see again this paper in more detail. – user75691 Jul 9 '14 at 17:50
• Also, check out this very nice paper of Agol, Groves, and Manning: arxiv.org/pdf/1405.0726.pdf . He discusses general graph of groups in Section 3. – Khalid Bou-Rabee Jul 9 '14 at 17:53
• @user75691: See also Sageev's lecture notes math.utah.edu/pcmi12/lecture_notes/sageev.pdf – Seirios Jul 10 '14 at 9:45

What you can prove is that if you take a $2$-complex $X$ with fundamental group $A\ast B$ and you take a finite subcomplex $K$ of a covering $X'$ of $X$ such that the fundamental group of $K$ maps under the covering to a subgroup of $A\ast B$ which is closed in the profinite topology, then the restriction of the covering map to $K$ can be extended to a finite sheeted covering. Moreover, you can do it so that the Kurosh decomposition of your finitely generated closed subgroup is compatible with the Kurosh decomposition of the finite index subgroup (e.g., the conjugates of factors are contained in conjugates of factors and the free part of the smaller group is a free factor of the free part of the larger group).

A topological proof in the case $A,B$ are subgroup separable (but the proof can be made to work under the weaker assumption above) is given in our paper http://www.sciencedirect.com/science/article/pii/S0022404902002104

An equivalent formulation in the language of Bass-Serre theory can be found in the paper of Ribes and Zalesskii PROFINITE TOPOLOGIES IN FREE PRODUCTS OF GROUPS

http://www.worldscientific.com/doi/abs/10.1142/S0218196704001992http://www.worldscientific.com/doi/abs/10.1142/S0218196704001992

I suppose a great resource for Bass Serre trees manipulations, including foldings of Bass Serre trees, is Rylee Lyman's thesis. Although the approach concerns the general case (not especially free products) and I can't find the exact result you are looking for, the techniques displayed are (I suppose) strongly related to your question.