On free decompositions of verbally closed subgroups in free products of finite groups

Andrey M. Mazhuga

doi:10.1515/jgth-2016-0058

Article Publicly Available

On free decompositions of verbally closed subgroups in free products of finite groups

Andrey M. Mazhuga

Published/Copyright: January 12, 2017

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From the journal Journal of Group Theory Volume 20 Issue 5

Abstract

For a subgroup of a free product of finite groups, we obtain necessary conditions (on its Kurosh decomposition) for it to be verbally closed.

1 Introduction

Algebraically closed objects play an important part in modern algebra. In the present paper, we study verbally closed subgroups in free products of finite groups. First of all we establish some terminology.

For groups H and G we write H⩽G if H is a subgroup of G. Let X={xi∣i∈ℕ} be a countable infinite set of variables, and let F⁢(X) be the free group with basis X. An equation with variables x1,…,xn∈X and coefficients from H is an arbitrary expression of the form w⁢(x1,…,xn,H)=1, where w⁢(x1,…,xn,H) is a word in the alphabet X∪X-1∪H (in other words, w(x1,…,xn,H) lies in the free product F⁢(X)∗H of F⁢(X) and H). If the left-hand side of the equation does not depend on coefficients from H, we will omit H in its expression: w⁢(x1,…,xn)=1. We say that an equation w⁢(x1,…,xn,H)=1 has a solution in G if there are some elements gi∈G, i=1,…,n such that w⁢(g1,…,gn,H)=1 in G.

A subgroup H is called algebraically closed in G if for every finite system of equations S={wi⁢(x1,…,xn,H)=1∣i=1,…,m} with coefficients from H the following holds: if S has a solution in G, then it has a solution in H. The following notion of verbally closed subgroups was introduced by V. Roman’kov and A. Myasnikov in [11].

Definition 1.

A subgroup H of G is called verbally closed if for any w∈F⁢(X) and h∈H an equation w⁢(x1,…,xn)⁢h-1=1 has a solution in H if it has a solution in G.

Verbally closed subgroups have a connection to research on the verbal width of elements in groups. For w=w⁢(x1,…,xn)∈F⁢(X) and a group G by w⁢[G] we denote the set of all w-elements in G, i.e.

w⁢[G]={w⁢(g1,…,gn)∣g1,…,gn∈G}.

The verbal subgroupw⁢(G) is the subgroup of G generated by w⁢[G]. Therefore H is a verbally closed subgroup in G if and only if w⁢[H]=w⁢[G]∩H for every w∈F⁢(X). The w-widthlw,G⁢(g) of an element g∈w⁢(G) is the minimal natural number n such that g is a product of nw-elements or their inverses in G; the width of w⁢(G) is the supremum of the widths of its elements. Usually, it is very hard to compute the w-length of a given element g∈w⁢(G) or the width of w⁢(G). For more details we refer to D. Segal’s book [15].

Another important notion for us is a retract.

Definition 2.

A subgroup H of a group G is called a retract of G if there is a homomorphism (termed retraction) φ:G→H which is identical on H. Equivalently, a retraction φ:G→G is a homomorphism such that φ2=φ and a retract is the image of a retraction.

Groups algebraically closed in the class of all groups (a group H is algebraically closed in a class of groups 𝔎 if and only if H is algebraically closed in every extension H⩽G with G∈𝔎) were introduced by W. R. Scott in [14]. They have been studied thoroughly in the 1970s and 1980s, see for references article [8] and books [3, 4]. On the other hand, we do not know much about algebraic closedness of subgroups in particular groups.

There are several papers devoted to the study of retracts of groups. It is known that the class of retracts of a non-abelian free group is much larger than its class of free factors (see, for instance, [9]). In [2] J. Bergman proved that the intersection of finitely many retracts of a free group is again a retract. At the same time, the question remains open whether it is true that if F is a free group of finite rank, R a retract of F, and H a subgroup of F of finite rank, then the intersection H∩R is a retract of H (see [2, Question 19]). An element g∈G is called a test element (for recognizing automorphisms) of a group G if, whenever φ⁢(g)=g for an endomorphism φ of G, it follows that φ is an automorphism. In [16] E. C. Turner obtained the following result and characterized test elements of a free group as elements that do not belong to any proper retract.

Theorem.

If φ:F→F is an endomorphism of the finitely generated free group F, then the subgroup

Hφ=⋂n=0∞φn⁢(F)

is a retract of F; Hφ is a proper retract precisely when φ fails to be an automorphism. If φ is a monomorphism, then Hφ is a free factor.

The generalization of this theorem for hyperbolic groups was proved in [12].

Very little is known in general about verbally closed subgroups of a given group G. For instance, the following questions are still open for most non-abelian groups:

Is there an algebraic description of verbally closed subgroups of G?
Is the class of verbally closed subgroups of G closed under intersections?
Does there exist the verbal closure vcl⁡(H) (vcl⁡(H) is the least (relative to inclusion) verbally closed subgroup of G containing H) of a given subgroup H of G?

Now we shall briefly discuss the relationship between retracts, algebraically closed and verbally closed subgroups of groups. In general, the property of being verbally closed is essentially weaker than being algebraically closed (see examples of verbally but not algebraically closed subgroups in 2-nilpotent torsion-free groups in [1]). It is easy to prove (see, for instance, [11]) that, if H is a retract of G, then H is algebraically (and verbally) closed in G. If G is a finitely presented group, and H is a finitely generated subgroup of G, then H is an algebraically closed subgroup of G if and only if H is a retract of G (see [11]). It was proved in [11] and [13] that a subgroup H of a free or free nilpotent group G of finite rank is verbally closed if and only if H is a retract of G. This implies that the properties of being verbally closed, being algebraically closed, and being a retract are equivalent for subgroups of free or free nilpotent groups of finite rank. Also in [11] the following result was obtained.

Theorem.

If F is a free group of finite rank, then the following holds:

There is an algorithm to decide if a given finitely generated subgroup of F is verbally (algebraically) closed or not.
There is an algorithm to find a basis of vcl⁡(H) for a given finitely generated subgroup H of F.

Except for free, some nilpotent, and finite groups, the structure of verbally closed subgroups remains unknown. The main problem in the study of verbal closeness is that it is exceedingly difficult to apply Definition 1 straightforwardly.

The following theorem was obtained by A. Kurosh [6].

Theorem.

Let

G=∗α∈ΛAα

be a free product of groups Aα and let H be a subgroup of G. Then H is a free product

H=F∗∗βBβ,w

here F is a free group such that F∩Aαg={1} for all α∈Λ, g∈G and each subgroup Bβ is conjugate in G to a subgroup of Aα for some α.

Such free decompositions of a subgroup H of G we call Kurosh free decompositions (of a subgroup H).

The main goal of this article is to study the structure of verbally closed subgroups of free products of finitely many finite groups. For a subgroup of this type of group we obtain necessary conditions (on its Kurosh decomposition) for it to be verbally closed (Theorem 1). To prove the first part of Theorem 1, we use an elementary technique combined with some tricks (Lemma 4). To prove its second part, we use a bit of Bass–Serre theory (Lemma 7). To prove that these conditions are not sufficient, we use Lee’s C-test words (the proof of Theorem 3).

Now we fix some notation that will be used in this text.

Denote by hg⁢n, n∈ℤ, an element g⁢hn⁢g-1 (similarly, Hg=g⁢H⁢g-1), and by [g,h] a commutator of the form g⁢h⁢g-1⁢h-1. A cyclic group of order n generated by an element a will be denoted as 〈a〉n, i.e. 〈a〉n≃ℤ/n⁢ℤ (similarly, 〈a〉∞≃ℤ). Let #⁢G denote the cardinality of a set G. Sometimes we use an abridged notation of the type x¯=x1,…,xn or even xi′¯=(xi,1′,…,xi,n′) rewriting, for example, an equation w⁢(x1,…,xn)⁢h-1=1 as w⁢(x¯)⁢h-1=1.

2 The main result

The following theorem is the main result of this article.

Theorem 1.

Let H be a subgroup of G=G1∗⋯∗Gn, where 1<#⁢Gi<∞ for i=1,…,n, with a Kurosh free decomposition:

(2.1)H=F∗∗i=1n(∗jHi,jgi,j),

where F is a free group and Hi,j≠{1} is a subgroup of Gi for all j. If H is a verbally closed subgroup in G, then the following holds:

F is the trivial group.
For any two subgroups Hi,j1 and Hi,j2 from decomposition (2.1) we have: if there are elements fi, gi∈Gi and natural numbers k1, k2 such that
fik1∈Hi,j1 𝑎𝑛𝑑 fik2∈Hi,j2gi,
then either fik1=1 or fik2=1.

The second assertion of the theorem means that subgroups Hi,j1 and Hi,j2gi cannot simultaneously intersect the same cyclic subgroup nontrivially. For example, the following condition can be easily deduced from the second assertion of the theorem:

For any two subgroups Hi,j1 and Hi,j2 from decomposition (2.1) we have: for an element gi∈Gi the intersection Hi,j1∩Hi,j2gi is trivial.

The following examples and corollaries demonstrate how the second assertion of this theorem can be applied.

Example 1.

Let H=〈a〉2∗〈bc〉2 be a subgroup of

G=G1∗G2=〈a,b∣a2,b2,(a⁢b)3〉∗〈c∣c2〉≃S3∗ℤ2.

Then H is not a verbally closed subgroup in G since for g1=b⁢a∈G1 we have 〈a〉2∩〈bg1〉2=〈a〉2∩〈a〉2≠{1} and it is the contradiction to the second assertion of the theorem.

Example 2.

Let H=〈a〉2∗〈bc〉3 be a subgroup of

G=G1∗G2=〈a,b∣a2,b3,[a,b]〉∗〈c∣c2〉≃ℤ6∗ℤ2.

Then H is not a verbally closed subgroup in G since for f1=a⁢b∈G1 we have (a⁢b)3=a∈〈a〉2 and (a⁢b)2=b2∈〈b〉3 but neither (a⁢b)3=1 nor (a⁢b)2=1.

Corollary 1.

If H is a verbally (algebraically) closed subgroup of G=∗i=1nGi, 1<#⁢Gi<∞, i=1,…,n, then it is a finitely generated group.

Proof.

Let H be a verbally closed subgroup of G with decomposition (2.1). Then F={1} by the first assertion of Theorem 1. Since all free factors Hi,jgi,j in this decomposition are finite groups, we need to prove that there are only finitely many of them. If we assume the contrary, then (since the set of all subgroups of all groups Gi, i=1,…,n, is finite) there are indices i, j1 and j2 such that Hi,j1=Hi,j2. But it means (by the second assertion of Theorem 1) that H cannot be verbally closed in G. ∎

Applying this corollary and the fact that a finitely generated algebraically closed subgroup of a finitely presented group is a retract ([11]) we can conclude the following.

Corollary 2.

For subgroups of G=G1∗⋯∗Gn, 1<#⁢Gi<∞, i=1,…,n, the properties of being algebraically closed and being a retract are equivalent.

Conditions (1) and (2) of the theorem are necessary for verbal closedness but not sufficient. Indeed, it is clear that they hold for the subgroup H=〈a〉n∗〈bc〉n of G=〈a,b∣an,bn,[a,b]〉∗〈c∣cn〉≃(ℤn×ℤn)∗ℤn, n⩾2 or n=∞. The goal of the two following theorems is to prove that if n=2 or n=∞, then H is not a verbally closed subgroup in G.

Theorem 2.

The subgroup H=〈a〉2∗〈bc〉2 of the group

G=〈a,b∣a2,b2,[a,b]〉∗〈c∣c2〉≃(ℤ2×ℤ2)∗ℤ2

is not verbally closed.

Proof.

Since 〈a,b∣a2,b2,[a,b]〉∗〈c|c2〉≃〈a,b,c∣a2,b2,c2,[a,bc]〉, it is sufficient to prove that H^=〈a〉2∗〈b〉2 is not a verbally closed subgroup in the group G^=〈a,b,c∣a2,b2,c2,[a,bc]〉.

Let us consider the equation

(2.2)(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(a⁢b)2.

This equation has a solution x=a, y=b, z=c in G^. We show that this equation has no solution in H^=〈a〉2∗〈b〉2.

Any element of H^ can be expressed as (b⁢a)k or (b⁢a)k⁢a where k is an integer. It means that, if we prove that any substitution of the form x=(b⁢a)k⁢aε1, y=(b⁢a)t⁢aε2, z=(b⁢a)s⁢aε3, where k,t,s∈ℤ and ε1,ε2,ε3∈{0,1}⊂ℤ is not a solution of equation (2.2), then we prove that this equation has no solution in H. Consider the following eight possible cases and evaluate the left-hand side of the equation:

If x=(b⁢a)k, y=(b⁢a)t, z=(b⁢a)s, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)6⁢(k+t).
If x=(b⁢a)k⁢a, y=(b⁢a)t, z=(b⁢a)s, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)-6⁢t.
If x=(b⁢a)k, y=(b⁢a)t⁢a, z=(b⁢a)s, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)6⁢k.
If x=(b⁢a)k, y=(b⁢a)t, z=(b⁢a)s⁢a, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)6⁢(k+t).
If x=(b⁢a)k⁢a, y=(b⁢a)t⁢a, z=(b⁢a)s, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)4⁢(-k+t-s).
If x=(b⁢a)k⁢a, y=(b⁢a)t, z=(b⁢a)s⁢a, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)6⁢t.
If x=(b⁢a)k, y=(b⁢a)t⁢a, z=(b⁢a)s⁢a, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)6⁢k.
If x=(b⁢a)k⁢a, y=(b⁢a)t⁢a, z=(b⁢a)s⁢a, then
(x3⁢[x,yz]⁢y3)2⁢[x,yz]3=(b⁢a)4⁢(-k+s).

It is clear that for any k, t and s the value of the left-hand side cannot be equal to (a⁢b)2 in H^. ∎

Theorem 3.

The subgroup H=〈a〉∞∗〈bc〉∞ of the group

G=〈a,b∣[a,b]〉∗〈c∣∅〉≃(ℤ×ℤ)∗ℤ

is not verbally closed.

Proof.

A word wr⁢(x1,…,xm) is called a C-test word in m letters for the free group Fr of rank r⩾2 if for any two tuples (g1,…,gm) and (g1′,…,gm′) of elements of Fr the following holds: if wr⁢(g1,…,gm)=wr⁢(g1′,…,gm′)≠1 in Fr, then there is an element s∈Fr such that gi′=gis, i=1,…,m. In [5] Ivanov introduced and constructed first C-test words for Fr in m letters for any r,m⩾2. In [7] Lee constructed for each r,m⩾2 a C-test word wr⁢(x1,…,xm) for Fr with the additional property that wr⁢(g1,…,gm)=1 if and only if the subgroup of Fr generated by g1,…,gm is cyclic. We refer to such words as Lee words.

Since G=〈a,b∣[a,b]〉∗〈c∣∅〉≃G^=〈a,b,c∣[a,bc]〉, it is sufficient to prove that H^=〈a〉∞∗〈b〉∞ is not a verbally closed subgroup in the group G^=〈a,b,c∣[a,bc]〉. For this purpose consider the equation

(2.3)w2⁢(x1,x2,[x1,x2x3])=w2⁢(a,b,1),

where w2⁢(x1,x2,x3) is a Lee word for F2 in three letters. Equation (2.3) has a solution x1=a, x2=b, x3=c in G^. We show that this equation has no solution in H^=F2⁢(a,b).

First, we note that since the subgroup 〈a,b〉=F2⁢(a,b) is not cyclic, by the property of Lee words we have w2⁢(a,b,1)≠1 in H^. If xi=hi, i=1,2,3, is a solution for (2.3) in H^, then (by the definition of Lee words) there is an element s∈H^ such that h1=as, h2=bs and [h1,h2h3]=1s. It is clear that these three equalities cannot be satisfied simultaneously in H^ for any s. ∎

3 The proof of the main result

One can easily see that under homomorphisms images of verbally closed subgroups are not such in general. Indeed, consider the group 〈a〉2∗〈b〉2∗〈c〉2 and its subgroup H=〈c〉2. Since H is a free factor of G, it is verbally closed in G (actually, it is a retract in G). Let H¯ be the image of H under the natural homomorphism onto a factor group G/R, where R is the normal closure of an element [a,b]⁢c. The equation [x,y]⁢c¯=1¯ has a solution x=a¯, y=b¯ in G/R (a¯, b¯ and c¯ are the images of a, b and c, respectively). But this equation has no solution in H¯ since it is a cyclic group and c¯≠1¯ in H¯. But under some additional restrictions the verbal closedness of subgroups is preserved for their homomorphic images.

Lemma 1 ([13]).

Let the kernel K=ker⁡φ of an epimorphism φ:G→G¯ be a verbal subgroup in G. If H is a verbally closed subgroup in G, then H¯=φ⁢(H) is a verbally closed subgroup in G¯.

Proof.

Suppose that an equation

(3.1)w⁢(x¯)⁢h¯-1=1¯,

where h¯∈H¯ has a solution x¯=g¯¯ in G¯ (recall that x¯ and g¯¯ are the abridged notations for tuples of the form (x1,…,xs) and (g¯1,…,g¯s), respectively). We show that equation (3.1) has a solution in H¯.

Let g¯ be a preimage of the tuple g¯¯ and let h be a preimage of h¯ in G. Then in G we have the equality

w⁢(g¯)⁢h-1=k,

where k∈K.

Suppose that K is generated by a set

V={fα⁢(g1,…,gnα)=fα⁢(gα¯)∣g1,…,gnα∈G}.

We fix an expression k=k⁢(fα1⁢(gα1¯),…,fαm⁢(gαm¯)), where fαi⁢(gαi¯)∈V, i=1,…,m, of the element k and set up the following equation:

k⁢(fα1⁢(y1¯),…,fαm⁢(ym¯))-1⁢w⁢(x¯)⁢h-1=1.

This equation has a solution x¯=g¯, yi¯=gαi¯, i=1,…,m in G. Since H is a verbally closed subgroup in G, this equation has a solution x¯=h¯, yi¯=hαi¯, i=1,…,m in H. Since K is a verbal subgroup in G, we have

k⁢(fα1⁢(hα1¯),…,fαm⁢(hαm¯))∈K.

From this we can conclude that the tuple φ⁢(h)¯ is a solution of (3.1) in H¯. ∎

Corollary 3.

If H is a verbally closed subgroup in G and φ is an automorphism of G, then φ⁢(H) is a verbally closed subgroup in G.

Lemma 2.

If H is a verbally closed subgroup in G and H^ is a verbally closed subgroup in H, then H^ is a verbally closed subgroup in G, i.e. verbal closedness is transitive.

Proof.

Let w⁢(x¯)=h^, h^∈H^ be an equation which has a solution in G. Since h^∈H and H is a verbally closed subgroup in G, this equation has a solution in H. Since H^ is a verbally closed subgroup in H, we can conclude that the equation has a solution in H^. ∎

Lemma 3.

If H=H1∗H2 is a verbally closed subgroup in G, then H1 and H2 are verbally closed subgroups in G.

Proof.

The subgroup H1 is a retract of H (since it is a free factor of H), thus it is a verbally closed subgroup in H. By Lemma 2, H1 is a verbally closed subgroup in G (the same arguments hold for H2). ∎

Notice that if subgroups H1 and H2 are verbally closed in G, then the subgroup H1∗H2 may not be verbally closed. Indeed, the subgroups H1=〈a〉2 and H2=〈b〉2 are retracts in 〈a,b,c∣a2,b2,c2,[a,bc]〉, but 〈a〉2∗〈b〉2 is not a verbally closed subgroup in 〈a,b,c∣a2,b2,c2,[a,bc]〉 (see Theorem 2 for a proof).

To prove the first assertion of Theorem 1, we need to establish the following lemma.

Lemma 4.

If Fr is the free subgroup of rank r⩾1 in the group G=G1∗⋯∗Gn, 1<#⁢Gi<∞, i=1,…,n, then Fr is not verbally closed in G.

Proof.

For each group Gi, i=1,…,n we fix a set Si={s1i,…,stii} of generators for it. In this case S=⋃i=1nSi={s1,…,st}, where t=t1+⋯+tn, is a generating set of G.

If r=1, we have F1=〈f〉∞ for some element f∈G. Fix an expression f=si1⁢si2⁢⋯⁢sim of the element f, where all sij are elements of S. Let p be a prime number such that p>max⁡({#⁢G1,…,#⁢Gn}); then the equation

x1p⁢x2p⁢⋯⁢xmp=f

has a solution xj=sijkj, j=1,…,m, in G (if sij∈Gl, then kj is a natural number such that kj⁢p≡1(mod#⁢Gl)). Since the left-hand side of the equation has the form fp⁢q, q∈ℤ, for any substitution xj=fnj, j=1,…,m, we can conclude that this equation has no solution in F1=〈f〉∞.

If r⩾2 (possibly r=∞), then Fr=〈f〉∞∗Fr-1 for some f∈G. If Fr is a verbally closed subgroup in G, then (by Lemma 3) 〈f〉∞ is a verbally closed subgroup in G. But this contradicts the result of the preceding paragraph. ∎

To prove the second assertion of Theorem 1, we need to establish the following lemma.

Lemma 5.

Let G be a group of the type G1∗⋯∗Gn, 1<#⁢Gi<∞, i=1,…,n. If H1 and H2 are subgroups of a free factor Gi and there are natural numbers k1, k2 and an element f∈Gi such that fk1∈H1, fk1≠1 and fk2∈H2, fk2≠1, then a subgroup of the form H1∗H2g, g∈G, is not verbally closed in G.

To prove this lemma, we need some auxiliary facts.

The following lemma is well known.

Lemma 6.

Each automorphism of a directed tree has either a fixed vertex or an invariant line. If an automorphism has no fixed edges, then it has only one invariant line.

Proof.

Let f be an automorphism of a directed tree T which fixes no vertices. Let us assume

m=infx∈V⁢(T)⁡(ρ⁢(x,f⁢(x))),

where ρ⁢(x,y) is the standard distance between vertices x and y and V⁢(T) is the set of all vertices of the tree T.

Since the distance ρ⁢(x,f⁢(x)) is a natural number for all x∈V⁢(T), there is a vertex y∈V⁢(T) such that ρ⁢(y,f⁢(y))=m. Let z be the middle point of the segment [y,f⁢(y)]; then f⁢(z) is the middle point of the segment [f⁢(y),f2⁢(y)] and

ρ⁢(z,f⁢(z))⩽ρ⁢(z,f⁢(y))+ρ⁢(f⁢(y),f⁢(z))=m.

Due to our choice of the vertex y this inequality must be an equality. But the equality ρ⁢(z,f⁢(z))=m means that the point f⁢(y) is located between the points z and f⁢(z), and therefore between the points y and f2⁢(y). It follows that all points fn⁢(y), n∈ℤ are located at the same line (which is an invariant line for the automorphism f).

Now we give a sketch of the proof of the second part of this lemma. Suppose that an automorphism f has two invariant lines a and b (we still surmise that f has no fixed vertices). In theory, there are four possibilities: the intersection of these lines is the empty set, a vertex, a segment or a beam.

If the intersection of the lines a and b is the empty set, then there is a unique segment c which connects these lines. When f acts on T, the ends of this segment must remain on their invariant lines, but it means that each edge of the segment c is fixed. This contradicts our assumption.

The same arguments hold mutatis mutandis for the remaining three cases. ∎

Definition 3.

Any element A of a group G=G1∗⋯∗Gn, where 1<#⁢Gi<∞ for i=1,…,n, can be expressed as the reduced product A=g1⁢⋯⁢gs, where each element gi lies in some free factor Gk. The norm (or syllable length) |A| of an element A is the number of syllables in its reduced product.

Lemma 7.

Let A be a cyclically reduced word of infinite order in the group G=G1∗⋯∗Gn, 1<#⁢Gi<∞, i=1,…,n, and let g∈G be such that the elements A and Ag do not commute in G. If D is a cyclically reduced word in the conjugacy class of an element AN1⁢Ag⁢N2, where N1,N2∈N, N1,N2⩾2, then the following estimate holds:

|D|>(N1+N2-4)⁢|A|.

Proof.

It is well known (see, e.g., [10]) that every free product G=G1∗⋯∗Gn of groups Gi, i=1,…,n can be realized as a group of symmetries on some tree T. Furthermore, the stabilizers of the vertices of this tree under the action of the group G1∗⋯∗Gn are conjugates of Gi for some i and stabilizers of the edges are trivial. We fix such a representation of G.

Since A is an element of infinite order, the automorphism of the tree T which corresponds to the action of A has no fixed points, therefore (by Lemma 6) this automorphism has a unique invariant line (the same is true for Ag). Hereafter we identify the elements of G with the automorphisms of the tree T that correspond to the actions of these elements.

Let α be the invariant line of the element A and let β be the invariant line of the element Ag. Notice that under the action of the elements A and Ag on the tree T the vertices of their invariant lines α and β are moved by the same distance (which is |A|). We show that the length |I| of the intersection I=α∩β is less than 2⁢|A|. Indeed, if |I|⩾2⁢|A|, then there is a vertex u∈V⁢(I) such that either A2⁢A2⁢g∈St⁡(u) (where St⁡(w) is the stabilizer of a vertex w under the action) or A-2⁢A2⁢g∈St⁡(u) (this is true due to the fact that the elements A and Ag move their invariant lines by the same distance and a point u can be selected so that A2⁢g⁢u∈V⁢(I)). Consider, for example, the first case (the same arguments hold mutatis mutandis for the second case). If A2⁢A2⁢g∈St⁢(u), then A⁢Ag∈St⁢(u) and A⁢Ag∈St⁢(Ag⁢u) (to see it easily look at the vertices u, Ag⁢u and A2⁢g⁢u in Figure 1), but the last means that under the action of the element A⁢Ag the segment [u,Ag⁢u] is fixed. Since under the action the stabilizers of the edges are trivial the element A⁢Ag must be the identity of G but this is the contrary to the condition of the lemma.

$Figure 1 Intersection of the invariant lines α and β of the elements AN1{A^{N_{1}}} and Ag⁢N2{A^{gN_{2}}}. Arrows at the ends of lines indicate the directions in which the elements AN1{A^{N_{1}}} and Ag⁢N2{A^{gN_{2}}} shift the lines.$

Figure 1

Intersection of the invariant lines α and β of the elements AN1 and Ag⁢N2. Arrows at the ends of lines indicate the directions in which the elements AN1 and Ag⁢N2 shift the lines.

Now we can establish the estimate. To do this we establish the estimate on what minimum distance the element AN1⁢Ag⁢N2 can move a vertex v of the tree T. It is not difficult to see that the minimal shift is possible in the following case (this case is depicted in Figure 1). The invariant lines α and β intersect in a segment I=[w1,w2] of length 2⁢|A|-1 and vertices v and Ag⁢N2⁢v are located on invariant line β outside the segment I and on opposite sides of this segment; the ends (vertices w2 and AN1⁢w2) of the segments which connect the vertices Ag⁢N2⁢v and AN1⁢Ag⁢N2⁢v with the invariant line α and belong to this line lie on α on opposite sides of the segment I. In this case the distance between vertices v and AN1⁢Ag⁢N2⁢v is

|[v,AN1⁢Ag⁢N2⁢v]|=|[v,w1]|+|[w1,AN1⁢w2]|+|[AN1⁢w2,AN1⁢Ag⁢N2⁢v]|=|[v,Ag⁢N2⁢v]|-|[w2,Ag⁢N2⁢v]|-|I|+|[w2,AN1⁢w2]|-|I|+|[AN1⁢w2,AN1⁢Ag⁢N2⁢v]|=|[v,Ag⁢N2⁢v]|+|[w2,AN1⁢w2]|-2⁢|I|⩾N1⁢|A|+N2⁢|A|-2⁢(2⁢|A|-1)=(N1+N2-4)⁢|A|+2,

where we have used the following facts: |[AN1⁢w2,AN1⁢Ag⁢N2⁢v]|=|[w2,Ag⁢N2⁢v]|, |[v,Ag⁢N2⁢v]|=N2⁢|A| and |[w2,AN1⁢w2]|=N1⁢|A|. ∎

Now we can prove Lemma 5.

Proof.

Let H1∗H2g be a subgroup of G. For the elements f and g we fix expressions f=f⁢(s1,…,st)=f⁢(s¯) and g=g⁢(s1,…,st)=g⁢(s¯), where si∈S for i=1,…,t (S is the generation set of G which has been described above during the proof of Lemma 4). Let N=1+∏i=1n#⁢Gi; then the equation

f⁢(x1,…,xt)k1⁢N⁢g⁢(x1,…,xt)⁢f⁢(x1,…,xt)k2⁢N⁢g⁢(x1,…,xt)-1

(3.2)=fk1⁢g⁢fk2⁢g-1

with variables x1,…,xt has a solution xi=si, i=1,…,t, in the group G. We denote H2g as H2′ and (using abridged notations f⁢(x1,…,xt)=f⁢(x¯) and g⁢(x1,…,xt)=g⁢(x¯)) rewrite equation (3.2) in the following form:

(3.3)f⁢(x¯)k1⁢N⁢g⁢(x¯)⁢f⁢(x¯)k2⁢N⁢g⁢(x¯)-1=a⁢b,

where a=fk1∈H1, a≠1, and b=g⁢fk2⁢g-1∈H2′, b≠1. We show that equation (3.3) has no solution in H1∗H2′. Let a tuple h¯=(h1,…,ht) of elements of H1∗H2′ be a solution of the equation. Thus, in H1∗H2′ we have the following equality:

(3.4)f⁢(h¯)k1⁢N⁢g⁢(h¯)⁢f⁢(h¯)k2⁢N⁢g⁢(h¯)-1=a⁢b.

We consider three cases.

(1) Suppose that the element f⁢(h¯) has a finite order in H1∗H2′. Then either f⁢(h¯)∈H1s or f⁢(h¯)∈H2′⁣s for some element s∈H1∗H2′. To be more specific, let f⁢(h¯) be an element of H1s, then the image of equality (3.4) under the natural epimorphism φ:H1∗H2′→H1∗H2′/[H1,H2′]≃H1×H2′ has the following form:

f¯k1⁢h¯1⁢f¯k2⁢h¯1-1=a¯⁢b¯,

where f¯,h¯1∈H1. But the last equality cannot hold in H1×H2′ since

f¯k1⁢h¯1⁢f¯k2⁢h¯1-1∈H1 and a¯⁢b¯∉H1.

(2) Suppose that the elements f⁢(h¯) and g⁢(h¯)⁢f⁢(h¯)⁢g⁢(h¯)-1 have infinite order and commute in H1∗H2′. In this case there is an element w∈H1∗H2′ such that f⁢(h¯)=wm1 and g⁢(h¯)⁢f⁢(h¯)⁢g⁢(h¯)-1=wm2 for some integer numbers m1 and m2 (indeed, in this case the subgroup 〈f⁢(h¯),g⁢(h¯)⁢f⁢(h¯)⁢g⁢(h¯)-1〉 is an abelian group of infinite order and in view of the Kurosh subgroup theorem it ought to have the form 〈w〉∞ for some element w). Thus, we can rewrite equality (3.4) as

w(k1⁢m1+k2⁢m2)⁢N=a⁢b.

But this is impossible in H1∗H2′ since N⩾2 and the element ab is not a proper power in H1∗H2′.

(3) Suppose that the elements f⁢(h¯) and g⁢(h¯)⁢f⁢(h¯)⁢g⁢(h¯)-1 have infinite order and do not commute in H1∗H2′. In this case there are natural numbers m1, m2 and a cyclically reduced word A∈H1∗H2′ such that

f⁢(h¯)=Af1⁢m1 and g⁢(h¯)⁢f⁢(h¯)⁢g⁢(h¯)-1=Af2⁢m2

for some elements f1,f2∈H1∗H2′. Then it is straightforward to see that in this case equality (3.4) can be rewritten as

(3.5)Ak1⁢m1⁢N⁢Af1-1⁢f2⁢k2⁢m2⁢N=(a⁢b)f1-1,

where the elements Ak1⁢m1⁢N and Af1-1⁢f2⁢k2⁢m2⁢N (and therefore the elements A and Af1-1⁢f2) do not commute in H1∗H2′.

By Lemma 7 the left-hand side of (3.5) is an element in the conjugacy class of a cyclically reduced word with norm greater than (k1⁢m1⁢N+k2⁢m2⁢N-4)⁢|A|. Since the right-hand side of this equality is an element in the conjugacy class of the cyclically reduced word with norm two and ((k1⁢m1+k1⁢m2)⁢N-4)⁢|A|⩾((1⋅1+1⋅1)⁢(1+4)-4)⋅2=12 (in this evaluation we used the facts: #⁢Gi>1 for all i, n>1, k1,k2,m1,m2∈ℕ and N=1+∏i=1n#⁢Gi), we obtained the contradiction. ∎

Now we are ready to prove Theorem 1.

Proof.

Let H be a verbally closed subgroup in G=G1∗⋯∗Gn, 1<#⁢Gi<∞, i=1,…,n, with Kurosh decomposition (2.1).

(1) If H is a verbally closed subgroup in G, then (by Lemma 3) each of its free factors is a verbally closed subgroup in G. In view of Lemma 4 a free subgroup can be verbally closed in G if and only if it is the trivial group.

(2) Suppose that for some index i in decomposition (2.1) there are subgroups Hi,j1gi,j1 and Hi,j2gi,j2 and an element fi∈Gi such that

fik1∈Hi,j1,fik1≠1,fik2∈Hi,j2gi,fik2≠1

for some element gi∈Gi and natural numbers k1 and k2. By Lemma 3 the subgroup Hi,j1gi,j1∗Hi,j2gi,j2 is verbally closed in G, therefore (in accordance with Corollary 3) the subgroup

Hi,j1∗Hi,j2gi,j1-1⁢gi,j2

is also verbally closed in G. But this is in contradiction to Lemma 5. ∎

Communicated by Alexander Olshanskii

Funding source: Russian Foundation for Basic Research

Award Identifier / Grant number: 15-01-05823

Funding statement: The work of the author was supported by the Russian Foundation for Basic Research (project no. 15-01-05823).

Acknowledgements

The author thanks A. A. Klyachko for many useful conversations and many useful remarks. Also the author is grateful to the anonymous referee for useful remarks.

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Received: 2016-7-10

Revised: 2016-11-4

Published Online: 2017-1-12

Published in Print: 2017-9-1

Articles in the same Issue

https://doi.org/10.1515/jgth-2016-0058