Handlebody subgroups in a mapping class group
aa r X i v : . [ m a t h . G T ] M a r Handlebody subgroups in a mapping classgroup
Mladen Bestvina, Koji Fujiwara ∗ September 24, 2018
Abstract
Suppose subgroups
A, B < M CG ( S ) in the mapping class group ofa closed orientable surface S are given and let h A, B i be the subgroupthey generate. We discuss a question by Minsky asking when h A, B i ≃ A ∗ A ∩ B B for handlebody subgroups A, B . Let V be a handlebody and S = ∂V the boundary surface. We havean inclusion of mapping class groups, M CG ( V ) < M CG ( S ). This sub-group is called a handlebody subgroup of M CG ( S ). The kernel of the map M CG ( V ) → Out ( π ( V )) is denoted by M CG ( V ).If M = V + ∪ S V − is a Heegaard splitting of a closed orientable 3-manifold,we have two handlebody subgroups Γ ± = M CG ( V ± ) < M CG ( S ) with S = ∂V ± . Minsky [9, Question 5.1] asked Question 1.1.
When is h Γ + , Γ − i < M CG ( S ) equal to the amalgamation Γ + ∗ Γ + ∩ Γ − Γ − ? Let C ( S ) be the curve graph of S and D ± ⊂ C ( S ) the set of isotopy classesof simple curves in S which bound disks in V ± . The Hempel/Heegaard dis-tance of the splitting is defined to be equal to d ( D + , D − ) = min { d C ( S ) ( x, y ) | ∗ The first author was supported by the National Science Foundation grant 1308178.The second author is supported in part by Grant-in-Aid for Scientific Research (No.23244005, 15H05739) ∈ D + , y ∈ D − } . The group M CG ( S ) acts on C ( S ) by isometries. Thestabilizer subgroups of D ± in M CG ( S ) are Γ ± . If the Hempel distance issufficiently large, depending only on S ( > S is atleast two [11]), then Γ + ∩ Γ − is finite [16].The following is the main result. It gives a (partial) negative answer toQuestion 1.1. Theorem 5.1.
For the closed surface S of genus g + 1 , g ≥ and for any N > there exists a Heegaard splitting M = V + ∪ S V − so that Γ + ∩ Γ − istrivial, h Γ + , Γ − i is not equal to Γ + ∗ Γ − , and d ( D + , D − ) ≥ N . We will prove Theorem 5.1 by constructing an example. To explain theidea we first construct a similar example in a certain group action on asimplicial tree (Theorem 4.1), then imitate it for the action of
M CG ( S ) on C ( S ).By contrast, for the subgroups M CG ( V ± ) = Γ ± < Γ ± , Ohshika-Sakuma[17] showed Theorem 1.2. If d ( D + , D − ) is sufficiently large (depending on S ), Γ ∩ Γ − is trivial and h Γ , Γ − i = Γ ∗ Γ − . That Γ ∩ Γ − is trivial follows from the fact that Γ , Γ − are torsion free(attributed to [19, proof of Prop 1.7] in [17]).Here is an alternative proof, suggested by Minsky, that Γ < M CG ( S ) istorsion free for a handlebody V . Let f ∈ Γ be a torsion element. Since f hasfinite order, we have a conformal structure on S invariant by f . Moreover,since f extends to V , by the classical deformation theory of Kleinian groupsdeveloped by Ahlfors, Bers, Kra, Marden, Maskit, and Sullivan, we havea unique hyperbolic structure on V whose conformal structure at infinityis the prescribed one. Since the conformal structure is f -invariant, so is thehyperbolic structure. Moreover, since f acts trivially on π ( V ), each geodesicin V is invariant by f . This implies that f is identity on V , hence f is trivial. Acknowledgements.
We would like to thank Yair Minsky for useful com-ments. 2
Preliminaries
Let S be a closed orientable surface. The mapping class group M CG ( S ) of S is the group of orientation preserving homeomorphisms modulo isotopy.The curve graph C ( S ) has a vertex for every isotopy class of essential simpleclosed curves in S , and an edge corresponding to pairs of simple closed curvesthat intersect minimally.It is a fundamental theorem of Masur and Minsky [13] that the curvegraph is δ -hyperbolic. Moreover, they show that an element F acts hyper-bolically if and only if F is pseudo-Anosov, and that the translation length trans ( F ) = lim d ( x , F n ( x )) n , x ∈ C ( S )of F is uniformly bounded below by a positive constant that depends only on S . It follows that F has an invariant quasi-geodesic, called an axis denotedby axis ( F ), whose quasi-geodesic constants depend only on S .A subset A ⊂ X in a geodesic space is Q - quasi-convex if any geodesic in X joining two points of A is contained in the Q -neighborhood of A . If S is the boundary of a handlebody V , then the set D ⊂ C ( S ) of curves thatbound disks in V is quasi-convex (i.e. Q -quasi-convex for some Q ), [14].The stabilizer of the set D in M CG ( S ) is M CG ( V ), the mapping classgroup of the handlebody V , i.e., the group of isotopy classes of diffeomor-phisms of V .Given a Q -quasi-convex subset X in C ( S ), we define the nearest pointprojection C ( S ) → X . The nearest point projection is not exactly a map,but a coarse map , since for a given point maybe there is more than onenearest point, but the set of such points is bounded in diameter, and thebound depends only on δ and Q , but not on X .In this paper we often take axis ( F ) of a pseudo-Anosov element F as X .We may take any F -orbit instead of axis ( F ). Two pseudo-Anosov elements F, G are independent (i.e., h F, G i is not virtually cyclic) if and only if thenearest point projection of axis ( E ) to axis ( F ) has a bounded image (cf. [13],[4]). In this section we discuss the acylindricity of a group action. This is a keyproperty to prove Theorem 1.2. 3cylindricity was introduced by Sela for group actions on trees and ex-tended by Bowditch [5]. Suppose G acts on a metric space X . The actionis acylindrical if for given R > L ( R ) and N ( R ) such that forany points v, w ∈ X with | v − w | ≥ L , there are at most N elements g ∈ G with | v − g ( v ) | , | w − g ( w ) | ≤ R . (Here | x − y | denotes the distance d ( x, y ).)Bowditch [5] showed that the action of M CG ( S ) on C ( S ) is acylindrical.The following criterion will be useful. Lemma 3.1.
Suppose G acts on a simplicial tree X . If the cardinality of theedge stabilizers is uniformly bounded then the action is acylindrical.Proof. Assume that every edge stabilizer contains at most K elements. Sup-pose an integer R >
L >> R . We will show that if | v − w | ≥ L then there are at most (2 R + 1) K elements g with | v − gv | , | w − gw | ≤ R .Indeed, let [ v, w ] be the geodesic from v to w , and [ v, w ] ′ and [ v, w ] ′′ be its subsegments after removing the R -neighborhood of v, w , and the 2 R -neighborhood of v, w , respectively. Then by the assumption, g ([ v, w ] ′′ ) ⊂ [ v, w ] ′ . Moreover, for an edge E ⊂ [ v, w ] ′′ near the midpoint, g ( E ) is con-tained in [ v, w ] ′′ and the distance between E and g ( E ) is at most R . Nowfix such E . Then there are elements h , · · · , h n ∈ G ′ with n ≤ R + 1,where h = 1, such that for any concerned element g , there exists h i with h i g ( E ) = E . But since the stabilizer of E contains at most K elements, thereare at most nK ≤ (2 R + 1) K distinct choices for g .To explain the background we quote a main technical result from [17] (wewill not use this result). Theorem 3.2.
Let a group G act acylindrically on a δ -hyperbolic space X .Then for a given Q > , there exists M > with the following property. Let A, B ⊂ X be Q -quasi-convex subsets, and G A < stab G ( A ) , G B < stab G ( B ) torsion-free subgroups. If d X ( A, B ) ≥ M then(1) G A ∩ G B is trivial.(2) h G A , G B i = G A ∗ G B . Applying the theorem to the action of
M CG ( S ) on C ( S ) with A = D + , B = D − , and G A = Γ < Γ + = stab ( A ), G B = Γ − < Γ − = stab ( B ), weobtain Theorem 1.2: if d ( D + , D − ) is sufficiently large, depending only on S ,then Γ ∩ Γ − is trivial and h Γ , Γ − i = Γ ∗ Γ − .To explain the difference between the torsion-free setting of [17] and ours,we review the proof of Theorem 3.2. We start with an elementary lemma.4 emma 3.3. Let X be a δ -hyperbolic space and A, B ⊂ X be Q -quasi-convexsubsets. Let γ be a shortest geodesic between A and B . Then,(1) for any x ∈ γ with both d ( x, A ) , d ( x, B ) > Q + 2 δ , and for any shortestgeodesic τ between A and B , we have d ( x, τ ) ≤ δ .(2) Suppose f is an isometry of X with f ( A ) = A, f ( B ) = B . Then forany x ∈ γ with both d ( x, A ) , d ( x, B ) > Q + 2 δ , we have d ( x, f ( x )) ≤ δ .Hence for any x ∈ γ , we have d ( x, f ( x )) ≤ Q + 8 δ .(3) Suppose f is an isometry of X with f ( A ) = A . For x ∈ X \ A let σ be ashortest geodesic from x to A . For an integer N > assume d ( x, A ) ≥ Q + 4 δN and d ( x, f ( x )) ≤ δ . Then for any point y ∈ σ with Q + 2 δ
0, we say that A i is K-terminal if for any other A j , A k and any shortest geodesic γ between A j , A k , the distance between A i and γ is at least K . Theorem 3.4.
Let a group G act acylindrically on a δ -hyperbolic space X .Then for a given Q > , there exists K > with the following property.Let A , · · · , A n ⊂ X be Q -quasi-convex subsets, and G A i < stab G ( A i ) besubgroups for all i .Assume that there exists a subset A ′ i containing A i for each i so that:(a) for every finite order element = a ∈ G A i we have CF ix ( a ) ⊂ A ′ i foreach i ;(b) d X ( A ′ i , A ′ j ) ≥ K for all pairs i = j ; and(c) each A ′ i is K -terminal.Then(1) G A i ∩ G A j is trivial for all i = j .(2) h G A , · · · , G A n i = G A ∗ · · · ∗ G A n . The proof is a slight variation of the proof of Theorem 3.2 and is omitted.If G A i does not contain any non-trivial elements of finite order, then we just7ut A i = A ′ i . Our counterexamples will have the property that A ′ ∩ A ′ = ∅ (namely, the two coarse fixed sets intersect although A and A are far away.cf. Claim 1 in the proof of Theorem 3.2 (2) when f has infinite order, where CF ix ( f ) is contained in the L -neighborhood of A ).If the sets A ′ , · · · , A ′ n satisfy properties (b) and (c) above for a constant K , we say that they are K - separated. We will show that Theorem 3.2 does not hold if we do not assume that G A and G B are torsion-free. We construct a counterexample in the action of M CG ( S ) on C ( S ) (Theorem 5.1).To explain the idea we start with a counterexample when X is a simplicialtree. The key geometric feature is that, if we keep the previous notations, γ and a ( γ ) may stay close along an arbitrarily long segment if a has finiteorder (each point on that segment does not move very much by a ). Theorem 4.1.
There exists an acylindrical group action on a simplicialtree X by a group G such that for any number N > there exist vertices v, w ∈ X with | v − w | ≥ N such that stab G ( v ) ∩ stab G ( w ) is trivial and h stab G ( v ) , stab G ( w ) i is not equal to the free product stab G ( v ) ∗ stab G ( w ) .Proof. We first construct an example with N = 2. Start with abelian groups A, B with non-trivial torsion elements a ∈ A and b ∈ B , for example, A, B ≃ Z / Z .Define the group G = A ∗ h a i ( h a i × h b i ) ∗ h b i B and let T be the Bass-Serre tree of this graph of groups decomposition.There are two vertices v, w in T at distance two whose stabilizers are A and B . The intersection A ∩ B is trivial in G since h a i ∩ h b i is trivialin h a i × h b i . On the other hand, h A, B i = G is not equal to A ∗ B since G is the quotient of the free product A ∗ B by the relation ab = ba . Thegeometric reason for why h a, b i is not equal to h a i ∗ h b i is that F ix ( a ) and F ix ( b ) intersect non-trivially in T .The action on T is acylindrical by Lemma 3.1 since the edge stabilizer isa conjugate of h a i or h b i . 8 xv t ( v ) ts ( x ) s ( x ) tsts ( x ) tst ( v ) st ( v ) sts ( x ) h b i tsA ′ EBC h a i h a i h a i h a i h a i Figure 2:To produce an acylindrical action that works for all
N > Z = h t i and the subgroup h a i × h b i in G . One can write the new group as A ∗ h a i {h a i × h b i × h t i} ∗ h b i B Further, add a new element s to A with a relation sa = as to get A ′ = A × h s i and set C = h a i × h b i × h t i and G ′ = A ′ ∗ h a i C ∗ h b i B This is a two edge decomposition.In the Bass-Serre tree of this decomposition, consider the “fundamentaldomain”, i.e. the subtree spanned by two vertices v, w at distance two withstabilizers A ′ and B respectively. Let x be the vertex between them withstabilizer C . See Figure 2.Now consider the ray based at x that contains the vertices x , t ( v ), ts ( x ), tst ( v ), tsts ( x ) , · · · . The stabilizer of every edge on this ray is h a i since both t and s commute with a .So, the intersection of B , the vertex group of w , and any of the vertexgroups along the ray except for C is h a i ∩ h b i = 1.But for each n > h C ( ts ) n , B i < G ′ is not equal to C ( ts ) n ∗ B since a ∈ C ( ts ) n and b ∈ B generate h a i × h b i and not h a i ∗ h b i .The action of G ′ is acylindrical by Lemma 3.1 since any edge stabilizer isa conjugate of h a i or h b i . 9 Example on C ( S ) and proof of theorem We will prove the main theorem.
Theorem 5.1.
For the closed surface S of genus g + 1 with g ≥ and forany N > there exists a Heegaard splitting M = V + ∪ S V − so that Γ + ∩ Γ − is trivial, h Γ + , Γ − i is not equal to Γ + ∗ Γ − , and d ( D + , D − ) ≥ N . We will need two properties of pseudo-Anosov elements to prove the the-orem (Lemma 5.2, Lemma 5.4).
Let S be a closed surface and F a pseudo-Anosov mapping class on S . The elementary closure of F is the subgroup E ( F ) of M CG ( S ) that consists ofmapping classes preserving the stable and unstable foliations of F . Equiva-lently, E ( F ) is the centralizer of F in M CG ( S ). The group E ( F ) contains aunique finite normal subgroup N ( F ) such that E ( F ) /N ( F ) is infinite cyclic.Note that E ( F k ) = E ( F ) and N ( F k ) = N ( F ) for every k = 0.If S ′ → S is a regular cover with deck group ∆ and if F : S → S is a pseudo-Anosov mapping class with N ( F ) = 1, then we certainly have N ( F ′ ) ⊇ ∆ for any lift F ′ : S ′ → S ′ of any power of F , but strict inclusionmay hold. It is an interesting question whether one can construct F so thatequality holds for all regular covers. We call such F prime and we discuss aconstruction of prime pseudo-Anosov mapping classes in Section 6. For ourpurposes we need quite a bit less. Lemma 5.2.
Suppose F : S → S is a pseudo-Anosov mapping class whosestable and unstable foliations have two singular points, one of order p , theother of order q , with both p, q odd and relatively prime. Let S ′ → S be adouble cover with deck group h a i and F ′ a lift of a power of F to S ′ . Then N ( F ) = 1 and N ( F ′ ) = h a i .Proof. We first argue that N ( F ) = 1. Suppose g ∈ N ( F ). Then g can berepresented by a homeomorphism, also denoted g : S → S , that preservesboth measured foliations. In particular, g is an isometry in the associated flatmetric on S with cone type singularities. The homeomorphism g fixes bothsingular points and satisfies both g p = 1 and g q = 1, since an isometry thatfixes a nonempty open set is necessarily the identity. Since p, q are relativelyprime it follows that g = 1. 10e now argue that N ( F ′ ) = h a i . We have h a i < N ( F ′ ). Let g ∈ N ( F ′ ),then g : S ′ → S ′ is a finite order homeomorphism that preserves the lift ofstable and unstable foliations of F . Composing with a if necessary we mayassume that g fixes both p -prong singularities. Arguing as above, we see that g p = 1. Since g fixes both q -prong singularities, similarly we have g q = 1and since ( p, q ) = 1 we have g = 1. We showed N ( F ′ ) = h a i . Corollary 5.3.
Pseudo-Anosov mapping classes as in Lemma 5.2 exist inevery genus ≥ .Proof. Write 4 g = p + q where p, q are relatively prime odd numbers. Forexample, we can take p = 2 g − q = 2 g +1. By the work of Masur-Smillie[15] F as above exists. Suppose V is a handlebody and S its boundary. Let D ⊂ C ( S ) be the set ofcurves that bound disks in V . Denote by L ⊂ PML ( S ) the closure of D ,viewed as a subset of PML ( S ). Then L is nowhere dense in PML ( S ) [12],and its complement Ω is called the Masur domain .Hempel [10] found that if the stable lamination of a pseudo-Anosov ele-ment F is in Ω then lim n →∞ d C ( S ) ( D, F n ( D )) = ∞ . We say a pseudo-Anosovelement F : S → S is Hempel for D if the nearest point projection of D to axis ( F ) is a bounded set. F is Hempel if and only if the end points of axis ( F ) are in Ω, [3]. Onthe other hand, both endpoints of axis ( F ) are in L if and only if axis ( F )is contained in a K -neighborhood of D for some K > D and axis ( F ) are quasi-convex subsets in the δ -hyperbolic space C ( S ) (cf. [3]).Since L is nowhere dense in PML ( S ) and the set of pairs of endpoints( λ + , λ − ) of pseudo-Anosov mapping classes is dense in PML ( S ) ×PML ( S ),there is a pseudo-Anosov element F whose stable and unstable laminationsare not in L , so that F is Hempel.Masur found a condition in terms of the intersection number for a curveto be in D [12, Lemma 1.1] and used it to prove L is nowhere dense [12,Theorem 1.2]. The following lemma is proved using his ideas. Lemma 5.4.
Let V ′ → V be a double cover between handlebodies with thedeck group h a i , S ′ = ∂V ′ , S = ∂V , and D ′ ⊂ C ( S ′ ) , D ⊂ C ( S ) the set ofcurves that bound disks in V ′ , V , respectively. f the genus of S is ≥ , then M CG ( S ) contains a pseudo-Anosov element F such that:(i) F is Hempel for D ,(ii) F lifts to F ′ : S ′ → S ′ and F ′ is Hempel for D ′ ,(iii) N ( F ′ ) = h a i .Proof. Let Ω be the Masur domain for V and Ω ′ for V ′ . We first find alamination Λ on S that is in Ω such that its lift Λ ′ on S ′ is also in Ω ′ . Choosea pants decomposition of S using curves in D and a lamination Λ ∈ PML ( S )whose support intersects each pair of pants in this decomposition in 3 (non-empty) families of arcs connecting distinct boundary components (so thereare no arcs connecting a boundary component to itself). In the proof of [12,Theorem 1.2] Masur shows that Λ ∈ Ω (for example, take the curve β in hisproof as Λ). This is done by verifying the conditions in Lemma 1.1 for β with respect to the pants decomposition in the last two paragraphs of theproof of Theorem 1.2. Now the lift Λ ′ of Λ to S ′ satisfies the same conditionwith respect to the lifted pants decomposition (it lifts since our covering isbetween handlebodies and the boundary curves bound disks), so we haveΛ ′ ∈ Ω ′ .Choose a pseudo-Anosov homeomorphism G : S → S both of whose fixedpoints in PML ( S ) are close to Λ and in particular they are in Ω since Ω isopen. The lift G ′ of G (or its power) to S ′ similarly has endpoints close toΛ ′ and in particular in Ω ′ . It follows that both G and G ′ are Hempel.To finish the proof we need to arrange that G has the extra property(iii). Let H : S → S be an arbitrary pseudo-Anosov mapping class thatsatisfies the assumption of Lemma 5.2. Such H exists by Corollary 5.3. Then F = G n HG − n also satisfies the assumptions, and hence also conclusion ofLemma 5.2 for any n > n > F is Hempel, and similarly, the lift F ′ has an axis whose endpoints close to Λ ′ , therefore F ′ is Hempel. We prove Theorem 5.1 by constructing an example.
Proof of Theorem 5.1.
Let H ≃ Z / Z + Z / Z with generators a , a , and let V ′ → V be a normal cover between handlebodies with the deck group H . If12 ≥ V , then the genus of V ′ is 4 g −
3. Let D ′ ⊂ C ( S ′ ) bethe set of curves in S ′ = ∂V ′ that bound disks in V ′ . We have two doublecovers S ′ → S ′ /a i . Let D i ⊂ C ( S ′ /a i ) be the set of curves in S ′ /a i thatbound disks in V ′ /a i . Put S i = S ′ /a i . The genus of S i is 2 g −
1. Let Q be a common quasi-convex constant for D ′ , D , D , and δ the hyperbolicityconstant of C ( S ′ ).Using Lemma 5.4, take a pseudo-Anosov element F i on S i that is Hempelfor D i such that the lift F ′ i of F i to S ′ is also Hempel for D ′ , and that N ( F ′ i ) = h a i i . Note that F ′ , F ′ are independent pseudo-Anosov elements on S ′ since their elementary closures are different. In particular, the projectionof axis ( F ′ ) to axis ( F ′ ) is bounded, and vice versa.Note that a i ∈ stab ( D ′ ) since a i ∈ H . Set D ′ i = F ′ Ni ( D ′ ) for N > a i ∈ stab ( D ′ i ) since F ′ i centralizes a i .Form the Heegaard splitting V ′ + ∪ S ′ V ′− such that D + = D ′ , D − = D ′ ⊂C ( S ′ ). The surface S ′ is fixed but the splitting depends on N . We will arguethis is a desired splitting.Set Γ i = stab ( D ′ i ) < M CG ( S ′ ). In other words, Γ = Γ + , Γ = Γ − in theHeegaard splitting convention. Since a i ∈ Γ i and a a = a a , h Γ , Γ i is notthe free product of Γ , Γ .To prove the theorem we are left to verify d ( D ′ , D ′ ) → ∞ as N → ∞ and Γ ∩ Γ = 1 for any large N > Lemma 5.5. d ( D ′ , D ′ ) → ∞ as N → ∞ .Proof. We claim that there is a constant A such that for any N > d ( D ′ , D ′ ) ≥ ( trans ( F ′ ) + trans ( F ′ )) N − A. Let π denote the projection to axis ( F ′ ), and π the projection to axis ( F ′ ).As we said they are coarse maps but we pretend they are maps for simplicity.Also, we pretend that both axis ( F ′ ) , axis ( F ′ ) are geodesics.Let L be a common bound of the diameter of the sets π ( D ′ ), π ( axis ( F ′ )), π ( D ′ ) and π ( axis ( F ′ )). Then L is a bound of π ( D ′ ) and π ( D ′ ) for all N >
0. Choose points q ∈ π ( D ′ ), q ∈ π ( D ′ ), r ∈ π ( axis ( F ′ )) and r ∈ π ( axis ( F ′ )).Now assume N > π ( axis ( F ′ )) and π ( D ′ ) are far apart,and also π ( axis ( F ′ )) and π ( D ′ ) are far apart (compared to L and δ ). Itsuffices to show the above inequality under this assumption.Let y ∈ D ′ and y ∈ D ′ be any points, and put x = π ( y ) , x = π ( y ). Then, by a standard argument using δ -hyperbolicity, the piecewise13 ′ D ′ D ′ γr r axis ( F ′ ) axis ( F ′ ) π ( D ′ ) π ( D ′ ) π ( axis ( F ′ )) π ( D ′ ) π ( axis ( F ′ )) π ( D ′ ) q q F ′ N ( q ) F ′ N ( q ) y x y x Figure 3: If
N > π ( axis ( F ′ )) and π ( D ′ ) are farapart on axis ( F ′ ), Also, π ( axis ( F ′ )) and π ( D ′ ) are far apart on axis ( F ′ ).As a consequence any shortest geodesic γ between D ′ and D ′ must enter abounded neighborhood of each of those four projection sets, and the segmentin γ near axis ( F ′ i ) is almost fixed by a i pointwise.geodesic [ y , x ] ∪ [ x , r ] ∪ [ r , r ] ∪ [ r , x ] ∪ [ x , y ] is a quasi-geodesic withuniform quasi-geodesic constants that depends only on L and δ . See Figure3. Hence the Hausdorff distance between the piecewise geodesic and thegeodesic [ y , y ] is bounded (the bound depends only on L and δ ).It follows that there is a constant C such that d ( D ′ , D ′ ) ≥ d ( y , x ) + d ( x , r ) + d ( r , r ) + d ( r , x ) + d ( x , y ) − C ≥ d ( x , r ) + d ( r , x ) − C ≥ d ( F ′ N ( q ) , r ) − L + d ( F ′ N ( q ) , r ) − L − C . On the other hand since q , r ∈ axis ( F ′ ) and q , r ∈ axis ( F ′ ) there is a constant B such that for all N > d ( F ′ N ( q ) , r ) + d ( F ′ N ( q ) , r ) ≥ ( trans ( F ′ ) + trans ( F ′ )) N − B .Combining them we get a desired estimate with A = 2 L + B + C .We note that any geodesic joining a point in D ′ and a point in D ′ passesthrough a bounded neighborhood of each of F ′ N ( q ) , r , r , F ′ N ( q ) providedthat N > L and δ . SeeFigure 3. In the argument we did not use that D ′ (as well as D ′ , D ′ ) arequasi-convex.To argue Γ ∩ Γ = 1, we will need the following lemma from [4, Propo-sition 6]. This is a consequence of the fact that F is a “WPD element”.14 emma 5.6. Let F be a pseudo-Anosov mapping class on a hyperbolic sur-face S . There is a constant M > such that for any g ∈ M CG ( S ) thediameter of the projection of g ( axis ( F )) to axis ( F ) in C ( S ) is larger than L , then g ∈ E ( F ) . Lemma 5.7. Γ ∩ Γ = 1 for any large N > .Proof. By Lemma 5.5 choose N large such that d ( D ′ , D ′ ) is very large com-pared to δ and L . Let γ be a shortest geodesic from D ′ to D ′ . Then as wenoted after the proof of Lemma 5.5, γ passes through the bounded neighbor-hood of each of F ′ N ( q ) , r , r , F ′ N ( q ).Now let f ∈ Γ ∩ Γ . Then we have d ( x, f ( x )) ≤ Q + 8 δ for any x ∈ γ byLemma 3.3. Since all of F ′ N ( q ) , r , r , F ′ N ( q ) are in bounded distance from γ we conclude each of those four points is moved by f a bounded amount(the bound depends only on δ, L, Q ).But since F ′ N ( q ) , r are contained in axis ( F ′ ) and are far apart fromeach other for any large N >
0, we find f ( axis ( F ′ )) has a long ( > M )projection to axis ( F ′ ), hence f ∈ E ( F ′ ) by Lemma 5.6. By the same reason f ∈ E ( F ′ ). We conclude f ∈ E ( F ′ ) ∩ E ( F ′ ). But since F ′ and F ′ areindependent, f must be a torsion element, so f ∈ N ( F ′ ) ∩ N ( F ′ ). By Lemma5.2, f ∈ h a i i ∩ h a i = 1. We showed the lemma.We proved the theorem. In view of Lemma 5.2 we introduce a property that looks interesting for itsown sake. We say a pseudo-Anosov mapping class F is prime if its sta-ble/unstable foliations are not lifts of any foliations of a (possibly orbifold)quotient of S .If F is prime then E ( F ) is cyclic and N ( F ) = 1. Indeed, if N ( F ) = 1then the two foliations lift from S/N ( F ), with N ( F ) realized as a group ofisometries of S using Nielsen realization. Moreover, Lemma 6.1. (cf. Lemma 5.2) Suppose S ′ → S is a finite cover with theDeck group ∆ . Let F be a prime pseudo-Anosov element on S and F ′ a liftof a power of F to S ′ . Then N ( F ′ ) = ∆ . roof. It is clear that ∆ < N ( F ′ ). If the inclusion is strict, then the stableand unstable foliations of F can be obtained by pulling back from S ′ /N ( F ′ ) = S/ ( N ( F ′ ) / ∆). So we have a contradiction.Note that if we have a prime pseudo-Anosov element on S , we can useLemma 6.1 instead of Lemma 5.2 in the proof of Lemma 5.4 and Theorem5.1. We will give a construction of prime pseudo-Anosov elements when thegenus of S is 3, so this will also prove the theorem for the genus 5 case.Recall that if a, b, c, d are 4 vectors in R then the cross ratio is[ a, b ; c, d ] = [ a, c ][ b, d ][ a, d ][ b, c ]where [ x, y ] = x y − x y for x = ( x , x ) , y = ( y , y ). We do not defineit when one of [ a, c ] , [ b, d ] , [ a, d ] , [ b, c ] is 0. The cross ratio is invariant un-der changing signs and scaling individual vectors and applying matrices in SL ( R ). It follows that for any flat structure on the torus the cross ratio forthe vectors in the directions of four distinct closed geodesics is (well-definedand) rational.A singular Euclidean structure (or just a flat structure) on a surface S is good if the cone angle is a multiple of π at each singularity. A geodesicsegment connecting two singular points, or a closed geodesic is good if theangle along the geodesic at each singular point is a multiple of π .The developing map ^ S − Σ → R defined on the universal cover of thecomplement of the cone points will take a good geodesic to a straight line, ora line segment. So, for any four good geodesics, the cross ratio for the fourdirections, if they are distinct, is well-defined.Next, if S ′ → S is a branched cover between good flat structures, thenthe cross ratio of four good geodesics in S ′ is equal to the cross ratio oftheir images in S , simply because S, S ′ have the “same” developing map. Inparticular, all cross ratios between good geodesics on a torus or a sphere with4 cone points are rational, and to prove that a particular good flat surfaceis not commensurable with a torus it suffices to produce four good geodesicswhose cross ratio is irrational. Lemma 6.2.
Suppose F : S → S is a pseudo-Anosov homeomorphism suchthat:(1) the stable foliation of F has two singular points x and y , with p and q prongs respectively, and with p and q distinct odd primes, and
2) a flat structure on S determined by the stable and unstable foliations hasfour good closed geodesics with the cross ratio of their (distinct) directionvectors a, b, c, d ∈ R irrational.Then F is prime. We note that there is a 2-parameter family of flat structures determinedby the two foliations; they depend on the choice of the transverse measurein a projective class on each foliation. However, since scaling and lineartransformations do not change the cross ratio, the assumption is independentof these choices.
Proof.
Let p, q, F be as in the statement. Now suppose π : S → S ′ is abranched cover of degree d > F = π − F ′ . The local degree of π at x is either 1 or p . It cannot be 1, since at any other preimage of π ( x ) thesingularity would have to have kp prongs, and there aren’t any. Thus at x the map is modeled on z z p , and similarly at y it looks like z z q . Thereare now two cases. Case 1. π ( x ) = π ( y ).It follows that the other points that map to π ( x ) have 2 prongs and sothe map there has local degree 2. Thus d is odd and away from the images ofsingular points the foliation F ′ is regular (since otherwise d would have to beeven). Thus there are ( d − p ) / π ( x ), and deleting theseand the same for the q -prong singularity we get that the Euler characteristicof S − π − ( { π ( x ) , π ( y ) } ) is(2 − g ) − ( d − p ) / − ( d − q ) / − − d So the Euler characteristic of the quotient S ′ minus 2 singular points is − R P , which does not support anypseudo-Anosov homeomorphisms (e.g. the curve complex is finite, see [20]).On the other hand, let ˜ S be a finite cover of the punctured S so that theinduced cover to R P minus two points is regular. Some power of F liftsto ˜ F on ˜ S , and since the cover is regular, a further power of ˜ F descends to R P (since each element, a , of the Deck group leaves the stable and unstablefoliations invariant, so that ( a ˜ F a − ) N = ˜ F N for some N >
0, so a and ˜ F N commute), contradiction. Case 2. π ( x ) = π ( y ).Again the other points that map to π ( x ) = π ( y ) are regular and the maphas local degree 2, so there are d − ( p + q )2 such points. Here d is even and we17ay have some number, say k ≥
0, of 1-prong singularities z , · · · , z k in thequotient, with each singularity having d preimages where local degree is 2.Now we have that the Euler characteristic of S − π − ( { π ( x ) , z , · · · , z k } ) is(2 − g ) − d − ( p + q )2 − − kd − ( k + 1) d k is odd and the Euler characteristic of the quotient S ′ minus the singularpoints is − k +12 . So, the Euler characteristic of S ′ is − k +12 +( k +1) = k +12 . Theonly possibilities are k = 1 and k = 3, and the quotient is twice punctured R P or 4 times punctured S . The first possibility is ruled out as in Case 1.In the second case the good flat structure on S descends to a good flatstructure on S with 4 singular points, then lifts to a flat structure on thebranch double cover T , with four closed geodesics such that the cross ratioof the four direction vectors is [ a, b ; c, d ] that is not rational, contradiction.Indeed, using the same notation as in Case 1, a power of F lifts to ˜ F on˜ S that regularly covers S minus 4 points. We lift the flat structure on S and the stable and unstable foliations of F to ˜ S . Then their regular leavesare straight lines. Each deck transformation preserves the foliations, so thatit is an isometry of ˜ S , and that the good flat structure on ˜ S , with cone angleat each singular point at least 2 π , descends to a good flat structure of S minus 4 points (and the angle at each puncture is π ). We obtain a goodflat structure on S with four good closed geodesics and the cross ratio is[ a, b ; c, d ]. Also, the cross ratio will not change when we take a double coverthat is a flat torus, contradiction. Remark 6.3.
Regarding the assumption (1), if g is the genus of S , by anEuler characteristic count we must have p + q = 4 g . Conversely, the Goldbachconjecture predicts that every even integer > can be written as a sum oftwo primes. When the integer is ≥ and divisible by 4, the two primes arenecessarily distinct and odd. For example,
12 = 5 + 7 satisfies the Goldbachconjecture. The work of Masur-Smillie [15] shows that if g ≥ and g = p + q then the surface S admits a pseudo-Anosov homeomorphism whose stable andunstable foliations have two singular points, one of order p , the other of order q . Example 6.4.
We now construct an explicit example in genus 3 satisfyingthe assumption of Lemma 6.2. We take p = 5 , q = 7 . Consider the flatsquare tiled surface S pictured below. Edges labeled by the same letter are o be identified. If the edges are on opposite sides of the parallelogram theyare identified by a translation, and otherwise by a rotation by π . The squaretiling of R induces one on the surface S . There are two cone points, withcone angles π and π respectively. So, S has a good flat structure. a b a b x ye f e x f yz z α α α α α α α ′ α ′ α ′ α ′ α ′ α ′ α ′ Figure 4: The square tiled surface. The round vertex has 7 prongs and thesquare vertex has 5. The surface has a good flat structure.
We will use Thurston’s construction of pseudo-Anosov homeomorphisms[7, Theorem 14.1] to construct F . The lines bisecting the squares form threehorizontal and three vertical geodesics. The matrix N of intersection num-bers, where the jk entry is the intersection number i ( α j , α ′ k ) , is N = The largest eigenvalue of
N N t is µ = 5 . . . . satisfying the minimalpolynomial µ − µ + 5 µ − . Then one can choose the lengths andheights of the squares (making them into rectangles, which gives a tiling of ) so that the twist in the horizontal multicurve is given by the matrix (cid:18) µ / (cid:19) and the twist in the vertical multicurve by the matrix (cid:18) − µ / (cid:19) This means that the product of the first and the inverse of the second is (cid:18) µ / (cid:19) (cid:18) µ / (cid:19) = (cid:18) µ µ / µ / (cid:19) = A whose trace is µ . So this product is pseudo-Anosov and its dilatationis the larger, λ , of the eigenvalues of A and the eigenvector is t (1 , σ ) with µ / = (1 − σ ) /σ and λ = 1 /σ .The heights and widths of the rectangles are coordinates of the µ -eigenvectors V of N N t and V ′ of N t N . We compute V = µ − µ + 1 − µ + 11 µ − = v v v and V ′ = µ − / N t V = µ − / − µ + 6 µ − µ − µ + 2 µ − µ + 1 = v ′ v ′ v ′ To show that F is prime it suffices to find 4 closed geodesics whose slopeshave irrational cross ratio. We take a = (1 , , b = (0 , , c = ( − v ′ , v + v ) , d = ( − v ′ − v ′ , v + v ) , where c and d connect second, respectively third,vertex on the lower left side in the figure with the upper right vertex. They aregood closed geodesics on S based at the round vertex (to compute the angleat the round vertex, it helps first to identify the two edges labeled by z ). Thecross ratio is [ a, c ][ b, d ][ a, d ][ b, c ] = ( v + v )( v ′ + v ′ )(2 v + v ) v ′ = ( µ − µ + 2) µ ( µ − µ + 3)( µ − µ + 2) = µµ − µ + 3= 13 µ − µ + 10 , which is irrational (for the last equality use µ − µ + 5 µ − ). Invariable generation
In this section we discuss another application of Theorem 3.2. For this weneed the version stated in Theorem 3.4.
Following Dixon [1] a group G is invariably generated by a subset S of G if G = h s g ( s ) | s ∈ S i for any choice of g ( s ) ∈ G, s ∈ S . The group G is IG if it is invariably generated by some subset S in G , or equivalently, if G isinvariably generated by G . G is FIG if is is invariably generated by somefinite subset of G . Kantor-Lubotzky-Shalev [2] prove that a linear group isFIG if and only if it is finitely generated and virtually solvable.Gelander proves that every non-elementary hyperbolic group is not IG[8]. We generalize this result to acylindrically hyperbolic groups. A group G is acylindrically hyperbolic if it admits an acylindrical action on a hyper-bolic space and G is not virtually cyclic [18]. Examples are non-elementaryhyperbolic groups, M CG ( S g,p ) except for the genus g = 0 and the numberof punctures p ≤
3, and
Out ( F n ) with n ≥ Theorem 7.1. If G is an acylindrically hyperbolic group, then G is not IG. In other words, G contains a proper subgroup such that any element in G is conjugate to some element in the subgroup. δ -hyperbolic spaces Suppose G acts on a δ -hyperbolic space X .For g ∈ G , define its minimal translation length by min ( g ) = inf p ∈ X | p − g ( p ) | It is a well-known fact that if min ( g ) ≥ δ then g is hyperbolic. (To beprecise, we assume δ > L > X ( g, L ) = { x ∈ X || gx − x | ≤ L } M ( g ) = X ( g, min ( g ) + 1000 δ ) M ( g ) is a g -invariant non-empty set. If g is hyperbolic, then M ( g ) is con-tained in a Hausdorff neighborhood of an axis of g, axis ( g ). This is an easyexercise and we leave it to the reader. Lemma 7.2. If g has a bounded orbit, then min ( g ) ≤ δ .Proof. This is well known too. Let Z be the set of centers of the orbit of apoint x by g . Z is invariant by g and its diameter is at most 6 δ , therefore iscontained in X ( g, δ ). Lemma 7.3. If L ≥ min ( g ) + 1000 δ , then X ( g, L ) is δ -quasi-convex.Proof. We define a function on X as follows: t g ( x ) = | x − g ( x ) | . Case 1 . min ( g ) ≤ δ .Fix p ∈ X ( g, δ ) ⊂ X ( g, L ). Suppose x ∈ X ( g, L ) is given. We will showthat [ p, x ] is contained in the 20 δ -neighborhood of X ( g, L ). If | x − p | ≤ δ ,then by triangle inequality, t g ( z ) ≤ δ for every z ∈ [ p, x ], so that [ p, x ] ⊂ X ( g, L ). So assume | x − p | > δ . To compute t g ( z ) for z ∈ [ p, x ], draw atriangle ∆ for p, x, g ( x ), and let c ∈ [ p, x ] be a branch point of this triangle,i.e., the distance to each side of ∆ from c is at most δ . Since | p − g ( p ) | ≤ δ , t g ( z ) ≤ δ for any point z ∈ [ p, c ]. For z ∈ [ c, x ], t g ( z ) is roughly equal to2 d ( c, z ), with an additive error at most 20 δ . Also it is roughly maximal at x on [ p, x ]. (Imagine the case that X is a tree and p = g ( p ).) It follows that[ p, x ] is contained in the 20 δ -neighborhood of X ( g, L ).Now suppose another point y ∈ X ( g, L ) is given. Then [ x, y ] is containedin the δ -neighborhood of [ p, x ] ∪ [ p, y ], so that [ x, y ] is contained in the 21 δ -neighborhood of X ( g, L ). Case 2 . min ( g ) ≥ δ .Then g is hyperbolic. To simplify the argument, let’s assume that there is ageodesic axis for g . Then for any x ∈ X , | t g ( x ) − { trans ( g ) + 2 d ( x, axis ( g )) }| ≤ δ To see this let x ′ ∈ axis ( g ) be a nearest point from x . Then the Hausdorffdistance between [ x, g ( x )] and [ x, x ′ ] ∪ [ x ′ , g ( x ′ )] ∪ [ g ( x ′ ) , g ( x )] is at most 5 δ ,and the estimate follows.It follows from the above estimate that if x ∈ X ( g, L ), then [ x, x ′ ] iscontained in the 20 δ -neighborhood of X ( g, L ). For y ∈ X ( g, L ), let y ′ ∈ xis ( g ) be a nearest point from another point y to axis ( g ). Then [ x, x ′ ] ∪ [ x ′ , y ′ ] ∪ [ y ′ , y ] is contained in the 20 δ -neighborhood of X ( g, L ). But since[ x, y ] is contained in the 5 δ -neighborhood of [ x, x ′ ] ∪ [ x ′ , y ′ ] ∪ [ y ′ , y ], [ x, y ] iscontained in the 25 δ -neighborhood of X ( g, L ).The argument for the case that g has only a quasi-geodesic (with uniformquasi-geodesic constants depending only on δ ) as an axis is similar and weonly need to modify the constants in the argument. We omit the details.Lemma 7.3 implies Lemma 7.4. M ( g ) is δ -quasi-convex. Let ∂X denote the boundary at infinity of X . For a quasi-convex subset Y ⊂ X , let ∂Y ⊂ ∂X be the boundary at infinity of Y . Lemma 7.5. If p ∈ ∂M ( g ) ⊂ ∂X , then g ( p ) = p .Proof. Let γ be a geodesic ray from a point in M ( g ) that tends to p . Thenthe ray is contained in the 10 δ -neighborhood of M ( g ). So every point of theray is moved by g by a bounded amount, therefore p is fixed by g .When f is hyperbolic, the subgroup of elements in G that fix each pointof ∂ ( axis ( f )) is called the elementary closure of f , denoted by E ( f ). Lemma 7.6.
Assume the action of G is acylindrical on X . If g fixes onepoint in ∂ ( axis ( f )) , then g ∈ E ( f ) .Proof. Let γ be a half of axis ( f ) that tends to the point fixed by g . Then | x − gx | is bounded for x ∈ γ . Assume that γ tends to the direction that f translates axis ( f ) (otherwise, we let N <
N > x ∈ γ , | f − N gf N ( x ) − x | is bounded. Now by acylindricity (apply it to x, y ∈ γ that are far from each other), there are only finitely many possibilities for f − N gf N , so g commutes with a nontrivial power of f . So g moves each pointin axis ( f ) by a bounded amount, therefore g ∈ E ( f ). Lemma 7.7.
Assume f is hyperbolic on X . If g E ( f ) , then π axis ( f ) M ( g ) is bounded.Proof. Suppose not. Let p be a point in ∂ ( axis ( f )) that the projection of M ( g ) tends to. 23e claim p ∈ ∂M ( g ). This is because since both axis ( f ) and M ( g ) arequasi-convex, a half of axis ( f ) to the direction of p is contained in a boundedneighborhood of M ( g ).So, by Lemma 7.5 g ( p ) = p , and by Lemma 7.6 g is in E ( f ), a contradic-tion.We will use the following result. It follows from the assumption that G contains a “hyperbolically embedded subgroup” that is non-degenerate, i.e.,proper and infinite, see Theorem 1.2 in [18].By a Schottky subgroup
F < G we mean a free subgroup such that anorbit map F → X is a quasi-isometric embedding. Proposition 7.8 ([6, Theorem 6.14]) . Suppose the action of G is acylindricaland G is not virtually cyclic.Then G contains a unique maximal finite normal subgroup K and a Schot-tky subgroup F so that for every nontrivial f ∈ F any element g ∈ E ( f ) iseither contained in K or has a nontrivial power that commutes with f . If K is trivial, E ( f ) is cyclic. When the order of g is N < ∞ , we define M M ( g ) = ∪ For any x ∈ X − M M ( g ) and any non-trivial h ∈ < g > , wehave | h ( x ) − x | > δ . Remark 7.10. Although we will not use this fact, we observe that if g hasfinite order N , then the set M M ( g ) is δ -quasi-convex. This is because g has an orbit whose diameter is at most δ (see the proof of Lemma 7.2), and M M ( g ) = ∪ Suppose there is a Schottky free subgroup F < G such thatany non-trivial f ∈ F is hyperbolic and E ( f ) is cyclic. et S = { g , g , g , · · · } be a (finite or infinite) set of non-trivial elementsin G .Then for any given K > there is a set S ′ = { g ′ i } in G such that • g i and g ′ i are conjugate for each i . • For any n > , the sets M ( g ′ i ) , · · · , M ( g ′ n ) are K -separated, and more-over, this property holds if we replace M ( g ′ k ) with M M ( g ′ k ) when g ′ k have finite order.Proof. We first prepare a sequence of elements in F that we will use toconjugate g i to g ′ i . Take two elements a, b ∈ F that produce a free subgroupof rank two. Put f i = ab i , i ≥ x ∈ X . Given a constant L > 0, if we choose P sufficiently large,then for any n > 0, and any P i ≥ P , the following points are L -separated: x, f P ( x ) , f P ( x ) , · · · , f P n n ( x ) . This is an easy consequence of the property such that the embedding of theCayley graph of F in X using the orbit of the point x is quasi-isometric tothe image.Note that the subsets in the above are L -separated if we replace the point x by a bounded set, possibly taking a larger constant for P .To define g ′ , g ′ , g ′ , · · · , choose a sequence1 = n < n < n < · · · such that for each i , g i E ( f n i ). This is clearly possible. Then by Lemma7.7, the projection of M ( g i ) to axis ( f n i ) is bounded. Moreover, if the order of g i is N < ∞ , then < g i > ∩ E ( f n i ) = 1, therefore the projection of M M ( g i ) to axis ( f n i ) is bounded since each M ( g ni ) , < n < N has a bounded projection.Now take a sequence of sufficiently large constants L i > 0, depending onthe given constant K , and put g ′ i = f L i n i g i f − L i n i . Then for each n > 0, thesets M ( g ′ ) , · · · , M ( g ′ n ) are K -separated since M ( g ′ i ) = f L i n i ( M ( g i )). Also wecan arrange so that the sets remain K -separated if we replace M ( g ′ i ) with M M ( g ′ i ) if the order of g ′ i are finite, maybe for larger constants L i . Proof of Theorem 7.1 . Case 1 . Assume G does not contain any non-trivial finite normal subgroup.25y Proposition 7.8, there is a Schottky subgroup F < G such that anynon-trivial element f ∈ F is hyperbolic and E ( f ) is cyclic.Let K > Q = 100 δ . Let C = { g , g , · · · } be a set of all conjugacy classes of G except for the class for 1.Apply lemma 7.11 to the set C and the constant K and obtain a new set C ′ = { g ′ i } . For each i > M ( g ′ i ) is a non-empty, g ′ i -invariant, 100 δ -quasi-convex subset.Now, for each n > 1, the assumptions (b) and (c) of Theorem 3.4 aresatisfied by the subgroups < g ′ i >, ≤ i ≤ n and the sets M ( g ′ i ) , ≤ i ≤ n . If g ′ k has finite order, then take M ( g ′ k ) ⊂ M M ( g ′ k ) as the desired neighborhood.Then by Lemma 7.11 they are K -separated, which implies (b) and (c).We claim that (a) holds for g ′ k that has finite order. But for any point x ∈ X − M M ( g ′ k ), and any non-trivial h ∈ < g ′ k > , we have | h ( x ) − x | ≥ δ (Lemma 7.9). This implies (a).It now follows from Theorem 3.4 that the subgroup generated by g ′ , · · · , g ′ n is the free product < g ′ > ∗ · · · ∗ < g ′ n > for each n > C ′ is a finite set, { g ′ , · · · , g ′ n } . Then < g ′ > ∗ · · · ∗ < g ′ n > must be a proper subgroup of G (therefore G is not IG) sinceotherwise G contains infinitely many conjugacy classes, a contradiction (byour assumption, G is not virtually cyclic).Second, we assume that C ′ is an infinite set in the following. To argue bycontradiction, assume that G is generated by C ′ .In < g ′ > ∗ < g ′ > ∗ < g ′ > , it is easy to choose elements g ′′ , g ′′ , g ′′ suchthat each g ′′ i is conjugate to g ′ i , and the subgroup generated by g ′′ , g ′′ , g ′′ is aproper subgroup of < g ′ > ∗ < g ′ > ∗ < g ′ > .Define C ′′ from C ′ by replacing g ′ , g ′ , g ′ by g ′′ , g ′′ , g ′′ . C ′′ contains all non-trivial conjugacy classes of G . We claim that the subgroup, G , generated by C ′′ is a proper subgroup in G (so G is not IG). To see that, define a quotienthomomorphism from G to the group H = < g ′ > ∗ < g ′ > ∗ < g ′ > bysending all g ′ i , i > G is a proper subgroup in H ,so G is a proper subgroup in G . Case 2 . Assume that G contains a non-trivial finite normal subgroup.Let K be the maximal finite normal subgroup in G . Then G ′ = G/K doesnot contain any non-trivial finite normal subgroup. Moreover G ′ is acylindri-cally hyperbolic group. Probably this fact is well known to specialists, andwe postpone giving an argument till the end (Proposition 7.12).By Case 1, G ′ contains a proper subgroup H ′ that contains all conjugacy26lasses of G ′ . Let H < G be the pull-back of H ′ by the quotient map G → G ′ . Then H is a proper subgroup that contains all conjugacy classes of G , therefore G is not IG. Proposition 7.12. Let G be an acylindrically hyperbolic group and N < G a finite normal subgroup. Then G ′ = G/N is an acylindrically hyperbolicgroup.Proof. By assumption G acts on a hyperbolic space X such that the actionis acylindrical and G is not virtually cyclic. Moreover we may assume thatthe action is co-compact. In fact, we may assume that X is a Cayley graphwith a certain generating set, which is maybe infinite, [18, Theorem 1.2].We will produce a new G -graph Y from X such that the kernel of theaction contains N and that Y and X are quasi-isometric. This is a desiredaction for G ′ .For each N -orbit of a vertex of X , we assign a vertex of Y . Note that G is transitive on the set of N -orbits of vertexes of X , so G acts transitivelyon the vertex set of Y . Now join two vertices of Y if the distance of thecorresponding N -orbits in X is 1. Y is a connected G -graph and it is easyto check that Y and X are quasi-isometric (by the obvious map sending avertex x of X to the vertex of Y corresponding to the orbit of x ), so that Y is hyperbolic, and that the G -action on Y is acylindrical. By construction, N acts trivially on Y . References [1] J.D. Dixon, Random sets which invariably generate the symmetricgroup, Discrete Math. 105 (1992) 2539.[2] William M. Kantor, Alexander Lubotzkyb, Aner Shalev. Invariablegeneration of infinite groups. to appear in Journal of Algebra.[3] Aaron Abrams, Saul Schleimer, Distances of Heegaard splittings.Geom. 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