On the intersection of tame subgroups in groups acting on trees
Konstantinos Lentzos, Mihalis Sykiotis

TL;DR
This paper investigates the intersection properties of tame subgroups within groups acting on trees, providing bounds on the complexity of their intersections, especially in cases where subgroups act freely on edges.
Contribution
It introduces bounds for the complexity, including Kurosh rank, of intersections of tame subgroups in groups acting on trees, under specific conditions.
Findings
Bounds for Kurosh rank of subgroup intersections.
Results for subgroups acting freely on edges.
Insights into subgroup complexity in tree actions.
Abstract
Let be a group acting on a tree with finite edge stabilizers of bounded order. We provide, in some very interesting cases, upper bounds for the complexity of the intersection of two tame subgroups and of in terms of the complexities of and . In particular, we obtain bounds for the Kurosh rank of the intersection in terms of Kurosh ranks and , in the case where and act freely on the edges of .
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On the intersection of tame subgroups in groups acting on trees
Konstantinos Lentzos and Mihalis Sykiotis
Abstract
Let be a group acting on a tree with finite edge stabilizers of bounded order. We provide, in some very interesting cases, upper bounds for the complexity of the intersection of two tame subgroups and of in terms of the complexities of and . In particular, we obtain bounds for the Kurosh rank of the intersection in terms of Kurosh ranks and , in the case where and act freely on the edges of .
1 Introduction
In 1954, Howson [9] showed that the intersection of two finitely generated subgroups and of a free group is also finitely generated and provided an upper bound for the rank of in terms of and . The Hanna Neumann conjecture, proved independently by Friedman [8] and Mineyev [12] in 2011, says that , where is the reduced rank of a free group .
For free products the situation is analogous. Let be a group. The Kurosh rank, denoted , of a free product decomposition of is defined to be the number of (non-trivial) factors . By the Kurosh subgroup theorem, any subgroup of inherits a free product decomposition , where each is non-trivial and conjugate to a subgroup of a free factor of and is a free group. The (subgroup) Kurosh rank of of with respect to the above splitting of , is the sum , which we again denote by . The reduced Kurosh rank of is defined to be .
Free products also have the Howson property, in the following sense: if , are subgroups of of finite Kurosh rank, then also has finite rank (see [15, Theorem 2.13 (1)] for a proof). In [11], Ivanov proved that if is torsion free, then . It is shown in [1], that if is right-orderable, then the coefficient can be replaced by .
The problem of finding bounds for the “rank” of the intersection of subgroups in free products and more generally in groups satisfying the Howson property has also been considered in [14, 4, 10, 6, 7, 16, 17, 2].
In this paper, we obtain, under appropriate hypotheses, bounds for the complexity of the intersection of tame subgroups in groups acting on trees with finite edge stabilizers.
Let be a group acting on a (simplicial) tree without inversions. A vertex of is called (-) degenerate if for some edge incident to . The corresponding vertex of the quotient graph is also called degenerate. Let be a subgroup of . We denote by the rank of the fundamental group of and by the set of -non-degenerate vertices of . The complexity of with respect to is defined to be the sum , if contains hyperbolic elements, and otherwise. The reduced complexity of with respect to , is defined as . The subgroup of is called tame if either fixes a vertex, or contains a hyperbolic element and the quotient graph is finite, where is the unique minimal -invariant subtree of . By [15, Theorem 2.13], if each edge stabilizer is finite, then the intersection of two tame subgroups , of is again tame. In the case where fixes a vertex, we obviously have .
Finitely generated subgroups are examples of tame subgroups. In the case of free products, finite Kurosh rank implies tameness (see Lemma 2.3) and the complexity of a non-trivial subgroup is exactly its Kurosh rank (see section 2 for more details). Our first main result is the following.
{restatable*}
thmfirstThmOne Let be a group acting on a tree with finite quotient and finite stabilizers of edges and let , be tame subgroups of such that does not fix a vertex of .
If and do not contain degenerate vertices of valence two, then
[TABLE]
where N=\max\big{\{}|G_{x}\cap HK|\,:\,x\in ET\big{\}} and M=\max\{M_{H},M_{K}\}\leq\max\big{\{}|G_{x}|\,:\,x\in ET\big{\}}. 2. 2.
Suppose and satisfy the following property: for each -degenerate (resp. -degenerate) vertex of , the stabilizer (resp. ) stabilizes each edge in the star of . Then
[TABLE]
In particular, if act freely on the edges of , then
[TABLE]
In the special case where both and act freely on , the above inequality was proved by Zakharov in [16].
Now let be the free product of the amalgamated free product of ’s with a finite amalgamated subgroup and , such that is normal in each . Following Dicks and Ivanov [6], we define a_{3}(G_{i}/A)=\min\big{\{}|\Gamma|\,:\,\Gamma\textrm{ is a subgroup of }G_{i}/A\textrm{ with }|\Gamma|\geq 3\big{\}} and \theta(G_{i}/A)=\Big{\{}\frac{a_{3}(G_{i}/A)}{a_{3}(G_{i}/A)-2}\Big{\}}\in[1,3], where .
We represent as the fundamental group of a graph of groups , where is the wedge of copies of (one copy for each factor ) and a bouquet of circles (one for each free generator of ). To each copy of and to the wedge point we associate the group , and to each circle we associate the trivial group. To each of the remaining vertices we associate a factor . Let be the corresponding universal tree.
{restatable*}
thmfirstThmTwo Let be the free product of the amalgamated free product of ’s with a finite amalgamated subgroup and , such that is normal in each . We consider the natural action of on defined above. Suppose that and are tame subgroups (with respect to ) of which act freely on the edges of . Then is tame and
[TABLE]
where and N=\max\big{\{}|gAg^{-1}\cap HK|\,:\,g\in G\big{\}}.
As an immediate corollary we obtain the main result of [11] mentioned above.
It should be noted that the arguments in the proof of Theorem 1, work in a slightly more general setting as well. Thus, with essentially the same proof, we obtain Theorem 3.7 (see also Remark 3.5): If , are tame subgroups of a free product with a finite and normal amalgamated subgroup , then where and is defined as above for .
After posting the first version of this paper on the arXiv, the authors learned from A. Zakharov that he, in collaboration with S. Ivanov, had also recently obtained (unpublished) upper bounds for the Kurosh rank of the intersection of free product subgroups in groups acting on trees with finite edge stabilizers.
Acknowledgements. We are grateful to Dimitrios Varsos for many useful discussions and comments. We are also grateful to the anonymous referee for careful reading of the manuscript and pointing out a mistake in an earlier version.
2 Preliminaries
To fix our notation, we first recall the definition of a graph.
Definition 2.1**.**
A graph consists of a (nonempty) set of vertices , a set of edges , a fixed-point free involution () and a map . The vertex is called the initial vertex of the edge . The terminal vertex of is defined by .
Throughout, let be a group acting on a (simplicial) tree (without inversions, i.e. for any and ). By Bass-Serre theory, for which we refer to [5, 13], this is equivalent to saying that is the fundamental group of the corresponding graph of groups . If , we denote by the -orbit of and by its stabilizer. An element is elliptic if it fixes a vertex of and hyperbolic otherwise. If is a subgroup of containing a hyperbolic element, then there is a unique minimal -invariant subtree which is the union of the axes of the hyperbolic elements of .
We recall that a subgroup of is called tame if either fixes a vertex, or contains a hyperbolic element and the quotient graph is finite. By [15, Prop. 2.2], the subtree is a “core” for the action of on in the sense that , i.e. . From this it follows that the complexity of a tame subgroup is finite.
Finitely generated subgroups of are examples of tame subgroups, since a finitely generated group acting by isometries on , either fixes a point of or else contains a hyperbolic isometry and the quotient graph is finite.
Remark 2.2*.*
We note that if the -stabilizer of each edge is finite and there is a bound on their orders, then any subgroup of consisting of elliptic elements fixes a vertex of ([15, Lem. 2.5]).
If we restrict attention to subgroups of that act edge-freely on , then the Kurosh rank of (with respect ) is defined to be the complexity of .
Let be a free product and a subgroup of . By the Kurosh subgroup theorem, , where for each , ranges over a set of double coset representatives in and is a free group intersecting each conjugate trivially. The (subgroup) Kurosh rank of with respect the above free product decomposition of , denoted by , is the sum , where is the number of all non-trivial factors . Note that the Kurosh rank of is the number of non-trivial factors .
It is not difficult to verify that the numbers , depend only on and the given free product decomposition of . In fact, if is any -tree corresponding to the given decomposition of , then the Kurosh rank of with respect to is equal to the Kurosh rank of the associated free product decomposition of coming from the action of on . Thus, if is non-trivial, then .
Lemma 2.3**.**
Let be a group acting on a tree and a subgroup of that act edge-freely on . If , then is tame.
Proof.
It suffices to consider the case when contains a hyperbolic element. Let be the natural projection given by . Since , there are finitely many vertices of with non-trivial group and finitely many edges of such that is a maximal tree of . Let be the finite subgraph of consisting of and all geodesics in between endpoints of the and . We claim that is connected. To see this, let be a reduced path connecting vertices of such that no edge of lies in . Then is contained in the complement of . Since contains the edges , each component of is a tree, and it is not difficult to see that intersects in only one vertex. It follows that there is an index such that . This means that for some and hence fixes the initial vertex of . From the construction of , is degenerate and therefore , which contradicts the choice of .
Thus, is a connected -invariant subgraph of . It follows that . We conclude that is finite, being a subgraph of . ∎
3 Proofs of the main results
Let be a graph and a vertex of . The star of , denoted , is the set of edges of with initial vertex , i.e. . The valence or degree of in , denoted , is the number of edges in the star of .
Lemma 3.1**.**
Let be a group acting on a tree , let be a tame subgroup of containing hyperbolic elements and let be the graph obtained from by attaching a loop at each -non-degenerate vertex. Then
[TABLE]
where the sum is taken over all vertices of .
Proof.
The reduced rank of a graph is equal to the number of its (geometric-oriented) edges minus the number of its vertices. The minimality of implies that each vertex of of valence one is -non-degenerate. Therefore, every vertex of has valence at least two. Now an easy calculation shows that the sum \sum\big{(}\deg_{\widetilde{X}}([v]_{H})-2\big{)}, over all vertices of , is equal to 2. By construction of , we have which completes the proof. ∎
Lemma 3.2**.**
Let be a group acting on a tree and let and be subgroups of such that . Suppose that and contain hyperbolic elements and that is a vertex of . We consider the graph map given by .
* (provided that they are finite).* 2. 2.
If, moreover, is -degenerate and stabilizes each edge in , then the restriction is an embedding.
Proof.
Suppose that and are two edges in the star of with . Then there are and such that , and . It follows that and thus . Hence . Now, if is an edge in the star of with , then as before and for some and . If we assume further that , then and so . This means that each fiber of the restriction (on stars) has at most elements, and the first assertion follows.
Now, if stabilizes each edge in , then stabilizes and therefore . ∎
In view of this lemma, we define .
The following is our first main result.
\firstThmOne
Proof.
Since does not fix a vertex, it follows from Remark 2.2 that , and contain hyperbolic elements. Let , , be the minimal subtrees of invariant under , , , respectively. Let and be the natural projections (defined as in Lemma 3.2). We consider the map given by . By [3, Proposition 8.7], each fiber , where is an edge or a vertex, has exactly elements. It follows that for each edge the fiber has at most elements.
For convenience we simplify notation by setting , and . As in Lemma 3.1, we construct graphs , and , by attaching a loop at each non-degenerate vertex of , and , respectively.
- By Lemma 3.1, it suffices to show that
[TABLE]
For any pair of vertices , we will show that
[TABLE]
from which (1) follows. The rest of the proof follows similar arguments to those given in [4], [6] and [11]. Let be the vertices of . Since the fiber of any edge of contains at most edges, we have
[TABLE]
We consider three cases depending on whether or not and are degenerate.
Case 1. Suppose that is -non-degenerate and is -non-degenerate. Then , while is equal to or . Hence
[TABLE]
Case 2. Exactly one of , , say , is degenerate. Then each is -degenerate as well, and thus , and . Also, by Lemma 3.2, for each we have .
If , then
[TABLE]
where the last inequality follows because .
On the other hand, if , then
[TABLE]
Case 3. Finally, suppose that , are degenerate in , , respectively. Then each vertex is -degenerate as well and , , . Moreover, by Lemma 3.2, . Suppose that and hence (the other case is handled in the same way).
If , then
[TABLE]
On the other hand, if , then
[TABLE]
Thus, in each case we have
[TABLE]
Since , or equivalently, \deg_{Y}(a)\leq 3\big{(}\deg_{Y}(a)-2\big{)}, it follows that
[TABLE]
This completes the proof of part 1) of the theorem.
- To prove the second part, again by Lemma 3.1, it suffices to show that
[TABLE]
for each pair of vertices . Proceeding exactly as before, we distinguish three cases. In Case 1, where both and are non-degenerate, we get the same inequality. In Cases 2 and 3, by Lemma 3.2 (2), we can now use instead of . Thus in Cases 2 and 3, we obtain respectively (from 4-5 and 7) the inequalities
[TABLE]
and
[TABLE]
It remains only to consider the case when both and are degenerate (in which case we are in Case 3) and , where is the vertex of maximal degree. If , then too, and inequality 8 follows since, by Lemma 3.2 (2), for each . ∎
Corollary 3.3**.**
([16, Theorem 1]) Let be a group acting on a tree with finite quotient and finite stabilizers of edges and let , be finitely generated subgroups of which intersect trivially each vertex stabilizer (and hence they are free groups). Then is finitely generated and
[TABLE]
where N=\max\big{\{}|G_{x}\cap HK|\,:\,x\in ET\big{\}}.
Corollary 3.4**.**
Let be a group acting on a tree with finite quotient, finite stabilizers of edges and infinite vertex stabilizers. If and are subgroups of finite index in , then
[TABLE]
Proof.
If the -stabilizer of every vertex is infinite and both and are of finite index in , then each vertex stabilizer is also infinite under the action of or (being of finite index in the corresponding -stabilizer) and thus Cases 2 and 3 do not occur. ∎
Following [6], given a group , we define a_{3}(G)=\min\big{\{}|\Gamma|\,:\,\Gamma\textrm{ is a subgroup of }G\textrm{ with }|\Gamma|\geq 3\big{\}} and \theta(G)=\Big{\{}\frac{a_{3}(G)}{a_{3}(G)-2}\Big{\}}\in[1,3], where .
In the sequel, we prove that if act freely on the edges, then the coefficient in the above theorem can be replaced by a number , where , by imposing some extra hypotheses on the structure of .
Let , , be a family of groups together with a group , let be a family of monomorphisms and let be the amalgamated free product of ’s with amalgamated subgroup (with respect to ). We can think of each as an inclusion. Let be a free group and let be the free product of and . We construct a graph of groups with fundamental group as follows. The graph consists of a wedge of open edges (i.e. one for each factor and distinct endpoints and , ), together with a wedge of loops , one for each free generator of , attached at a vertex with vertex group . To each edge we associate the group , to each loop we associate the trivial group and to each vertex we associate the group . We denote by the corresponding universal tree.
\firstThmTwo
Proof.
We proceed as in the proof of Theorem 1. With the notation of that proof, we have to prove that
[TABLE]
for any pair of vertices (recall that denotes the vertices of ). Since , act freely on the edges of , it follows, by Lemma 3.2 (2), that we can use instead of . Suppose first that at least one of and is non-degenerate. Then the arguments of Cases 1, 2 of the proof of Theorem 1 apply to show that
[TABLE]
and inequality 11 holds. Thus it suffices to consider the case where is -degenerate and is -degenerate (i.e. Case 3 in the proof of Theorem 1). In this case we have , and , while by Lemma 3.2 (2), . For each , choose a vertex, , of , so that . Note that all lie in the same -orbit. There are two subcases to consider.
(i) and are in the same -orbit, i.e. for some . Notice that . If one of the vertices or has valence , then (each) has valence as well and inequality 11 is obvious. If both and have valence at least , then
[TABLE]
(ii) and are in the same -orbit for some , i.e. there exists such that . As before, we may assume that both and have valence at least .
If , then
[TABLE]
Suppose now that . Let be a set of representatives of left cosets of in . From the construction of , the stars of different vertices of are disjoint, while each edge in the star of , where , is of the form and its terminal vertex is . It follows that there is such that and thus . If we write as , where and , then . It follows that there exists a subset of such that Star([w_{i}]_{\Gamma})=\big{\{}[g_{i}g_{\lambda}e_{j}]_{\Gamma}\,:\,g_{\lambda}\in\mathcal{R}^{i}_{\Gamma}\big{\}} and . In particular, and .
Fix . For any , let be the subset of consisting of all pairs such that \big{(}[g_{i}g_{\lambda}e_{j}]_{H},[g_{i}g_{\mu}e_{j}]_{K}\big{)} is the image under of some edge in the star of in . Let denote the natural epimorphism. Note that the restriction of on is a bijection.
We will show that \phi(C_{k})=\big{\{}\big{(}\phi(g_{\lambda}),\phi(g_{\mu})\big{)}:\,(g_{\lambda},g_{\mu})\in C_{k}\big{\}} is a single-quotient subset of , in the terminology of [6], i.e. that the product is constant for all pairs . Suppose that and are edges in the star of and that \pi(y_{t})=\big{(}[g_{i}g_{\lambda(t)}e_{j}]_{H},[g_{i}g_{\mu(t)}e_{j}]_{K}\big{)}, . We want to show that . From the above analysis, we can write for some , , and thus \pi(y_{t})=\big{(}[g_{k}g_{s(t)}e_{j}]_{H},[g_{k}g_{s(t)}e_{j}]_{K}\big{)}. It follows that \big{(}[g_{k}g_{s(1)}e_{j}]_{H},[g_{k}g_{s(1)}e_{j}]_{K}\big{)}=\big{(}[g_{i}g_{\lambda(1)}e_{j}]_{H},[g_{i}g_{\mu(1)}e_{j}]_{K}\big{)}, and that \big{(}[g_{k}g_{s(2)}e_{j}]_{H},[g_{k}g_{s(2)}e_{j}]_{K}\big{)}=\big{(}[g_{i}g_{\lambda(2)}e_{j}]_{H},[g_{i}g_{\mu(2)}e_{j}]_{K}\big{)}. Hence there are , and such that
[TABLE]
By normality of in , the stabilizer of any edge in the star of is equal to . Therefore our assumption that is degenerate implies that and . Now, from the first two equalities above we deduce that
[TABLE]
while from the last two
[TABLE]
The above relations imply that and . Thus, , from which it follows that .
Our aim is to apply [6, Corollary 3.5], which requires pairwise-disjoint, single-quotient subsets. Note that if the intersection is nonempty, then there are edges and in and , respectively, such that . Thus, for each , we choose a subset of with maximum such that the restriction of on the union is an injection. In particular, they are pairwise-disjoint. Since the inverse image of any edge of under contains at most elements, we have |Star_{X}(v_{1})|+\cdots+|Star_{X}(v_{n})|\leq N\big{(}|F_{1}|+\cdots+|F_{n}|\big{)}. If denotes the subset of corresponding to edges of , then are pairwise-disjoint. It follows that are pairwise-disjoint, single-quotient subsets of and [6, Corollary 3.5] applies to show that
[TABLE]
Finally
[TABLE]
This completes the proof. ∎
Remark 3.5*.*
The analogous theorem with the same proof is valid for fundamental groups of graphs of groups defined as follows. The subject graph is the same as the one defined previously (prior to Theorem 1). To the terminal vertex of we associate the group , to the common initial vertex of ’s we associate the finite group , and to each open edge we associate a subgroup of normally embedded in such that for some (this means that the “central” vertex is -degenerate and thus in Case (i) of the proof). To each loop we associate the trivial group. We need normality of in in order to make the natural map a homomorphism (and thus the same arguments in Case (ii) work equally well to this more general setting).
As a corollary, we obtain the main result of Ivanov in [11] (in fact our proof can be slightly modified to generalize [6, Theorem 6.3] as well).
Corollary 3.6**.**
Suppose that , are subgroups of a free product and , have finite Kurosh rank , . Then the intersection also has finite Kurosh rank and
[TABLE]
In particular, if is torsion-free (or more generally, every finite subgroup of has order at most ), then
[TABLE]
Proof.
By Lemma 2.3 and the comments preceding it, the subgroup Kurosh rank is equal to the Kurosh rank with respect to (i.e. , where is as above) and finite Kurosh rank implies tameness. ∎
In the case of free products with a finite, normal subgroup amalgamated, we can use the same arguments to improve the bound for the complexity of the intersection of tame subgroups.
Let , , be a family of groups together with a group and let be the amalgamated free product of ’s with amalgamated subgroup (with respect to a family of monomorphisms, regarded as inclusions). We construct a tree of groups with fundamental group as usual. The tree consists of a wedge of open edges (one for each factor ) attached at a vertex (where ) with vertex group . To each edge we associate the group and to each vertex we associate the group . We denote by the corresponding universal tree.
Theorem 3.7**.**
Let be the amalgamated free product of ’s with a finite and normal amalgamated subgroup . We consider the action of on defined above. If and are tame subgroups (with respect to ) of , then is tame and
[TABLE]
where .
Proof.
The proof is exactly the same as the proof of Theorem 1. There are two things to note:
- (a)
The normality of in and the fact that is a -degenerate vertex imply that for each subgroup of the -stabilizer of the star of any -degenerate vertex of is equal to and therefore Lemma 3.2 (2) applies (i.e we can again use instead of to obtain inequality 12).
- (b)
Using the notation of the proof of Theorem 1, the relations 13 and 14 now give and . Since is the kernel of , we again conclude that .
∎
Remark 3.8*.*
In general, there are examples (see [11, 16]) showing that the bounds obtained in the previous two theorems are sharp.
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