aa r X i v : . [ m a t h . G R ] N ov ON A CONJECTURE OF KARRASS AND SOLITAR
BENJAMIN STEINBERG
Abstract.
We settle an old conjecture of Karrass and Solitar by proving thata finitely generated subgroup of a non-trivial free product G = A ∗ B has finiteindex if and only if it intersects non-trivially each non-trivial normal subgroup of G . This holds, more generally, for subgroups of finite Kurosh rank. Introduction
Karrass and Solitar deduced in [9], as an easy consequence of M. Hall’s theo-rem [5], that a finitely generated subgroup of a free group F has finite index if andonly if it intersects non-trivially every non-trivial normal subgroup of F . This resultwas then forgotten for a number of years until it was revisited by Arzhantseva, whoin her 1998 Ph. D. thesis proved a quantitative version of this result: if H is a finitelygenerated subgroup of infinite index in a free group F , then the normal closure of al-most any finite set of elements of F misses H . Her approach was via Stallings graphsand small cancellation theory. The result was published in [1]. In the proceedingsof a 1998 conference in Lincoln, Nebraska, another proof of the result of Karrassand Solitar was given by Ivanov and Schupp [7], who apparently were unaware ofthe result of Karrass and Solitar. The result of Karrass and Solitar does appearin the 1977 book of Lyndon and Schupp from twenty years earlier [11, Chapter 1,Proposition 3.17]. The proof of Ivanov and Schupp also used Stallings graphs andsmall cancellation theory. Shortly after, Kahrobaei [8] offered another simple proofusing Marshall Hall’s theorem.In a paper [10], published in the same year as [9], Karrass and Solitar conjecturedthat a finitely generated subgroup of a free product G of non-trivial groups has finiteindex if and only if it intersects non-trivially each non-trivial normal subgroup of G .We settle here this conjecture affirmatively. Our proof follows that of Arzhantsevaand of Ivanov and Schupp, but uses covering spaces instead of Stallings graphs andmakes use of an analogue of the Spelling Theorem, due to Duncan and Howie [4],instead of classical small cancellation theory. In fact, we prove more generally thefollowing theorem. Date : March 29, 2013; revised November 6, 2013.1991
Mathematics Subject Classification.
Key words and phrases.
Free product, Kurosh rank.This work was partially supported by a grant from the Simons Foundation (
Theorem 1.
Let G = A ∗ B be a free product of non-trivial groups A and B andlet H ≤ G be a subgroup of finite Kurosh rank. Then [ G : H ] < ∞ if and only if H ∩ N = { } for every normal subgroup { } 6 = N ⊳ G . We shall recall the notion of Kurosh rank in Section 2. Suffice it to say for ourpurposes that finitely generated subgroups have finite Kurosh rank. Consequently,the conjecture of Karass and Solitar is now a theorem.
Theorem 2.
Let G = A ∗ B be a free product of non-trivial groups A and B .Then a finitely generated subgroup of G has finite index if and only if it intersectsnon-trivially each non-trivial normal subgroup of G . Theorem 2 can be seen as a strong generalization of the theorem of B. Baumslag [2]that any finitely generated subgroup of a free product of non-trivial groups, whichcontains a non-trivial normal subgroup is of finite index (since in a free product ofnon-trivial groups, any two non-trivial normal subgroups intersect non-trivially).2.
The core -complex of a subgroup The purpose of this section is to associate to each subgroup H ≤ G = A ∗ B a2-complex Core( H ) that plays the role of the Stallings graph of a subgroup in thetheory of free groups. Several core-like constructions have been considered before inthe literature [3, 6, 12], but ours seems to be a new variation, in particular becauseit is a 2-complex.Fix groups A = { } 6 = B and let G = A ∗ B . If C is a group, put e C = C \{ } . Thefree monoid on a set X is denoted X ∗ . Sometimes, we will write ( x ) for x ∈ e A ∪ e B ifwe want to emphasize that we are viewing x as an element of ( e A ∪ e B ) ∗ rather thanas an element of G . A word γ in ( e A ∪ e B ) ∗ is called reduced if it contains no factorsof the form ( x )( y ) with both x, y ∈ e A or x, y ∈ e B . It is well known [11, Chapter IV,Theorem 1.2] that any element of G is represented by a unique reduced word andthat any word can be put into its reduced form using the following elementaryrewriting rules: ( x )( x − ) → , for x ∈ e A ∪ e B (1)( a )( a ′ ) → ( aa ′ ) , for a, a ′ ∈ e A , a ′ = a − (2)( b )( b ′ ) → ( bb ′ ) , for b, b ′ ∈ e B , b ′ = b − . (3)For γ ∈ ( e A ∪ e B ) ∗ , we write red( γ ) for the unique reduced word equivalent to itin G . Abusing notation, denote by red( g ) the unique reduced word representing g ∈ G . Let ℓ ( g ) be the length of the word red( g ). A word γ ∈ ( e A ∪ e B ) ∗ is said to be cyclically reduced if γ is reduced, or equivalently, if γ n is reduced for all n ≥
0. Letus say that g ∈ G is cyclically reduced if red( g ) is cyclically reduced. There is aninvolution on ( e A ∪ e B ) ∗ defined in the usual way: [( x ) · · · ( x n )] − = ( x − n ) · · · ( x − ).The involution preserves the properties of being reduced and of being cyclicallyreduced. N A CONJECTURE OF KARRASS AND SOLITAR 3 p p x x − Figure 1.
Rewriting rule of type (1)If C is a group, let K ( C ) be the 2-complex with a single vertex v , edge set e C (consisting of loops) and with 2-cells of the form v v xx − and v v v x yxy where x ∈ C and y = x − . Of course, π ( K ( C ) , v ) ∼ = C . Let K = K ( A ) ∨ K ( B )be the wedge product. Then π ( K ( A ) ∨ K ( B ) , v ) ∼ = G .If X is a 2-complex and V ⊆ X (0) is a set of vertices, then the subcomplex X [ V ] induced by V has 1-skeleton X [ V ] (1) consisting of V , together with all edges of X between vertices of V . A 2-cell c of X belongs to X [ V ] if the attaching map ∂c → X (1) has image contained in X [ V ] (1) . A subcomplex of the form X [ V ] forsome set of vertices V is called an induced subcomplex .Fix a subgroup H ≤ G for the remainder of this section and let ϕ : ( e K H , w ) → ( K, v )be the pointed covering map with ϕ ∗ ( π ( e K H , w )) = H . By a path in a 2-complex,we shall always mean an edge path. Each path in e K H is labelled by a word in( e A ∪ e B ) ∗ . Let us say that the path is reduced if its label is reduced. By usualcovering space theory, a path p is homotopy equivalent, rel base points, to a uniquereduced path red( p ), whose label, moreover, is the reduced form of the label of p . Lemma 3.
Let X ⊆ e K H be an induced subcomplex and let p be a path in X . Then red( p ) is also a path in X and p is homotopy equivalent, rel base points, to red( p ) inside of X .Proof. It suffices to show that the result of applying one of the rewriting rules (1)–(3) to the label of p is the label of a path contained in X , which is homotopic to p in X , rel base points. If p has a factorization p = p ef p where e, f are edges withrespective labels x, x − with x ∈ e A ∪ e B , then the path ef bounds a 2-cell c in e K H by construction of K and because ϕ is a covering map. See Figure 1. Since X is aninduced subcomplex, c belongs to X . Thus p is homotopy equivalent to p p in X ,rel base points. This takes care of rewriting rule (1).We now handle rewriting rule (2), as the case of (3) is analogous. Suppose p = p ef p where e, f are edges with respective labels a, a ′ ∈ e A with a ′ = a − . Let BENJAMIN STEINBERG p a a ′ aa ′ p Figure 2.
Rewriting rule of type (2) v be the initial vertex of e and w the terminal vertex of f . Because ϕ is a coveringand ( a )( a ′ )( aa ′ ) − bounds a 2-cell in K , there is an edge e ′ from v to w labelled by aa ′ . See Figure 2. As X is an induced subcomplex and v, w are vertices of X , theedge e ′ belongs to X and so p e ′ p is a path in X . Moreover, since ϕ is a covering ef ( e ′ ) − bounds a 2-cell c , which furthermore belongs to X because X is an inducedsubcomplex. It follows that p is homotopic to p e ′ p in X , rel base points. Thiscompletes the proof. (cid:3) As a corollary, we obtain that the restriction of the covering map ϕ to any inducedsubcomplex is π -injective. Corollary 4. If X ⊆ e K H is an induced subcomplex containing w , then ϕ ∗ : π ( X, w ) → H is injective.Proof. If q is a loop in X at w , then q is homotopic in X , rel base points, to red( q )by Lemma 3. Thus without loss of generality we may assume that q is reduced. Butthen ϕ ∗ ([ q ]) = 1 if and only if q is an empty path by the normal form theorem forfree products [11, Chapter IV, Theorem 1.2]. (cid:3) Let ∅ 6 = S ⊆ H . We say that a vertex v of e K H is in the span of S if there is anelement s ∈ S such that the lift of red( s ) at w passes through v . The subcomplexof e K H induced by the vertices in the span of S is denoted Span( S ) and is called thesubcomplex spanned by S . Note that w ∈ Span( S ) and Span( S ) is connected. Alsoobserve that if S is finite, then Span( S ) has finitely many vertices. Let us definethe core of e K H to be the subcomplex Core( H ) = Span( H ) spanned by H itself.The next proposition sets up a Galois correspondence between non-empty subsetsof H and induced subcomplexes of e K H containing the base point w . Proposition 5.
Let ∅ 6 = S ⊆ H and let X be an induced subcomplex of e K H con-taining w . Then Span( S ) ⊆ X if and only if S ⊆ ϕ ∗ ( π ( X, w )) .Proof. By construction S ⊆ ϕ ∗ ( π (Span( S ) , w )) and so the only if statement isclear. For the converse, if g ∈ S , then there is a loop p in X at w with ϕ ∗ ([ p ]) = g .But then Lemma 3 implies that red( p ), which is the lift of red( g ) to w , is containedin X . We conclude that Span( S ) ⊆ X . (cid:3) Taking S = H we obtain the following corollary. Corollary 6.
The
Core( H ) is the smallest induced subcomplex X of e K H containing w such that ϕ ∗ ( π ( X, w )) = H . N A CONJECTURE OF KARRASS AND SOLITAR 5
The next corollary shows that Core( H ) is determined by a generating set of H . Corollary 7.
Let ∅ 6 = S ⊆ H . Then Span( S ) = Span( h S i ) . In particular, if ∅ 6 = S generates H , then Core( H ) = Span( S ) .Proof. Trivially, Span( S ) ⊆ Span( h S i ). For the converse, observe that S ⊆ ϕ ∗ ( π (Span( S ) , w ))and hence h S i ⊆ ϕ ∗ ( π (Span( S ) , w )). Proposition 5 now yields that Span( h S i ) ⊆ Span( S ). (cid:3) Corollary 8.
Suppose that H is finitely generated. Then Core( H ) has finitely manyvertices. Finiteness of the vertex set of the core is all that we use in the proof of Theorem 1,so we now prove that Core( H ) is finite if and only if H has finite Kurosh rank. Werecall here the definition of Kurosh rank; see [3, 6, 12] for details.By the Kurosh theorem, we know that H ≤ G = A ∗ B has a free productdecomposition of the form H = (cid:20) ∗ i ∈ I s i H i s − i (cid:21) ∗ (cid:20) ∗ j ∈ J t j K j t − j (cid:21) ∗ F ( H ) (4)where H i ≤ A , for i ∈ I , K j ≤ B for j ∈ J , F ( H ) is a free group intersectingno conjugate of A or B , and the s i , t j ∈ G run over certain sets of double cosetrepresentatives of H \ G/A and H \ G/B , respectively. The
Kurosh rank r ( H ) of H ,which may be infinite, is defined (relative to the splitting G = A ∗ B ) by r ( H ) = rank( F ( H )) + { i ∈ I | H i = { }} + { j ∈ J | K j = { }} . One can show that this does not depend on any of the choices made in the Kuroshdecomposition [6, Lemma 4].There is an alternative description of the Kurosh rank in terms of Bass-Serretheory. Let T be the Bass-Serre tree for the free splitting G = A ∗ B . Then, becausethe edge stabilizers are trivial, there is a unique minimal H -invariant subtree T H for any subgroup H ≤ G . One has r ( H ) = rank( π ( H \ T H )) + { Hv ∈ H \ T H | G v = { }} where G v is the stabilizer of the vertex v ∈ T H . The Kurosh rank is finite preciselywhen H \ T H is finite. See [12] for details.We shall require just the “only if” statement of the following theorem; therefore,the other implication will only be sketched. Theorem 9.
Let H ≤ G = A ∗ B . Then H has finite Kurosh rank if and only if Core( H ) has finitely many vertices.Proof. Assume first that r ( H ) < ∞ and let (4) be its Kurosh decomposition. Let X be a finite generating set for F ( H ) and let I ′ = { i ∈ I | H i = { }} , J ′ = { j ∈ J | K j = { }} ; note that I ′ , J ′ are finite. Then Corollary 7 yieldsCore( H ) = Span X ∪ [ i ∈ I ′ s i H i s − i ∪ [ j ∈ J ′ t j K j t − j . BENJAMIN STEINBERG
Clearly X contributes only finitely many vertices to Core( H ). It suffices to showthat each s i H i s − i with i ∈ I ′ and t j K j t − j with j ∈ J ′ contributes finitely manyvertices to Core( H ). We handle only the first case, as the second is identical.Let q i be the lift of red( s i ) at w and let C i be the subcomplex of e K H inducedby the vertices visited by the path q i ; note that C i has finitely many vertices. Let1 = a ∈ H i ≤ A . Let p be the lift at w of the path red( s i ) a red( s i ) − in K . Then p is a loop at w (as s i as − i ∈ H ) and so by uniqueness of path lifting, p = q i eq − i where e is the edge out of the endpoint of q i labeled by a . In particular, e must bea loop edge. Thus p is contained in C i and, therefore, red( p ) is contained in C i byLemma 3. We conclude that the lift at w of the reduced form of each element of s i H i s − i is contained in C i and hence s i H i s − i contributes only finitely many verticesto Core( H ).That the finiteness of Core( H ) (0) implies r ( H ) < ∞ is immediate from the nextlemma. (cid:3) Lemma 10.
Let X ⊆ e K H be a connected induced subcomplex containing w with ϕ ∗ ( π ( X, w )) = H . Let Γ( X ) be the bipartite graph with V (Γ( X )) = π ( ϕ − ( K ( A )) ∩ X ) ∪ π ( ϕ − ( K ( B )) ∩ X ) E (Γ( X )) = { e v | v ∈ X (0) } where v e v is a bijection from X (0) to E (Γ( X )) . The edge e v connects therespective components of v in ϕ − ( K ( A )) ∩ X and ϕ − ( K ( B )) ∩ X .Then the equality r ( H ) = rank( π (Γ( X ))) + { Y ∈ V (Γ( X )) | π ( Y ) = { }} holds. In particular, if X has finitely many vertices, then Γ( X ) is finite and hence r ( H ) < ∞ .Proof. Corollary 4 shows that ϕ ∗ : π ( X, w ) → H is an isomorphism. The standardcovering space proof of the Kurosh theorem shows that X is homotopy equivalentto a graph of spaces with underlying graph Γ( X ) where the vertex space associatedto Y ∈ V (Γ( X )) is the space Y and the edge space associated to e v is the one-point space { v } with the obvious inclusions of edge spaces into vertex spaces. Thisgives a free product decomposition of π ( X, w ) which maps under ϕ ∗ to a Kuroshdecomposition of H . One has π (Γ( X )) ∼ = F ( H ), the H i are the p ∗ ( π ( Y )) with Y ∈ π ( ϕ − ( K ( A )) ∩ X ) and the K j are the p ∗ ( π ( Z )) with Z ∈ π ( ϕ − ( K ( B )) ∩ X )(with appropriate choices of base points). (cid:3) It is, in fact, not difficult to prove that Γ(Core( H )) ∼ = H \ T H where T H is theunique minimal H -invariant subtree of the Bass-Serre tree T for the splitting G = A ∗ B .The Kurosh rank is well known not to change under conjugation. We shall needthe second statement of the following corollary. Corollary 11. If H ≤ G = A ∗ B has finite Kurosh rank, then each conjugate gHg − of H has finite Kurosh rank. Moreover, if v is a vertex of Core( H ) and N A CONJECTURE OF KARRASS AND SOLITAR 7 g ∈ G is represented by the label of a reduced path in Core( H ) from v to w , then Core( gHg − ) ⊆ Core( H ) .Proof. Let v be the endpoint of the lift p of the reduced word representing g − beginning at w . Then ϕ ∗ : ( e K H , v ) → π ( K, v ) is the pointed covering with ϕ ∗ ( π ( e K H , v )) = gHg − . If h ∈ H and q is the reduced loop at w with ϕ ∗ ([ q ]) = h ,then p − qp is a loop at v with ϕ ∗ ([ p − qp ]) = ghg − . It follows that if X is thesubcomplex of e K H induced by the vertices of the path p together with the verticesof Core( H ), then ϕ ∗ ( π ( X, v )) = gHg − . Thus Core( gHg − ) ⊆ X by Corollary 6.But X has finitely many vertices by construction. Thus gHg − has finite Kuroshrank. Moreover, if v is a vertex of Core( H ) and p is chosen to be a reduced path inCore( H ), then X = Core( H ) in this case. This proves the second statement. (cid:3) The next proposition generalizes that a Stallings graph of a finitely generatedsubgroup of a free group is a Schreier graph if and only if the subgroup has finiteindex.
Proposition 12.
Let H ≤ G = A ∗ B be a subgroup of finite Kurosh rank. Thenthe index of H in G is finite if and only if Core( H ) = e K H .Proof. If Core( H ) = e K H , then since [ G : H ] is the cardinality of the vertex set of e K H ,we conclude that H has finite index. Suppose conversely that [ G : H ] = n < ∞ andlet v be a vertex of e K H . Chooses a reduced path p from w to v and let γ ∈ ( e A ∪ e B ) ∗ be the label of p . If γ is not cyclically reduced, then we can find c ∈ e A ∪ e B such that γc is cyclically reduced. Hence there is a cyclically reduced word β such that thelift of β at w passes through v . Now β n ! is a reduced word representing an elementof H . Thus v is a vertex of Core( H ). We conclude Core( H ) = e K H . (cid:3) The proof of Theorem 1
In order to prove Theorem 1 we shall need some technical notions and results.Let us assume that H ≤ G = A ∗ B is a subgroup of finite Kurosh rank. We retainthe notation of the previous section.We now associate to each g ∈ G a partial permutation of the finite set V of verticesof Core( H ). If v is a vertex of e K H and g ∈ G , then vg denotes the endpoint of the liftof red( g ) starting at v ; this is the right monodromy action of G on e K H . Let dom( g )to be the set of vertices v ∈ V such that the lift of red( g ) at v is contained in Core( H ).Define ran( g ) = { vg | v ∈ dom( g ) } and note that ρ g : dom( g ) → ran( g ) given by v vg is a bijection. Also observe that ρ g − is the inverse partial permutation to ρ g . Define rank( g ) = g ) = g ). Clearly, we have rank( g ) = rank( g − ).If γ ∈ ( e A ∪ e B ) ∗ , denote by γ the element of G represented by γ . Proposition 13.
The following properties hold. (a) If αβ ∈ ( e A ∪ e B ) ∗ is reduced, then dom( αβ ) ⊆ dom( α ) . (b) If αβ ∈ ( e A ∪ e B ) ∗ is reduced, then ran( αβ ) ⊆ ran( β ) . (c) If αβγ ∈ ( e A ∪ e B ) ∗ is reduced, then rank( αβγ ) ≤ rank( β ) . BENJAMIN STEINBERG
Proof.
By uniqueness of path lifting, if the lift of αβ at v is contained in Core( H ),then the same is true for the lift of α at v . This proves (a). Item (b) is dual. Thefinal claim follows by (a) and (b) because rank( αβγ ) ≤ rank( αβ ) ≤ rank( β ). (cid:3) Our next lemma follows from the construction of Core( H ). Lemma 14.
Let v ∈ V with v = w . Then there are reduced paths p A ( v ) and p B ( v ) from v to w such that the first edge of p A ( v ) is labelled by a letter from e A and thefirst edge of p B is labelled by a letter from e B .Proof. By construction of Core( H ), there is an element g ∈ H such that the lift q of red( g ) at w passes through v . Say that q = q q with the endpoint of q being v .Replacing g by g − , if necessary, we may assume that the last edge of q is labelledby an element of e A and the first edge of q is labelled by an element of e B . Then wemay take p A ( v ) = q − and p B ( v ) = q . (cid:3) If H ≤ G is of finite Kurosh rank and infinite index, then Core( H ) = e K H byProposition 12. Since e K H is connected, there must be an edge e of e K H such thatone endpoint of e belongs to Core( H ) and the other endpoint does not. Let us saythat H is well situated if there is an edge e at w whose other endpoint does notbelong to Core( H ). We introduce this notion only to simplify the argument, not forany essential purpose. Lemma 15.
Let H ≤ G be an infinite index subgroup of finite Kurosh rank. Thenthere is a conjugate of H which is well situated.Proof. Let e be an edge of e K H with one endpoint v in Core( H ) and one endpointoutside Core( H ). Then K = ϕ ∗ ( π ( e K H , v )) is a conjugate of H and by Corollary 11we have Core( K ) is a subcomplex of Core( H ) with base point v . Thus K is wellsituated. (cid:3) The following technical lemma is the key to our proof of the conjecture of Karassand Solitar.
Lemma 16.
Let H ≤ G be an infinite index subgroup of finite Kurosh rank whichis well situated. Then there is a cyclically reduced element g ∈ G with rank( g ) = 0 .Proof. Since H is well situated, there is an edge e at w whose endpoint is not inCore( H ). Without loss of generality, we may assume that e is labelled by a ∈ e A .We now show that there is an element g ∈ G of rank zero.Choose g ∈ G of minimum rank. Multiplying g on the right by an element of e B , if necessary, and using Proposition 13, we may assume without loss of generalitythat the last symbol in g is from e B . Suppose that rank( g ) > v ∈ dom( g ).Let p be the lift of red( g ) at v .There are two cases. Suppose first that vg = w . Then p ends at w and pe is areduced path with label ga . Thus v / ∈ dom( ga ). On the other hand, Proposition 13implies that dom( ga ) ⊆ dom( g ). We conclude rank( ga ) < rank( g ), a contradiction. N A CONJECTURE OF KARRASS AND SOLITAR 9
Next suppose that vg = w . Lemma 14 implies that there is a reduced path q from vg to w such that the first symbol of the label of q belongs to e A . Let g ′ bethe element of G represented by the label of q . Then red( gg ′ ) = red( g ) red( g ′ ) and ℓ ( g ′ ) ≥
1. Thus red( g ) is still a prefix of red( gg ′ a ) and hence dom( gg ′ a ) ⊆ dom( g )by Proposition 13. But red( gg ′ a ) labels the path red( pqe ) from v to w a / ∈ Core( H ).Therefore, v / ∈ dom( gg ′ a ) and so rank( gg ′ a ) < rank( g ), again a contradiction. Weconclude that rank( g ) = 0.If g is cyclically reduced, we are finished. Otherwise, by right multiplying g byan appropriate element of e A ∪ e B we obtain a cyclically reduced element, which hasrank zero by an application of Proposition 13. (cid:3) We are now ready to prove Theorem 1 and its corollary, Theorem 2.
Proof of Theorem 1.
Suppose H has finite index. Then it contains a non-trivialnormal subgroup, and in a free product any two non-trivial normal subgroups havenon-trivial intersection, cf. [9, pg. 211].Next suppose that H has infinite index and finite Kurosh rank. We must showthat there is a non-trivial normal subgroup N with H ∩ N = 1. Replacing H by aconjugate, we may assume that H is well situated by Lemma 15. By Lemma 16,there is a cyclically reduced element g of rank zero. Let N be the normal subgroupgenerated by g . We claim that H ∩ N = { } . First note that if 1 = z ∈ H ∩ N ,then red( z ) lifts to a loop at w in Core( H ). Thus it suffices to show that there isno z ∈ N \ { } such that red( z ) lifts to a loop in Core( H ) at some vertex.Suppose that z ∈ N \ { } is such that red( z ) lifts to a loop in Core( H ) at somevertex v . Then red( z ) = red( u ) red( z ′ ) red( u ) − with red( z ′ ) cyclically reduced.Clearly, red( z ′ ) lifts to a loop in Core( H ) at the vertex vu and so without loss ofgenerality we may assume that z is cyclically reduced. By [4, Theorem 3.1], there isa cyclic conjugate ζ of red( z ) containing a factor γ of length at least 3 ℓ ( g ) − > ℓ ( g )that is also a factor of a cyclic conjugate of red( g ) ± . Thus γ has red( g ) ± as a factor.Now rank( g − ) = rank( g ) = 0, and so rank( γ ) = 0 by Proposition 13. Therefore,rank( ζ ) = 0 (again by Proposition 13). But since z labels a loop in Core( H ) atsome vertex, the same is clearly true of its cyclic conjugate ζ . Thus rank( ζ ) > (cid:3) References [1] G. N. Arzhantseva. A property of subgroups of infinite index in a free group.
Proc. Amer.Math. Soc. , 128(11):3205–3210, 2000.[2] B. Baumslag. Intersections of finitely generated subgroups in free products.
J. London Math.Soc. , 41:673–679, 1966.[3] R. G. Burns, T. C. Chau, and S.-M. Kam. On the rank of intersection of subgroups of a freeproduct of groups.
J. Pure Appl. Algebra , 124(1-3):31–45, 1998.[4] A. J. Duncan and J. Howie. Spelling theorems and Cohen-Lyndon theorems for one-relatorproducts.
J. Pure Appl. Algebra , 92(2):123–136, 1994.[5] M. Hall, Jr. Coset representations in free groups.
Trans. Amer. Math. Soc. , 67:421–432, 1949.[6] S. V. Ivanov. On the Kurosh rank of the intersection of subgroups in free products of groups.
Adv. Math. , 218(2):465–484, 2008. [7] S. V. Ivanov and P. E. Schupp. A remark on finitely generated subgroups of free groups. In
Algorithmic problems in groups and semigroups (Lincoln, NE, 1998) , Trends Math., pages139–142. Birkh¨auser Boston, Boston, MA, 2000.[8] D. Kahrobaei. A simple proof of a theorem of Karrass and Solitar. In
Geometric methods ingroup theory , volume 372 of
Contemp. Math. , pages 107–108. Amer. Math. Soc., Providence,RI, 2005.[9] A. Karrass and D. Solitar. On finitely generated subgroups of a free group.
Proc. Amer. Math.Soc. , 22:209–213, 1969.[10] A. Karrass and D. Solitar. On finitely generated subgroups of a free product.
Math. Z. , 108:285–287, 1969.[11] R. C. Lyndon and P. E. Schupp.
Combinatorial group theory . Classics in Mathematics. Springer-Verlag, Berlin, 2001. Reprint of the 1977 edition.[12] M. Sykiotis. On subgroups of finite complexity in groups acting on trees.
J. Pure Appl. Algebra ,200(1-2):1–23, 2005.
Department of Mathematics, City College of New York, Convent Ave at 138th St.,New York, New York 10031
E-mail address ::