A non-Hopfian relatively hyperbolic group with respect to a Hopfian subgroup

We produce an example demonstrating that every finitely generated relatively hyperbolic group with respect to a collection of Hopfian subgroups need not be Hopfian. This answers a question of Osin \cite[Problem 5.5]{Osin} in the negative.


Introduction
Recall that a group G is Hopfian if every epimorphism G → G is an automorphism.Recall also that a group G is residually finite if for every g ∈ G \ {1}, there is some finite group P and an epimorphism ψ : G → P so that ψ(g) = 1.Inspirited by well-known questions about ordinary hyperbolic groups, Osin [8, Problems 5.5 and 5.6] asked the following questions.
• If a finitely generated group G is hyperbolic relative to a collection of Hopfian subgroups {H 1 , . . ., H m }, does it follow that G is Hopfian?• If a group G is hyperbolic relative to a collection of residually finite subgroups {H λ } λ∈Λ , does it follow that G is residually finite?Later, Osin [11] proved that the second question is equivalent to Gromov's famous open question of whether every hyperbolic group is residually finite.The Hopf property and the residual finiteness property have a close connection.In particular, Mal'cev [7] proved that every finitely generated residually finite group is Hopfian.Mal'cev's result provides a useful tool to prove that a certain finitely generated group is non-residually finite.The Hopf properties of torsion-free hyperbolic groups, toral relatively hyperbolic groups, hyperbolic groups with torsion, lacunary hyperbolic groups and finitely presented C ′ (1/6) or C ′ (1/4)-T (4) small cancellation groups were verified by many authors (see [2,5,12,13,15]).In contrast, Wise [16] constructed a non-Hopfian CAT (0)-group.
On the other hand, there is another property related to the Hopf property.A group G is called equationally noetherian if for every system of equations in G, there exists a finite subsystem that has the same set of solutions.It is well-known that every finitely generated equationally noetherian group is Hopfian.Reinfeldt and Weidmann [12] proved that every hyperbolic group is equationally noetherian.
Also, for relatively hyperbolic groups, Groves and Hull [6] proved that if a group G is hyperbolic relative to a collection of equationally noetherian subgroups, then G is itself equationally noetherian.However, it has been unknown up to the present whether every finitely generated group that is hyperbolic relative to a collection of Hopfian subgroups is Hopfian.
The main result of this paper is the following.This solves Osin's first question mentioned above in the negative.
Theorem 1.1.Let H 0 be the group given by the presentation and take successively two HNN-extensions from H 0 as follows: Next, form the free product H = H 2 * e, f | ∅ .Finally, letting a be an infinite cyclic group, take successively two multiple HNN-extensions from H * a as follows: Then G is a non-Hopfian group which is hyperbolic relative to the Hopfian subgroup H.In more detail, the following hold.
(i) K is hyperbolic relative to the subgroup H. (2) The subgroup H is non-residually finite.Indeed, for any finite group P and for any epimorphism ψ from H to P , ψ(c 3 ) = 1.The reason can be seen as follows.
From the defining relation t −1 st = s 3 of H, it follows that ψ(s) and ψ(s) 3 have the same order, so that the order of ψ(s) is relatively prime to 3, say m.Also from the defining relation s −1 bs = bc −3 of H, it follows that b −1 sb = sc 3 in H, so that ψ(sc 3 ) m = 1.Here, since ψ(s) and ψ(c) commute with each other, ψ(c 3 ) m = 1.On the other hand, since c 9 = 1 in H, ψ(c 3 ) 3 = 1, which together with ψ(c 3 ) m = 1 finally yields ψ(c 3 ) = 1.
This paper is organized as follows.In Section 2, we recall necessary definitions and known results to be used throughout this paper.The proof of Theorem 1.1 is contained in Sections 3-6.In Section 3, by using Osin's theorem concerning the unique maximal elementary subgroups of hyperbolic elements in relatively hyperbolic groups, we first prove that the free product H * a is hyperbolic relative to the collection of subgroups {H, a , ac }.And then by successively using Osin's combination theorem for relatively hyperbolic groups, we show that K is hyperbolic relative to H.In Section 4, again by using Osin's theorem about unique maximal elementary subgroups, we show that the peripheral structure of K can be extended to the collection of subgroups {H, u , v }.At this point, by using Osin's combination theorem twice, we obtain that G is hyperbolic relative to H.In Section 5, we show that G is non-Hopfian by constructing a particular surjective, but not injective, endomorphism of G. To be more precise, the endomorphism of G induced by the mapping b → b, c x → x and y → y is shown to be surjective but not injective.Finally, Section 6 is devoted to the proof of that H is Hopfian, in which Bass-Serre theory plays a crucial role.

Preliminaries
In this section, we recall necessary definitions, notation and known results to be used throughout this paper.
2.1.Relatively hyperbolic groups.In this paper, we adopt Osin's definition [9] among many equivalent definitions of relatively hyperbolic groups.
Let G be a group, H = {H λ } λ∈Λ a collection of subgroups of G, and X a subset of G. Suppose that X is a relative generating set for (G, H), namely, G is generated by the set λ∈Λ H λ ∪ X (for convenience, we assume that X = X −1 ).Then G can be regarded as the quotient group of the free product where the groups Hλ are isomorphic copies of H λ , and F (X) is the free group generated by X.Let H be the disjoint union For every λ ∈ Λ, we denote by S λ the set of all words over the alphabet Hλ \ {1} that represent the identity in F .Let S be the disjoint union Then we may describe G as a relative presentation with respect to the collection of subgroups {H λ } λ∈Λ , where R ⊆ F .If both the sets R and X are finite, relative presentation ( 4) is said to be finite and the group G is said to be finitely presented relative to the collection of subgroups H.For every word w over the alphabet X ∪ H representing the identity in the group G, there exists an expression with the equality in the group F , where R i ∈ R and f i ∈ F for i = 1, . . ., k.The smallest possible number k in a presentation of the form (5) is called the relative area of w and is denoted by Area rel (w).
Definition 2.1 (Relatively hyperbolic groups).A group G is said to be hyperbolic relative to a collection of subgroups H if G admits a relatively finite presentation (4) with respect to H satisfying a linear relative isoperimetric inequality.That is, there is a constant C > 0 such that for any cyclically reduced word w over the alphabet X ∪ H representing the identity in G, we have where w is the length of the word w.This definition is independent of the choice of the finite relative generating set X and the finite set R in (4).

2.2.
Unique maximal elementary subgroups.Suppose that G is hyperbolic relative to a collection of subgroups H = {H λ } λ∈Λ .Then we refer to the collection H as a peripheral structure of G, and any element in H as a peripheral subgroup of G.
An element is called hyperbolic if it has infinite order and it is not conjugate to any element of a peripheral subgroup of G. Due to Osin [10], there is a well-known example of subgroups which may be added to enlarge peripheral structures.).Let G be hyperbolic relative to a collection of subgroups H. Then for any hyperbolic element g ∈ G, G is hyperbolic relative to H ∪ {E(g)}, where E(g) is the unique maximal elementary subgroup containing g defined as follows: 2.3.Osin's combination theorem.We recall one of Osin's combination theorems for relatively hyperbolic groups.Earlier, Dahmani [3] proved the following combination theorem for finitely generated groups.In fact, applying Dahmani's combination theorem is sufficient for our purposes in this paper, but we introduce Osin's combination theorem in order to match with the definition of relatively hyperbolic groups stated above.Corollary 1.4]).Suppose that a group G is hyperbolic relative to a collection of subgroups H = {H λ } λ∈Λ .Assume in addition that there exists a monomorphism ι : H µ → H ν for some µ = ν ∈ Λ, and that H µ is finitely generated.Then the HNN-extension 2.4.Bass-Serre trees for HNN-extensions.We recall some basic concepts of Bass-Serre theory (see [1,14]).A graph of groups (A, X) consists of a connected graph X and a collection of groups indexed by the vertices and edges of X, and a family of monomorphisms from the edge groups to the adjacent vertex groups.For each spanning tree T in X, one can canonically associate a unique group, called the fundamental group and denoted π 1 (A, T ).Here, it turns out that the fundamental group π 1 (A, T ) is independent of the choice of a spanning tree T , so that we simply write π 1 (A) instead of π 1 (A, T ).The fundamental group π 1 (A) admits an orientation-preserving action on a tree Γ such that the quotient graph A/π 1 (A) is isomorphic to X.Such a tree is called a Bass-Serre tree of A.
On the other hand, given a graph of groups (A, X) with the fundamental group G ∼ = π 1 (A), where G is an HNN-extension, one can construct a Bass-Serre tree for G due to the following theorem (see, for example, [17]).This result plays an important role in the proof of Proposition 6.2.
Theorem 2.4 (Bass-Serre trees for HNN-extensions).Suppose that G * is an HNNextension of a group G with associated isomorphism ι between two subgroups H and K, that is,  Then Γ is a tree and G * acts on Γ without inversion by left multiplication such that the quotient graph Γ/G * is isomorphic to X, where X is the underlying graph of A.

Proof of Theorem 1.1(i)
Let H and K be the groups defined in the statement of Theorem 1.1.The aim of this section is to prove the relative hyperbolicity of K with respect to the subgroup H.
We start with the free product H * a , which is clearly hyperbolic relative to the collection of subgroups {H, a }.Recall that ( 6) Clearly, ac is a hyperbolic element in H * a seen as a relatively hyperbolic group with peripheral structure {H, a }.Moreover, we can prove the following Claim A. The unique maximal elementary subgroup E(ac) of H * a is precisely the infinite cyclic subgroup ac .
Proof.Suppose to the contrary that E(ac) \ ac = ∅.Among all such elements in E(ac) \ ac , we take an element in normal form, say f , with shortest syllable length.Here, by the syllable length, we mean the total number of syllables which are maximal subwords consisting entirely of letters from either H or a .For such f , clearly f (ac) ±n f −1 = (ac) n for some n ∈ N.Moreover, f satisfies the following.(i) f does not begin with ah for any 1 = h ∈ H, nor with c −1 a ′ for any 1 = a ′ ∈ a ; (ii) f does not end with ha −1 for any 1 = h ∈ H, nor with a ′ c for any 1 = a ′ ∈ a .The reason goes as follows.First, assume that f begins with ah for some 1 = h ∈ H, that is, f ≡ ahf 1 in normal form.Then it follows from the equality f (ac Here, since c −1 h ∈ H, the element c −1 hf 1 has shorter syllable length than f does.This is a contradiction to our choice of f .Next, assume that f begins with c −1 a ′ for some and so This means that aa ′ f 2 ∈ E(ac) \ ac .In addition, aa ′ f 2 has shorter syllable length than f does, since aa ′ ∈ a .This is also a contradiction to our choice of f .So (i) holds.
For (ii), note that f −1 ∈ E(ac) \ ac with the same syllable length as that of f .Also, if f ends with ha −1 for some 1 = h ∈ H, or with a ′ c for some 1 = a ′ ∈ a , then f −1 begins with ah −1 or with c −1 a ′ −1 .But then by the same argument applied to f in the proof of (i), we reach a contradiction.So (ii) holds.
But then the expression f (ac) ±n f −1 (ac) −n cannot represent the identity element in H * a by the normal form theorem for free products.This contradiction completes the proof of the claim.
The above Claim A together with Theorem 2.2 yields that H * a is hyperbolic relative to the collection of subgroups {H, a , ac }.Then due to Theorem 2.3, the group is hyperbolic relative to the subgroup H again by Theorem 2.3, completing the proof of Theorem 1.1(i).

Proof of Theorem 1.1(ii)
Let H, K and G be the groups defined in the statement of Theorem 1.1.The aim of this section is to prove the relative hyperbolicity of G with respect to the subgroup H.By the result of Section 3, K is relatively hyperbolic with peripheral structure {H}.
Since K is a multiple HNN-extension of H * a with stable letters u and v, the element u is clearly a hyperbolic element in K.Moreover, we can prove the following Claim B. The unique maximal elementary subgroup E(u) of K is precisely the cyclic subgroup u .Proof.To find E(u), view K as an HNN-extension with stable letter u of Suppose to the contrary that E(u) \ u = ∅.Among all such elements in E(u) \ u , we take an element, say f , in u-reduced form with minimal number of u ±1 .For such f , clearly f u ±n f −1 = u n for some n ∈ N.Moreover, f satisfies the following.
(i) f does not begin with hu for any h ∈ bacb −1 , nor with a ′ u −1 for any a ′ ∈ a ; (ii) f does not end with u −1 h for any h ∈ bacb −1 , nor with ua ′ for any a ′ ∈ a .The reason is as follows.First, assume that f begins with hu for some h ∈ bacb −1 , that is, f ≡ huf 1 (u-reduced).It then follows from the equality f u ±n f −1 = u n that But clearly a ′ f 1 has fewer number of u ±1 than f does, which is a contradiction to the choice of f .Next, assume that f begins with a ′ u −1 for some In addition, hf 2 has fewer number of u ±1 than f does, which is also a contradiction to the choice of f .Therefore, (i) holds.
For (ii), note that f −1 ∈ E(u) \ u with the same number of u ±1 as f has.Also, if f ends with u −1 h for some h ∈ bacb −1 , or with ua ′ for any a ′ ∈ a , then f −1 begins with h −1 u or with a ′ −1 u −1 .But then by the same argument applied to f in the proof of (i), we reach a contradiction.So (ii) holds.
But then the expression f u ±n f −1 u −n cannot not represent the identity element in K by Britton's Lemma.This contradiction completes the proof of the claim.
By Claim B together with Theorem 2.2, the peripheral structure K can be extended to {H, u }.In this point of view, v is a hyperbolic element in K, since v has infinite order, and since is not conjugate to any element of H nor to any element of u .Moreover, the unique maximal elementary subgroup E(v) of K is precisely the cyclic subgroup u .To see this, view K as an HNN-extension with stable letter v of M := H * a , u ≤ K, and then apply a similar argument as in the proof of Claim B. This together with Theorem 2.2 again, the peripheral structure of K can be extended further to the collection of subgroups {H, u , v }.

Proof of Theorem 1.1(iii)
Let G be the group defined in the statement of Theorem 1.1.The aim of this section is to prove that G is a non-Hopfian group by constructing a surjective, but not injective, endomorphism of G.
Let ψ be a unique homomorphism from a free group with basis {b, c, s, t, e, f, a, u, v, x, y} to G induced by the mapping Then it is easy to see that every defining relator in presentations ( 1)-(3b) is sent to the identity element in G by ψ.So ψ induces an endomorphism ψ of G.
We will show that ψ is surjective but not injective.To prove that ψ is surjective, it is sufficient to show that a, c, u, v ∈ im ψ.From the x-and y-relations in presentation (3b), it follows that u, v ∈ im ψ.This together with the v-relation in presentation (3a) yields that a ∈ im ψ, so that c ∈ im ψ from the u-relation in presentation (3a).Therefore, ψ is a surjective endomorphism of G.
which is obviously a contradiction.

Proof of Theorem 1.1(iv)
Let H be the group defined in the statement of Theorem 1.1.The aim of this section is to prove that H is a Hopfian group.Since the free product of two finitely generated Hopfian groups is also Hopfian (see [4]), it suffices to prove that H 2 is Hopfian.
Throughout this section, let ϕ be a surjective endomorphism of H 2 .We begin with the following Proposition 6.1.We may assume that ϕ(b) = b and ϕ(c) = c k with k ∈ Z.
So by replacing ϕ with the composition of ϕ and an appropriate inner automorphism of H 2 , we may assume that ϕ(H 0 ) ⊆ H 0 , so that ϕ(b), ϕ(c) ∈ H 0 .
First consider ϕ(c).Note that every element in H 0 can be written as This implies that the quotient group H 2 / c H 2 can be generated by two elements.But then, could be also generated by two elements, which is a contradiction.Hence ϕ(b) = 1, and thus ϕ(b) = bc m .Now let ρ be the unique homomorphism from a free group with basis {b, c, s, t} to H 2 induced by the mapping Then every defining relator in presentations ( 1)-( 2b) is sent to the identity element in H 2 by ρ.So ρ induces an endomorphism ρ of H 2 .Also, since m is an arbitrary integer, there exists an endomorphism ρ′ of H 2 defined by b → bc m , c → c, s → s and t → t.Then clearly ρ • ρ′ = id H 2 and ρ′ • ρ = id H 2 .This means that ρ is an automorphism of H 2 .By replacing further ϕ with the composition ρ • ϕ, we may finally assume that ϕ(b) = b and ϕ(c) = c k ′ with k ′ ∈ Z, as desired.Proposition 6.2.Under the assumption that ϕ(b) = b and ϕ(c) = c k with k ∈ Z, we may further assume that ϕ(s) = s ±3 p with p ∈ Z + ∪ {0} and that where h ≥ 1, ǫ i = ±1 and w i ∈ H 1 for every i = 1, . . ., h.
Proof.Let Γ be the Bass-Serre tree associated to H 2 viewed as an HNN-extension of H 1 (see Theorem 2.4).Denote by v the vertex labeled as the coset H 1 .Since ϕ(H 0 ) ⊆ H 0 ⊆ H 1 , clearly ϕ(H 0 ) fixes v.We shall show that ϕ(s) fixes v as well.Assume on the contrary that ϕ(s and since T is a tree, ϕ(s)ϕ(H 0 )ϕ(s) −1 fixes an edge in Γ, and thus it is a subgroup of some edge stabilizer of Γ.But since every edge stabilizer of Γ is conjugated to s which is an infinite cyclic group, and since ϕ(s)ϕ(H 0 )ϕ(s) −1 is finite, the only possibility is that ϕ(H 0 ) = {1}, which is impossible.Therefore, 3 , and hence This means that ϕ(s) fixes ϕ(t)v.Since ϕ(t)v = v, ϕ(s) fixes an edge in Γ.In particular, ϕ(s) fixes every edge on the geodesic [v, ϕ(t)v] in Γ.
Thus by replacing ϕ with the composition of ϕ and the inner automorphism of H 2 given by w 0 −1 , we may assume that ϕ where h ≥ 1, ǫ i = ±1 for every i = 1, . . ., h, and w ′ h , w j ∈ H 1 for every j = 1, . . ., h − 1.Then by replacing further ϕ with the composition ρ • ϕ, where ρ is the automorphism of H 2 (see the proof of Proposition 6.1) defined by ρ : b → bc −m , c → c, s → s and t → t, we may further assume that ϕ(b) = b and ϕ(c) = c k ′ , ϕ(s) ∈ H 1 and ϕ(t) as above.Now recall that ϕ(s) fixes every edge on the geodesic [v, ϕ(t)v] in Γ.In particular, ϕ(s) belongs to the stabilizer of the first edge on the geodesic [v, ϕ(t)v].Since ϕ(t) begins with the letter t or t −1 , the first edge on the geodesic [v, ϕ(t)v] is labeled as s or as t −1 s .So the stabilizer of the former is s and that of the latter is t −1 s t = s 3 .In either case, ϕ(s) ∈ s , that is, ϕ(s) = s r for some r ∈ Z.
Next assume that c −3k ′ = c −3k = c 3 .Let τ and τ ′ be the unique homomorphisms from a free group with basis {b, c, s, t} to H 2 induced by the mappings Again since every defining relator in presentations ( 1)-( 2b) is sent to the identity element in H 2 by both τ and τ ′ , there are endomorphisms τ and τ ′ of H 2 induced by τ and τ ′ , respectively.For such τ and τ ′ , clearly τ But from the defining relations in presentation (2a), we obtain that sbs −1 = bc 3 and that s ∓3 i bs ±3 i = b for any i ≥ 1. Combining these with (8) yields c 6 = 1 or c 3 = 1, contrary to the fact that c has order 9. Therefore, the only possibility to avoid a contradiction is that ϕ(s) = s, as desired.
Proposition 6.4.Under the assumption that ϕ , where ǫ i = ±1 and w i ∈ H 1 for every i = 1, . . ., h, we may further assume that ϕ(t Proof.Define the unique homomorphism ρ from a free group with basis {b, c, s, t} to H 2 induced by the mapping ρ : b → b, c → c, s → s and t → tw −1 h .Then every defining relator in prsentation (1)-(2b) is sent to the identity element in . So ρ induces an endomorphism ρ of H 2 .For the same reason, there exists the endomorphism ρ′ of H 2 defined by b → b, c → c, s → s and t → tw h .Then clearly ρ • ρ′ = id H 2 and ρ′ • ρ = id H 2 .Therefore, ρ is an automorphism of H 2 .By replacing ϕ with the composition ϕ • ρ, we may assume that ϕ(b) = b, ϕ(c) = c and ϕ(s) = s, and that ϕ(t) = t ǫ 1 w 1 • • • w h−1 t ǫ h , as desired.
Proof.From the defining relation t −1 st = s 3 in presentation (2b) together with the hypothesis, it follows that (9 By Britton's Lemma, this expression is not t-cyclically reduced.Assume that ǫ 1 = −1.Then the part t −ǫ 1 st ǫ 1 is already t-reduced.So the only t-reductions can be made successively starting from the part t ǫ h s −3 t −ǫ h .Assume that t-reductions can be made only until Here, if j ≥ 2, then by Britton's Lemma, equality (9) cannot hold, a contradiction.Also, if j = 1, then after making all t-reductions, t ǫ In particular, since ǫ 1 = −1, k is a multiple of 3.But then s k+1 = 1 from equality (9), contrary to the fact that s is an element of infinite order.

Proof of Claim.
Let us consider all possible t-reductions in each ϕ(t) δ i z i ϕ(t) δ i+1 .First, if δ i = −1 and δ i+1 = 1, it follows from the fact ǫ 1 = 1 that there is no t-reduction in ϕ(t) −1 z i ϕ(t), since z i / ∈ s in this case.Next, if either both δ i = −1 and δ i+1 = −1, or both δ i = 1 and δ i+1 = 1, it follows from our assumption z i / ∈ s in either case that there is no t-reduction in ϕ(t) ∓1 z i ϕ(t) ∓1 .Finally, assume that δ i = 1 and δ i+1 = −1.In this case, there can be t-reductions in ϕ(t)z i ϕ(t) −1 .Even if there are t-reductions in ϕ(t)z i ϕ(t) −1 , not all of t ±1 can be reduced.The reason is as follows.Suppose that all of t ±1 in ϕ(t)z i ϕ(t) −1 can be reduced.Then since the initial letter of ϕ(t) is t ǫ 1 with ǫ 1 = 1, the last t-reduction in ϕ(t)z i ϕ(t) −1 has the form tz ′ i t −1 , where z ′ i ∈ s 3 , and so ϕ(t)z i ϕ(t) −1 , after making all t-reductions, becomes a power of s.But then from the equality ϕ(t) −1 sϕ(t) = s 3 in H 2 , it follows that z i ∈ s 3 .This is a contradiction to the assumption that z(b, c, s, t) is t-reduced.
Therefore, only in the case where δ i = 1 and δ i+1 = −1, t-reductions can happen in ϕ(t) δ i z i ϕ(t) δ i+1 .But even in this case, not all of t ±1 in ϕ(t) δ i z i ϕ(t) δ i+1 can be reduced.Therefore, the assertion of Claim follows.
In view of Claim, in order for equality (10) to hold, we see that the only possibility is that ℓ = 1, so that z(b, c, s, ϕ(t)) = z 0 ϕ(t) δ 1 z 1 .Combining this with (10), we get (11) t Since ϕ(t) is written as a t-reduced form, the right-handed expression of ( 11) is already t-reduced.Then by Britton's Lemma, for the equality in (11) to hold, only one alphabet t occurs and at the same time no alphabet t −1 occurs in the righthanded expression.Here, since ϕ(t) = t ǫ 1 w 1 • • • w h−1 t ǫ h with ǫ 1 = 1, we see that this happens only when δ 1 = 1 and ϕ(t) = t, completing the proof of Proposition 6.5.
In conclusion, we obtain the following Corollary 6.6.The group H 2 is Hopfian.
Proof.Let ϕ be a surjective endomorphism of H 2 .Propositions 6.1-6.5 show that the composition of ϕ with appropriate automorphisms of H 2 becomes the identity function of H 2 , so that ϕ is indeed an automorphism of H 2 .This means that H 2 is Hopfian.

Theorem 2 . 2 (
[10, Theorem 4.3, Corollary 1.7] be a graph of groups consisting of a single loop-edge e, a single vertex v = o(e) = t(e), a vertex group G, an edge group H, and the boundary monomorphisms α e : H → G and ω e : H → G. Then the fundamental group of A is clearly isomorphic to G * .On the other hand, let Γ be a graph defined as follows.
(i) The vertex set V consists of all cosets in {xG | x ∈ G * }. (ii) The edge set E consists of all cosets in {xH | x ∈ G * }. (iii) The edge xH ∈ E connects xG and xtG.