Hyperbolically embedded virtually free subgroups of relatively hyperbolic groups
aa r X i v : . [ m a t h . G R ] M a y Hyperbolically embedded virtually free subgroupsof relatively hyperbolic groups
Yoshifumi Matsuda ∗† , Shin-ichi Oguni ‡ , Saeko Yamagata § Abstract
We show that if a group is not virtually cyclic and is hyperbolic relativeto a family of proper subgroups, then it has a hyperbolically embeddedsubgroup which contains a finitely generated non-abelian free group as afinite index subgroup.Keywords: relatively hyperbolic groups; hyperbolically embedded sub-groups; strongly relatively undistorted subgroups; almost malnormal sub-groups.2010MSC: 20F67; 20F65
The notion of relatively hyperbolic groups was introduced in [6] and has beenstudied by many authors (see for example [2], [4], [5] and [13]). In this paper weconsider relatively hyperbolic groups in accordance with a definition due to D.Osin [13, Definition 2.35]. Note that for certain cases (e.g. for finitely generatedgroups), this definition has several equivalent formulations (see for example [7,Sections 3 and 5]).In [14], D. Osin introduced the notion of hyperbolically embedded subgroupsof a relatively hyperbolic group.
Definition 1.1. ([14, Definition 1.4]) Let G be a group which is hyperbolic rel-ative to a family K of subgroups. A subgroup H of G is said to be hyperbolicallyembedded into G relative to K if G is hyperbolic relative to K ∪ { H } . ∗ Graduate School of Mathematical Sciences, University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo, 153-8914 Japan, [email protected] † Supported by the Global COE Program at Graduate School of Mathematical Sciences,the University of Tokyo, and Grant-in-Aid for Scientific Researches for Young Scientists (B)(No. 22740034), Japan Society of Promotion of Science. ‡ Department of Mathematics, Faculty of Science, Ehime University, 2-5 Bunkyo-cho, Mat-suyama, Ehime, 790-8577 Japan, [email protected] § Faculty of Education and Human Sciences, Yokohama National University, 240-8501 Yoko-hama, Japan, [email protected] G is infinite and hyperbolic relative to a family K of proper subgroups,then there exists a virtually infinite cyclic subgroup of G which is hyperbolicallyembedded into G relative to K (see [14, Corollaries 1.7 and 4.5]). Also if G istorsion-free, hyperbolic and not cyclic, then it contains a free subgroup of ranktwo which is quasiconvex and malnormal in G , that is, hyperbolically embeddedinto G relative to the empty family ∅ (see [9, Theorem C] and [2, Theorem 7.11]).In this paper we show the following (see also Theorem 5.1). Theorem 1.2.
Suppose that a group G is not virtually cyclic and is hyperbolicrelative to a family K of proper subgroups. Then there exists a finitely generatedand virtually non-abelian free subgroup of G which is hyperbolically embeddedinto G relative to K . Moreover if G is torsion-free, then it contains a freesubgroup of rank two which is hyperbolically embedded into G relative to K .We refer to [12, Theorems 1.4 and 1.5] for applications of this theorem tothe study of convergence actions of groups. We also refer to [11, Theorem 6.3]for another application. Remark 1.3.
After the first version of this paper appeared, the notion of ahyperbolically embedded subgroup was further generalized in [3]. It turns outthat we can prove a stronger version of Theorem 1.2 by using the argument inthe proof of [3, Theorem 6.14 (c)] (see [12, Appendix B] for details). In whatfollows, hyperbolically embedded subgroups which we consider are those in thesense of [14].In Section 2, we recall the fact that hyperbolically embedded subgroupsof a relatively hyperbolic group are characterized as strongly relatively undis-torted and almost malnormal subgroups. Strongly relatively undistorted freesubgroups of rank two of a relatively hyperbolic group are found in Section 3.In Section 4, we construct almost malnormal subgroups of a virtually free groupwith additional properties. Theorem 1.2 is proved in Section 5.
The strategy of our proof of Theorem 1.2 is based on Osin’s characterizationof hyperbolically embedded subgroups of relatively hyperbolic groups statedbelow.To state the characterization, we begin by introducing several definitions.Let G be a group. For a family K of subgroups of G , we put K = S K ∈ K K \ { } .A subset X of G is called a relative generating set of G with respect to K if G is generated by X ∪ K . The group G is said to be finitely generated relative to K if there exists a finite relative generating set of G with respect to K . When Z is a (possibly infinite) generating set of G , we denote by Γ( G, Z ) the Cayleygraph of G with respect to Z and by d Z the word metric with respect to Z .2 efinition 2.1. Let G be a group which is finitely generated relative to a family K of subgroups. A subgroup H of G is said to be strongly undistorted relativeto K in G if H is generated by some finite subset Y and for some finite relativegenerating set X of G with respect to K , the natural map ( H, d Y ) → ( G, d X ∪K )is a quasi-isometric embedding. Definition 2.2.
Let G be a group and H a subgroup of G . The subgroup H is said to be malnormal (resp. almost malnormal) in G if for every element g of G \ H , the intersection H ∩ gHg − is trivial (resp. finite). Theorem 2.3. ([14, Theorem 1.5]) Let G be a group which is hyperbolic relativeto a family K of subgroups. Then a subgroup H of G is hyperbolically embeddedinto G relative to K if and only if H is strongly undistorted relative to K andalmost malnormal in G .For finitely generated relatively hyperbolic groups, we have the followingcharacterization of strongly relatively undistorted subgroups. Definition 2.4. ([13, Definitions 4.9 and 4.11]) Let G be a group with a finitefamily K of subgroups. Suppose that G is generated by a finite set X . Asubgroup H of G is said to be quasiconvex relative to K in G if there exists aconstant σ ≥ h and h are elements of H and p is a geodesic from h to h in Γ( G, X ∪ K ), then for every vertex v on p , thereexists an element h of H such that d X ( v, h ) ≤ σ . A subgroup H of G is saidto be strongly quasiconvex relative to K in G if H is quasiconvex relative to K in G and for every element K of K and every element g of G , the intersection H ∩ gKg − is finite. Theorem 2.5. ([13, Theorem 4.13]) Let G be a group which is hyperbolicrelative to a finite family K of subgroups. Suppose that G is finitely generated.Then a subgroup H of G is strongly undistorted relative to K if and only if H is strongly quasiconvex relative to K . When a group G is hyperbolic relative to a family K of subgroups, a subgroupof G is said to be parabolic with respect to K if it is conjugate to a subgroup ofsome element of K . The main purpose of this section is to show the following. Proposition 3.1.
Let G be a group which is hyperbolic relative to a family K of proper subgroups and Γ a subgroup of G which is neither virtually cyclic norparabolic with respect to K . If Γ contains an element of infinite order, then itcontains a free subgroup F of rank two which is strongly undistorted relative to K in G .Proposition 3.1 yields the following corollary. Corollary 3.2.
Let G be a group which is not virtually cyclic and is hyperbolicrelative to a family K of proper subgroups. Then G contains a free subgroup F of rank two which is strongly undistorted relative to K in G .3 roof of Corollary 3.2 using Proposition 3.1. It follows from [14, Corollary 4.5]that the group G contains an element of infinite order. Hence the assertionfollows from Proposition 3.1.For the proof of Proposition 3.1, we prepare several lemmas.When a group G is hyperbolic relative to a family K of proper subgroups,an element g of G is said to be parabolic with respect to K if it is conjugate toan element of a subgroup of G which belongs to K . Otherwise g is said to behyperbolic with respect to K . Lemma 3.3.
Let G be a group which is hyperbolic relative to a family K ofproper subgroups and Γ a subgroup of G which is neither virtually cyclic norparabolic with respect to K . Suppose that either G is countable and K is finiteor Γ contains an element of infinite order. Then there exists an element h of Γwhich is of infinite order and hyperbolic with respect to K . Proof.
First suppose that G is countable and K is finite. Then we can considera geometrically finite convergence action of G on a compact metrizable spacesuch that the set of all maximal parabolic subgroups of the action is equal to thecollection of all conjugates of elements of K which are infinite (see for example[7, Definition 3.1]). Since Γ is neither virtually cyclic nor parabolic with respectto K , the restriction of this action to Γ is a non-elementary convergence action.Hence Γ contains an element h which is loxodromic with respect to this action(see [16, Theorem 2T]).Next suppose that Γ contains an element h of infinite order. We only haveto consider the case where h belongs to the conjugate gKg − for some element K of K and some element g of G . Since Γ is not parabolic with respect to K ,we can take an element γ of Γ \ gKg − . By [14, Lemma 4.4], there exists aninteger n such that the element γh n of Γ is of infinite order and hyperbolic withrespect to K . Lemma 3.4. ([14, Theorem 4.3 and Corollary 1.7]) Let G be a group which ishyperbolic relative to a family K of subgroups and h be an element of G whichis of infinite order and hyperbolic with respect to K . Then there exists a uniquesubgroup E ( h ) of G such that E ( h ) is virtually cyclic, contains h and maximalamong such subgroups of G . Moreover E ( h ) is hyperbolically embedded into G relative to K .The following lemma is shown by a similar argument in the proof of [10,Corollary 1.12]. Lemma 3.5.
Let G be a group generated by a finite set X and hyperbolic rela-tive to a finite family K of subgroups. Let a subgroup H of G be hyperbolicallyembedded into G relative to K . We denote the union K ∪ { H } by H . Supposethat a subgroup Q of G is strongly quasiconvex relative to H in G . Then thereexists a constant C ( Q, H ) ≥ R of H such that(a) Q ∩ H ⊂ R ; 4b) d X (1 , r ) ≥ C ( Q, H ) for every element r of R \ Q ;(c) the subgroup R is quasiconvex relative to K in G ,the natural homomorphism Q ∗ Q ∩ R R → G is injective and its image h Q ∪ R i isstrongly quasiconvex relative to K in G . Proof.
By [10, Theorem 5.12], there exists a constant C ( Q, H ) ≥ R of H satisfying above conditions (a)and (b), the natural homomorphism Q ∗ Q ∩ R R → G is injective, its image h Q ∪ R i is quasiconvex relative to H in G and for every element g of G and every element H ′ of H , the intersection h Q ∪ R i ∩ gH ′ g − is either finite or conjugate to R in h Q ∪ R i .Now we suppose that R satisfies above condition (c) and show that h Q ∪ R i is strongly quasiconvex relative to K in G .First we show that h Q ∪ R i is quasiconvex relative to K in G . By [10, Theorem1.1 (2)], it suffices to show that for every element g of G , the intersection h Q ∪ R i ∩ gHg − is quasiconvex relative to K in G . Every finite subgroup of G isautomatically quasiconvex relative to K in G . Since R is quasiconvex relative to K in G , it is well-known that every conjugate of R is also quasiconvex relativeto K in G . Thus the subgroup h Q ∪ R i is quasiconvex relative to K in G .Next we show that for every element g of G and every element K of K , theintersection h Q ∪ R i ∩ gKg − is finite. We have only to consider the case wherethere exists an element s of h Q ∪ R i such that h Q ∪ R i∩ gKg − is equal to sRs − .Then h Q ∪ R i ∩ gKg − is contained in s ( s − gK ( s − g ) − ∩ H ) s − . Since H ishyperbolically embedded into G relative to K , the intersection s − gK ( s − g ) − ∩ H is finite. Hence h Q ∪ R i ∩ gKg − is also finite.Thus h Q ∪ R i is strongly quasiconvex relative to K in G .By using the above lemmas, we prove the following, which implies Proposi-tion 3.1 for finitely generated groups. Lemma 3.6.
Let G be a group which is hyperbolic relative to a finite family K of proper subgroups and Γ a subgroup of G which is neither virtually cyclicnor parabolic with respect to K . Suppose that G is finitely generated. Then Γcontains a free subgroup F of rank two which is strongly quasiconvex relativeto K in G . Proof.
By Lemma 3.3, there exists an element h of Γ which is of infinite orderand hyperbolic with respect to K . We denote by H a subgroup E ( h ) of G givenby Lemma 3.4. We put H = K ∪ { H } . Then H consists of proper subgroupsof G and G is hyperbolic relative to H . Since H is virtually infinite cyclic, thesubgroup Γ is not parabolic with respect to H .Hence it follows from Lemma 3.3 that there exists an element q of Γ which isof infinite order and hyperbolic with respect to H . We denote by Q the infinitecyclic subgroup of Γ generated by q . By Lemma 3.4 and Theorem 2.5, thesubgroup Q is strongly quasiconvex relative to H in G . Since Q is torsion-free,this implies that the intersection Q ∩ H is trivial.5 K K K m L L L m Figure 1: The graph of groups G Let X be a finite generating set of G and C ( Q, H ) ≥ h is of infinite order, there exists a positive integer k suchthat d X (1 , h kn ) ≥ C ( Q, H ) for every integer n ∈ Z \ { } . We denote by R theinfinite cyclic subgroup of Γ generated by h k . Since R is a finite index subgroupof H , it is strongly quasiconvex relative to K in G by Theorem 2.5. Hence R satisfies conditions (a), (b) and (c) in Lemma 3.5. By Lemma 3.5, the subgroup h Q ∪ R i of Γ is a free group of rank two which is strongly quasiconvex relativeto K in G .For the proof of the general case, we need the following lemma which is ob-tained from a specialization of [13, Theorem 2.44] together with [13, Proposition2.49]. Lemma 3.7.
Let G be a group which is hyperbolic relative to a family K ofproper subgroups and X a finite relative generating set of G with respect to K .Then there exists a finite subfamily K = { K , . . . , K m } of K such that G splitsas the free product G = G ∗ ( ∗ K ∈ K \ K K ) , where G is the subgroup of G which is generated by K , . . . , K m and X .Moreover there exist a finitely generated subgroup Q of G and a family L = { L , . . . , L m } of subgroups of Q satisfying the following:(i) the finite relative generating set X is contained in Q and for every i ∈{ , . . . , m } , the subgroup L i is contained in K i ;(ii) the group G is isomorphic to the fundamental group of the graph ofgroups G drawn in Figure 1;(iii) the subgroup Q is hyperbolic relative to L , the set X is a relative generat-ing set of Q with respect to L and the natural map ( Q, d X ∪L ) → ( G, d X ∪K )is an isometric embedding, where we put L = S L ∈ L L \ { } . Lemma 3.8.
In the setting of Lemma 3.7, we have the following:(1) if no elements of K contain X , then L consists of proper subgroups of Q .(2) if a subgroup of Q is strongly quasiconvex relative to L in Q , then it isstrongly undistorted relative to K in G ;63) if a subgroup of Q is hyperbolically embedded into Q relative to L , thenit is hyperbolically embedded into G relative to K ; Proof. (1) This follows from condition (i) in Lemma 3.7.(2) This follows from Theorem 2.5 and condition (iii) in Lemma 3.7.(3) Suppose that a subgroup V of Q is hyperbolically embedded into Q relative to L . Then V is strongly quasiconvex relative to L in Q by Theorems2.3 and 2.5. By assertion (2), the subgroup V is strongly undistorted relativeto K in G .We claim that V is almost malnormal in G . Indeed, since V is hyperbolicallyembedded into Q relative to L , it is almost malnormal in Q . Hence it sufficesto show that if g is an element of G \ Q , then the intersection V ∩ gV g − isfinite. First suppose that g belongs to G \ G . Since G is a free factor of G ,the intersection G ∩ gG g − is trivial. Hence the intersection V ∩ gV g − is alsotrivial. Next suppose that g belongs to G \ Q . We denote by T the Bass-Serrecovering tree of the graph of groups G . Then the group Q is the stabilizer groupof a vertex v of T and we have gv = v . Since the intersection Q ∩ gQg − fixesboth v and gv , it fixes an edge of T . Hence Q ∩ gQg − is parabolic with respectto L . Since every element of L is contained in a element of K , the intersection Q ∩ gQg − is parabolic with respect to K . Since the subgroup V is stronglyquasiconvex relative to K in G , the intersection V ∩ ( Q ∩ gQg − ) is finite. Hencethe intersection V ∩ gV g − is also finite. Proof of Proposition 3.1.
Since Γ contains an element of infinite order, it followsfrom Lemma 3.3 that there exists an element h of Γ which is of infinite orderand hyperbolic with respect to K . Let E ( h ) be a subgroup of G given by Lemma3.4. Since Γ is not virtually cyclic, we can take an element γ of Γ \ E ( h ). ByLemma 3.4, every subgroup of G that contains { h, γ } is not virtually cyclic. Wetake a finite relative generating set X of G with respect to K which contains { h, γ } . Note that since h is hyperbolic with respect to K , no elements of K contain X .Let Q and L be given by Lemma 3.7. By Lemma 3.8 (1), the family L consists of proper subgroups of Q . Since Γ ∩ Q contains { h, γ } and each elementof L is contained in some element of K , the subgroup Γ ∩ Q is neither virtuallycyclic nor parabolic with respect to L .Since Q is finitely generated, it follows from Lemma 3.6 that Γ ∩ Q containsa free subgroup F of rank two which is strongly quasiconvex relative to L in Q . By Lemma 3.8 (2), the subgroup F is strongly undistorted relative to K in G . Remark 3.9.
We give an alternative proof of Corollary 3.2.Let F ′ be a free group of rank two with a free basis Y ′ and X a finite rel-ative generating set of G with respect to K . By [1, Theorem 1.1], there existsa quotient group G ′ of G and an embedding ι : F ′ → G ′ such that G ′ is hyper-bolic relative to { ψ ( K ) } K ∈ K ∪ { ι ( F ′ ) } , where ψ : G → G ′ denotes the naturalprojection. Since ι ( F ′ ) is a hyperbolic group, it follows from [13, Theorem 2.40]that G ′ is hyperbolic relative to { ψ ( K ) } K ∈ K . Hence ι ( F ′ ) is hyperbolically7mbedded into G ′ relative to { ψ ( K ) } K ∈ K . By Theorem 2.3, the natural map( F ′ , d Y ′ ) → ( G ′ , d ψ ( X ) ∪K ′ ) is a quasi-isometric embedding, where K ′ denotes S K ∈ K ψ ( K ) \ { } . We take a subset Y of G such that Y consists of two ele-ments and ψ ( Y ) is equal to ι ( Y ′ ). We denote by F the subgroup of G which isgenerated by Y . Then F is a free group of rank two. We can confirm that thenatural map ( F, d Y ) → ( G, d X ∪K ) is a quasi-isometric embedding. This finishesthe proof of Corollary 3.2. In this section, we show the following (compare with [9, Theorem 5.16] for thecase of non-abelian free groups of finite rank), which is necessary for the proofof Theorem 1.2.
Theorem 4.1.
Let M be a finitely generated and virtually non-abelian freegroup and let { M l | l ∈ { , . . . , n }} be a finite family of finitely generatedsubgroups of M of infinite index. Then there exists a proper subgroup V of M satisfying the following:(i) the subgroup V is finitely generated, virtually non-abelian free, and almostmalnormal in M ;(ii) for every l ∈ { , . . . , n } and every element m of M , the intersection mV m − ∩ M l is finite.For the proof of Theorem 4.1, we prepare two lemmas. Lemma 4.2.
If a proper subgroup H of an infinite group G is almost malnormalin G , then H is an infinite index subgroup of G . Proof.
Assume that H is a finite index subgroup of G . Then for every element g of G , the intersection H ∩ gHg − is a finite index subgroup of G and henceit is infinite. This contradicts the assumption that H is almost malnormal in G . Lemma 4.3.
Let F be a non-abelian free group of finite rank and let { H l | l ∈{ , . . . , n }} be a finite family of finitely generated subgroups of F of infiniteindex. Let U be a finite subgroup of Out( F ). We denote by π : Aut( F ) → Out( F ) the quotient map and put A = π − ( U ). Then there exists a propersubgroup H of F satisfying the following:(i) the subgroup H is a free subgroup of rank two and is malnormal in F ;(ii) for every l ∈ { , . . . , n } and every element a of A , the intersection H l ∩ a ( H )is trivial;(iii) for every element a of A , either a ( H ) is equal to H or the intersection H ∩ a ( H ) is trivial. 8 roof. We put U = { u i | i ∈ { , . . . , m }} and choose an element a i of π − ( u i )for each i ∈ { , . . . , m } . We denote by M the collection of all proper subgroups H ′ of F satisfying the following: • the subgroup H ′ is a free subgroup of rank two and is malnormal in F ; • for every i ∈ { , . . . , m } , every l ∈ { , . . . , n } and every element f of F ,the intersection a − i ( H l ) ∩ f H ′ f − is trivial.By [9, Theorem 5.16], the collection M is not empty. We remark that everyelement of M satisfies conditions (i) and (ii) in Lemma 4.3.For each i ∈ { , . . . , m } and each element H ′ of M , we put as follows: K i = { K ⊂ H ′ | K = { } and K = H ′ ∩ f a i ( H ′ ) f − for some f ∈ F } ; I ( H ′ ) = { i ∈ { , . . . , m } | K i = ∅} ; I ( H ′ ) = { i ∈ { , . . . , m } | K i = { H ′ }} ; I ( H ′ ) = { i ∈ { , . . . , m } | K i = ∅ and K i = { H ′ }} . Since every finitely generated subgroup of F is quasiconvex in F (see [15, Section2]), the subgroup a i ( H ′ ) as well as H ′ is quasiconvex and malnormal in F . By[2, Theorem 7.11], the group F is hyperbolic relative to { a i ( H ′ ) } . Since H ′ isquasiconvex in F , it follows from [10, Theorem 1.1 (1)] that H ′ is quasiconvexrelative to { a i ( H ′ ) } in F . By [7, Theorem 9.1], the collection K i has a finiteset of representatives of H ′ -conjugacy classes K i = { K i,j | j ∈ { , . . . , n i }} and H ′ is hyperbolic relative to K i . For every j ∈ { , . . . , n i } the subgroup K i,j isfinitely generated and malnormal in H ′ (see [13, Propositions 2.29 and 2.36]).For the proof of the lemma, it suffices to show that there exists an element H of M such that I ( H ) is empty. Indeed if H is such an element of M , thenfor every a ∈ A , either both H ∩ a ( H ) and H ∩ a − ( H ) are equal to H or theintersection H ∩ a ( H ) is trivial. If the former occurs, then a ( H ) is equal to H .Hence the subgroup H has the desired properties.Let H ′ be an element of M . We have only to consider the case where the set I ( H ′ ) is not empty. Then for every i ∈ I ( H ′ ) and every j ∈ { , . . . , n i } , thegroup K i,j is a proper malnormal subgroup of H ′ and hence it is of infinite indexin H ′ by Lemma 4.2. By [9, Theorem 5.16], there exists a proper subgroup H ′′ of H ′ satisfying the following: • H ′′ is a free subgroup of rank two and is malnormal in H ′ ; • for every i ∈ I ( H ′ ), every j ∈ { , . . . , n i } and every element h ′ of H ′ , theintersection K i,j ∩ h ′ H ′′ h ′− is trivial.Since H ′ belongs to M , the subgroup H ′′ also belongs to M .We claim that if i ∈ { , . . . , m } belongs to I ( H ′ ), then for every element f of F , the intersection H ′′ ∩ a i ( f H ′′ f − ) is trivial. Indeed, the intersection H ′ ∩ a i ( f H ′ f − ) is either trivial or conjugate to K i,j in H ′ for some j ∈ { , . . . , n i } . Ifthe former occurs, the claim obviously holds. If the latter occurs, the intersection H ′′ ∩ a i ( f H ′′ f − ) is conjugate to a subgroup of K i,j ∩ h ′ H ′′ h ′− for some j ∈ , . . . , n i } and some element h ′ of H ′ . By the choice of H ′′ , the intersection H ′′ ∩ a i ( f H ′′ f − ) is trivial.The above claim implies that I ( H ′ ) is contained in I ( H ′′ ). Since H ′′ is asubgroup of H ′ , the set I ( H ′ ) is also contained in I ( H ′′ ). Hence the union I ( H ′ ) ∪ I ( H ′ ) is a proper subset of the union I ( H ′′ ) ∪ I ( H ′′ ) if I ( H ′′ ) is notempty. By repeating this procedure if necessary, we can find an element H of M such that I ( H ) is empty.Now we are ready to prove Theorem 4.1. Given a subgroup H of a group G ,we put V G ( H ) = { g ∈ G | H ∩ gHg − is of finite index both in H and gHg − } . Proof of Theorem 4.1.
It follows from the assumption that M has a finite indexnormal subgroup F which is a non-abelian free group of finite rank. The actionof M on F by conjugations induces a homomorphism from M to Out( F ). Wedenote the image of this homomorphism by U . Since F is a finite index subgroupof M , U is a finite subgroup of Out( F ). For each l ∈ { , . . . , n } , we put H l = M l ∩ F . Since M l is finitely generated and F is a finite index subgroup of M , thisimplies that H l is also finitely generated. Since the subgroup M l is of infiniteindex in M and the subgroup F is of finite index in M , the subgroup H l is ofinfinite index in F and of finite index in M l .Therefore we can take a subgroup H of F given by Lemma 4.3. We put V = V M ( H ). By [8, Theorem 1.6], the group H is a finite index subgroup of V . Hence V is a finitely generated and virtually non-abelian free group. Bythe definition of U and condition (iii) in Lemma 4.3, for every element m of M ,either mHm − is equal to H or the intersection H ∩ mHm − is trivial. Hencefor every element m of M \ V , the intersection H ∩ mHm − is trivial. Since H isa finite index subgroup of V , this implies that V is almost malnormal in M . Bycondition (ii) in Lemma 4.3, for every l ∈ { , . . . , n } and every element m of M ,the intersection H l ∩ mHm − is trivial and hence the intersection M l ∩ mV m − is finite. We prove Theorem 1.2. As the case of Corollary 3.2, it suffices to show thefollowing in view of [14, Corollary 4.5].
Theorem 5.1.
Let G be a group which is hyperbolic relative to a family K ofproper subgroups and Γ a subgroup of G which is neither virtually cyclic norparabolic with respect to K . If Γ contains an element of infinite order, thenthere exists a finitely generated and virtually non-abelian free subgroup V of G which is hyperbolically embedded into G relative to K and contains V ∩ Γ asa finite index subgroup. Moreover if G is torsion-free, then Γ contains a freesubgroup of rank two which is hyperbolically embedded into G relative to K .10his generalizes a result due to I. Kapovich [9, Theorem C] for torsion-freehyperbolic groups. For the proof of Theorem 5.1, we show the following lemma. Lemma 5.2.
Let G be a group which is hyperbolic relative to a finite family K of proper subgroups and Γ a subgroup of G which is neither virtually cyclic norparabolic with respect to K . Suppose that G is finitely generated. Then thereexists a finitely generated and virtually non-abelian free subgroup V of G whichis hyperbolically embedded into G relative to K and contains V ∩ Γ as a finiteindex subgroup. Moreover if G is torsion-free, then Γ contains a free subgroupof rank two which is hyperbolically embedded into G relative to K . Proof.
By Lemma 3.6, the group Γ contains a free subgroup F of rank twowhich is strongly quasiconvex relative to K in G . We put M = V G ( F ). Since F is strongly quasiconvex relative to K in G , it follows from [8, Theorem 1.6] that F is a finite index subgroup of M . Hence M is strongly quasiconvex relativeto K in G and we have V G ( M ) = M . By [8, Corollary 8.7], there exists onlyfinitely many double cosets M gM in G such that gM g − is not equal to M and the intersection M ∩ gM g − is infinite. We denote the collection of suchdouble cosets by { M g l M | l ∈ { , . . . , n }} . For each l ∈ { , . . . , n } , we put M l = M ∩ g l M g − l .We claim that for each l ∈ { , . . . , n } , M l is of infinite index in both M and g l M g − l . Indeed, assume that this does not hold. Since g l does not belong to M and V G ( M ) is equal to M , the subgroup M l is of infinite index in either M or g l M g − l . By replacing g l by its inverse if necessary, we may assume that thesubgroup M l is of finite index in M and of infinite index in g l M g − l . It followsfrom [9, Lemma 6.6] that for every positive integer k , the subgroup M ∩ g kl M g − kl is of finite index in M and of infinite index in g kl M g − kl . Then for every positiveinteger p , the subgroup T pk =1 M ∩ g kl M g − kl is of finite index in M and hencethe intersection T pk =1 g kl M g − kl is infinite. By [8, Theorem 1.4], there exists apositive integer p such that g pl belongs to M . This contradicts the assumptionthat the subgroup M ∩ g pl M g − pl is of infinite index in g pl M g − pl .We also claim that for every l ∈ { , . . . , n } , the subgroup M l is finitely gener-ated. Indeed, since M is strongly quasiconvex relative to K in G , it follows fromTheorem 2.5 that the conjugate g l M g − l is also strongly quasiconvex relative to K in G . Hence M l is strongly quasiconvex relative to K in G (see for example[7, Theorem 9.8] and [13, Theorem 4.18]). By Theorem 2.5, the subgroup M l isfinitely generated.Hence it follows from Theorem 4.1 that there exists a finitely generated andvirtually non-abelian free subgroup V of M such that V is almost malnormalin M and mV m − ∩ M l is finite for every l ∈ { , . . . , n } and every element m of M . Since M contains a finitely generated free subgroup of finite index and V is a finitely generated subgroup of M , the subgroup V is undistorted in M ,that is, it is strongly undistorted relative to the empty family ∅ in M . Since M is strongly undistorted relative to K in G by Theorem 2.5, the subgroup V isstrongly undistorted relative to K in G .We claim that V is almost malnormal in G . Indeed, assume that V is not11lmost malnormal in G . Then there exists an element g of G \ V such that theintersection V ∩ gV g − is infinite. In particular the intersection M ∩ gM g − isalso infinite. Since V is almost malnormal in M , the element g belongs to G \ M .Then for some l ∈ { , . . . , n } and some elements m and m of M , the element g is equal to m g l m . Therefore the intersection V ∩ ( m g l m ) V ( m g l m ) − is infinite and hence the intersection m − V m ∩ g l M g − l is also infinite. Thiscontradicts the condition that for every l ∈ { , . . . , n } and every element m of M , the intersection mV m − ∩ M l is finite.Thus the subgroup V is strongly undistorted relative to K and almost mal-normal in G . By Theorem 2.3, the subgroup V is hyperbolically embedded into G relative to K .For the case where G is torsion-free, we can take a desired subgroup V of Γby applying [9, Theorem 5.16] to F instead of applying Theorem 4.1 to M inthe above argument. Proof of Theorem 5.1.
The proof is done in the same way as the proof of Propo-sition 3.1 by using Lemma 5.2 and Lemma 3.8 (3) instead of Lemma 3.6 andLemma 3.8 (2), respectively.
Acknowledgements
The authors would like to thank Professor Ilya Kapovich for his useful sugges-tion.