## 1 Useful facts and definitions [2, 10]

- –A semigroup
*S*is called*cancellative*if for all*a*,*b*,*c*∊*S*either of equalities*ca*=*cb*,*ac*=*bc*implies*a*=*b*. All semigroups below are assumed to be cancellative. - –A semigroup
*S*satisfies left (resp. right)*Ore condition*if for arbitrary*a*,*b*∊*S*there are*a*′,*b*′ ∊*S*such that*aa*′ =*bb*′ (*resp*.*a*′*a*=*b*′*b*). The Ore conditions can be defined as*aS*∩*bS*≠ ∅, (*resp*.*Sa*∩*Sb*≠∅). - –If a semigroup
*S*satisfies both Ore conditions it embeds in the group*G*of its fractions*G*=*SS*^{−1}=*S*^{−1}*S*. - –A subsemigroup
*S*of a group*G*is called*generating*if elements of*S*generate*G*as a group. The group of fractions of*S*is unique in the sense that if a group*G*contains*S*as a generating semigroup then*G*is naturally isomorphic to*S*^{−1}*S*=*SS*^{−1}. - –A group
*G*satisfies an*identity u*(*x*_{1},…,*x*)≡_{m}*v*(*x*_{1},…,*x*) if for every elements_{m}*g*_{1},…,*g*in_{m}*G*the equality*u*(*g*_{1},…,*g*) =_{m}*v*(*g*_{1},…,*g*) holds._{m} - –An identity in a semigroup has a form
*u*(*x*_{1},…,*x*)≡_{n}*v*(*x*_{1},…,*x*) where the words_{n}*u*and*v*are written without inverses of variables,*u*,*v*∊ ℱ. Such an identity in a group is called a semigroup identity or a positive identity. By the word “identity” we mean a non-trivial identity.

## 2 *GB*-Problem

*If a generating semigroup S of a group G* = *F*/*N satisfies an identity, then for every s*, *t* ∊ *S there exist s*′, *t*′ ∊ *S such that ss*′ = *tt*′ (*left Ore condition*), *G* = *SS*^{−1} *and F* = ℱℱ^{−1}*N*.

An *n*-variable identity *a*(*x*_{1},…,*x _{n}*)≡

*b*(

*x*

_{1},…,

*x*) in

_{n}*S*implies a 2-variable identity if we replace the

*i*-th variable by

*xy*. In view of the cancellation property, it can be written as

^{i}*xu*(

*x*,

*y*) =

*yv*(

*x*,

*y*). For

*x*=

*s*,

*y*=

*t*,

*s*′ :=

*u*(

*s*,

*t*),

*t*′ :=

*v*(

*s*,

*t*), we have

*ss*′ =

*tt*′. Since each

*g*∊

*G*is a product of elements in

*S*∪

*S*

^{−1}, and for every

*s*,

*t*∊

*S*,

*s*

^{−1}

*t*=

*s*′

*t*′

^{−1}, we obtain

*G*=

*SS*

^{−1}. Since by assumption

*G*=

*F*/

*N*, we have

*F*= ℱℱ

^{−1}

*N*. □

*Note, that if we use the last letters on both sides of the identity, we get right Ore condition and the equality G* = *S*^{−1}*S*.

In 1981 G. Bergman [3], [4] posed the following natural question mentioned later in a different form in [5, Question 11.1]. It asks **whether the group of fractions** *G* = *SS*^{−1} **of a semigroup** *S* **must satisfy every semigroup identity satisfied by** *S*.

Note that the behavior of the identities may depend on additional properties of the group. A group *G* is called **semigroup respecting** (**S-R** group) if all of the identities holding in any generating semigroup *S* of *G*, hold in G. It is shown that extensions of soluble groups by locally finite groups of finite exponent are *S*-*R* groups [6, Theorem C]. For more results see [2].

*We say that a semigroup identity a* ≡ *b is* **transferable** *if being satisfied in S*, *it is necessary satisfied in G* = *SS*^{−1}.

It is clear that the abelian identity is transferable. The nilpotent semigroup identities in *S* found by A. I. Mal’tsev [7] are transferable. The identities *x ^{n} y^{n}* =

*y*are transferable [8].

^{n}x^{n}So the Problem is **whether every semigroup identity is transferable**. This question we address as the *GB*- Problem. A group *G* = *SS*^{−1} where *S* satisfies a non-transferable identity we call *GB*-counterexample.

The first non-transferable identity and hence *GB*-counterexample (actually a family of them) was found in 2005 by *S*. Ivanov and A. Storozhev [9]. To speak of *GB*-counterexamples we suggest a new approach to the *GB*-Problem.

## 3 Another approach to the *GB*-Problem

Let *F* : = *F _{n}* be a free group and ℱ be a free semigroup with the unity, both freely generated by the set

*X*= {

*x*

_{1},

*x*

_{2},

*x*

_{3},…,

*x*}.

_{n}By [10, Construction 12.3], if *S* = ℱ/*ρ* is a semigroup and *G* = *F*/*N* is its group of fractions then there is a natural homomorphism *φ* : *F* → *F*/*N* such that ℱ^{φ} = ℱ/*ρ*, that is

*a*=

*b*in

*S*is defined by the pair (

*a*,

*b*)∊

*ρ*. We consider the set

*a*=

*b*in

*S*.

*Let S be a generating semigroup in a group G* = *F*/*N*, *A* = *N* ∩ ℱℱ^{−1} *and A ^{F} be the normal closure of the set A in F*.

*Then if S satisfies an identity, the following equality holds*

There is the natural homomorphism *F* → *F*/*A ^{F}*, such that ℱ→ ℱ/

*μ*, where the congruence

*μ*consists of pairs (

*a*,

*b*) for

*ab*

^{−1}∊

*A*∩ ℱℱ

^{F}^{−1}

*A*∩ ℱℱ

^{F}^{−1}=

*A*, we have

*μ*=

*ρ*. So

*S*is the generating semigroup in the group

*F*/

*A*. Then, if

^{F}*S*satisfies an identity, we have by Proposition 2.1

*F*/

*A*=

^{F}*SS*

^{−1}, which implies

*F*= ℱℱ

^{−1}

*A*. □

^{F}*Let S be a generating semigroup in a group G* = *F*/*N*, *and A* = *N* ∩ ℱℱ^{−1}. *If S satisfies an identity, then N* = *A ^{F}*.

Since *F* = ℱℱ^{−1} *A ^{F}*. and

*A*⊆

^{F}*N*, we have by Dedekind’s law

## 4 Positive endomorphisms in *F*

The words in ℱ are called *positive words*. We say that an endomorphism in *F* is a positive endomorphism if it maps generators to positive words *X* → ℱ The set of positive endomorphisms in *F* we denote by *End*^{+}, the set of all endomorphisms - by *End*. The positive endomorphisms act as endomorphisms of ℱ.

*Let G* = *SS*^{−1} = *F*/*N*. *The normal subgroup N is positively invariant if and only if S is a relatively free semigroup*.

If *N* is positively invariant subgroup in *F* then the set A = *N* ∩ ℱℱ^{−1} is also positively invariant. If *a* = *b* is a relation in *S*, then the word *ab*^{−1} is in *A*. Since *A* is positively invariant, the relation is the identity in *S*. So if *N* is positively invariant, then *S* is the relatively free semigroup.

Conversely, if *S* is a relatively free semigroup then each relation *a* = *b* in *S* is the identity, hence the set *A* is positively invariant and by Lemma 3.2, the normal subgroup *N* = *A ^{F}* is positively invariant. □

We recall now the following facts similar to those in ([11] 12.21, 12.22).

**Notation**. *Let w* := *ab*^{−1}∊ ℱℱ^{−1}. *By w ^{End} we denote the End-invariant subgroup* (

*fully invariant subgroup*)

*corresponding to the word w in F which is generated by all images of w under endomorphisms in F*.

By *w*^{End+} we denote the *End*^{+}-invariant subgroup (positively invariant subgroup) corresponding to the word *w* in *F* which is generated by all images of *w* under positive endomorphisms in *F*. This subgroup need not be normal.

*If G* = *SS*^{−1} = *F*/*N*, *then*:

*S satisfies an identity a*≡*b if*(*ab*^{−1})^{End+}⊆*N*.*G satisfies an identity a*≡*b if*(*ab*^{−1})^{End}⊆*N*.*G is an***S**-**R***group if***for each identity***ab*^{−1}:$$\begin{array}{}the\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}inclusion\phantom{\rule{thinmathspace}{0ex}}(a{b}^{-1}{)}^{En{d}^{+}}\subseteq N\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}implies\phantom{\rule{thinmathspace}{0ex}}(a{b}^{-1}{)}^{End}\subseteq N.\end{array}$$ *A semigroup identity a*≡*b is***transferable***if***for every***N*◃*F*:$$\begin{array}{}the\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}inclusion\phantom{\rule{thinmathspace}{0ex}}(a{b}^{-1}{)}^{En{d}^{+}}\subseteq N\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}implies\phantom{\rule{thinmathspace}{0ex}}(a{b}^{-1}{)}^{End}\subseteq N.\end{array}$$

*A semigroup identity a* ≡ *b is* **transferable** *if*

*If S is a relatively free semigroup satisfying only transferable identities, then its group of fractions is the relatively free group*.

If *S* is a relatively free semigroup defined by one identity *a* ≡ *b*, then *A* = (*ab*^{−1})^{End+} and by Lemma 3.2 the group of fractions of *S* is *G*_{0} := *F*/*N*_{0}, where

*For every semigroup S satisfying an identity a* ≡ *b its group of fractions G* = *F*/*N is a quotient group of G*_{0} = *F*/*N*_{0}.

## 5 *GB*-counterexamples and non-Hopfian groups

*A group G* = *SS*^{−1} = *F*/*N is a GB***-counterexample** *if S satisfies a* **non-transferable identity** *a* ≡ *b*. *By another words, if N contains* (*ab*^{−1})^{End+}, *but does not contain* (*ab*^{−1})^{End}.

The first and the only known family of the *GB*-counterexamples was found by S.V.Ivanov and A.M.Storozhev [9]. Their groups *G* = *SS*^{−1} do not satisfy any identity, while *S* satisfies a 2-variable non-transferable identity with sufficiently large parameters, similar to that introduced by A. Yu. Ol’shanskii in [12].

Our aim is to show which *GB*-counterexamples are non-Hopfian groups.

*A group G is called Hopfian if each epimorphizm G* → *G is the automorphism, or in other words, if G is not isomorphic to its proper quotient*.

*If G is a non-Hopfian group, there is K* ◃ *G such that G*≅ *G*/*K*. *Then for every quotient G*/*M we have*

*which implies that G*/

*M is the non-Hopfian group unless K*⊆

*M*.

It may help to see that the group of fractions *G* = *F*/*N* of a semigroup *S* is non-Hopfian since by Corollary 4.4, it is the quotient of *G*_{0} = *F*/*N*_{0}, the group of fractions of a relatively free semigroup.

So we are interested in the question: **when a relatively free semigroup satisfying at least one non-transferable identity, must have a non-Hopfian group of fractions?**

It is not known whether each non-transferable identity implies a 2-variable non-transferable identity. So we assume that *S* satisfies a non-transferable identity of the form *a*(*x*_{1},…,*x _{m}*)≡

*b*(

*x*

_{1},…,

*x*). We show that if

_{m}*S*is the

*n*-generator semigroup and

*n*>

*m*, then the group of fractions of

*S*is the non-Hopfian group. For this purpose we start with the following “Common denominator Lemma”.

*Let a generating semigroup S of a group G satisfy an identity. Then for all g*_{1}, *g*_{2},…,*g _{m}*

*in G there are s*

_{1},

*s*

_{2},…,

*s*,

_{m}*and r in S such that g*=

_{i}*s*

_{i}r^{−1},

*i*= 1, 2,…,

*m*.

In view of Proposition 2.1, *G* = *SS*^{−1}. So for *m* = 1 the statement is clear. Let *g _{i}* =

*t*

_{i}q^{−1}for

*i*≤

*m*−1 and

*g*=

_{m}*ab*

^{−1}. By left Ore condition there exist

*q*′,

*b*′ ∊

*S*such that

*bb*′. We denote

*r*: =

*bb*′,

*s*: =

_{i}*t*′ and

_{i}q*s*:=

_{m}*ab*′. Then

*Let G* = *F*/*N* = *SS*^{−1} *be an n*-*generator group of fractions of a relatively free semigroup S*, *satisfying a non-transferable identity a*(*x*_{1},…,*x _{m}*)≡

*b*(x

_{1},…,

*x*),

_{m}*n*>

*m*.

*Then G is the non-Hopfian group*.

In view of Lemma 4.1 we can assume that *N* is the normal, positively invariant subgroup in *F*, such that by Corollary 4.2,

*a*: =

*a*(

*x*

_{1},…,

*x*), and

_{m}*b*:=

*b*(

*x*

_{1},…,

*x*) are the words in ℱ.

_{m}Let *α* be an automorphism in the free group *F*, which maps *x*_{1}→ *x*_{1} and *x _{i}* →

*x*

_{i}x_{1}

^{−1}for

*i*> 1. Then

Note that *α*^{−1} maps *x _{i}* →

*x*

_{i}x_{1},

*i*> 1, hence

*α*

^{−1}is the positive endomorphism. Since

*N*is the positively invariant subgroup in

*F*, we have

*N*

^{α−1}⊆

*N*, and hence

*N*=

*N*. By assumption the word

^{α}*w*:=

*ab*

^{−1}, where

*w*:=

*w*(

*x*

_{1},

*x*

_{2}, …,

*x*)∊

_{m}*N*defines the identity

*a*≡

*b*in

*S*but not in

*G*, so the condition (1) holds. It means that for some words

*g*

_{2},

*g*

_{3},…,

*g*

_{m+1}in

*F*, the value of

*w*is not in

*N*:

*w*(

*x*

_{1},

*x*

_{2}, …,

*x*),

_{m}*N*must contain the word

*w*(

*x*

_{2},

*x*

_{3},…,

*x*

_{m+1}). Since we assumed that (2) is:

*N*=

^{α}*N*, we have

*g*

_{2},

*g*

_{3},…,

*g*

_{m+1}in

*G*there are

*s*

_{2},

*s*

_{3},…,

*s*

_{m+1}, and

*r*in

*S*such that

*g*=

_{i}*s*

_{i}r^{−1}. The map

*x*

_{1}→

*r*and

*x*→

_{i}*s*is positive. So since

_{i}*N*is positively invariant, we get

*w*(

*g*

_{2},

*g*

_{3},…

*g*

_{m+1})∊

*N*, which contradicts (3).

So the inclusion (2) is proper *N*⊊ *N ^{α}*. Then

*F*/

*N*is isomorphic to its quotient, which proves that

*G*=

*F*/

*N*is the non-Hopfian group. □

*The n*-*generator*(*n*> 2)*GB*-*counterexamples constructed in*[9]*by S. V.Ivanov and A.M. Storozhev are the non-Hopfian groups*.*An infinitely generated GB*-*counterexample must be the non-Hopfian group*.

## 6 Remark: When *A*^{F} = 〈*A*〉

^{F}

In the book [10], just after Theorem 12.10 there is a wrong statement saying that *it is shown in* [13] that *the right Ore condition implies that* 〈*A*〉 = *A ^{F}*. Our next Theorem describes the conditition for this equality to hold.

*Let G* = *F*/*N where N is positively invariant*, *A* = *N* ∩ ℱℱ^{−1}. *The equality* 〈*A*〉 = *A ^{F} holds if and only if G is a relatively free group of finite exponent*.

If 〈*A*〉 = *A ^{F}* then by Lemma 3.2,

*N*= 〈

*A*〉 is generated as a subgroup by elements from the set

*A*. Since

*N*is normal, together with

*ab*

^{−1}where

*a*,

*b*∊ ℱ, it must contain the word (

*ab*

^{−1})

^{b}=

*b*

^{−1}

*a*as a product of words

*st*

^{−1}∊

*A*. That is

*b*

^{−1}

*a*=

*s*,

*t*∊ ℱ, and hence

*N*is positively invariant,

*N*contains

*x*,

^{n}*n*= Σ

*,*

_{i}k_{i}*x*∊

*X*. Then

*N*contains

*x*

^{−n}. Now, since

*x*

^{−1}=

*x*

^{n−1}

*x*

^{−n}∊ ℱ

*N*, we have ℱ

^{−1}⊆ ℱ

*N*. Then by Proposition 3.1 and Lemma 3.2,

*F*= ℱ

*N*. Hence each endomorphism

*ϕ*∊

*End*coincides modulo

*N*with a positive endomorphism

*φ*∊ End

^{+}. If

*w*∊

*N*, then

*w*∊

^{ϕ}*w*=

^{φ}N*N*. So

*N*is fully invariant and must contain

*F*, which implies that

^{n}*G*is a relatively free group of finite exponent.

Conversely, let *x ^{n}* belong to

*N*. To show that

*A*= 〈

^{F}*A*〉 it suffices to check that for every

*ab*

^{−1}∊

*A*and each generator

*x*, the word

*x*

^{−1}(

*ab*

^{−1})

*x*is a product of elements in

*A*. So let

*ab*

^{−1}∊

*A*, then for each

*c*∊ ℱ, (

*ca*)(

*cb*)

^{−1}∊

*A*, hence (

*x*

^{n−1}

*a*)(

*x*

^{n−1}

*b*)

^{−1}∊

*A*. Now, since

*x*

^{± n}∊

*N*∩ ℱℱ

^{−1}=:

*A*, we get

*x*

^{−1}(

*ab*

^{−1})

*x*=

*x*

^{−n}(

*x*

^{n−1}

*a*)(

*x*

^{n−1}

*b*)

^{−1}

*x*∊〈

^{n}*A*〉. This implies that the subgroup 〈

*A*〉 is normal, which finishes the proof. □

Let *G* = *F*/*N* be a group constructed by S.V.Ivanov and A.M.Storozhev in [9]. Since *G* = *SS*^{−1} where *S* satisfies a positive identity, *N* is positively invariant and *S* satisfies the left Ore condition. However the equality 〈*A*〉 = *A ^{F}* does not hold because otherwise

*G*would have a finite exponent, which is not so.

The author is grateful to Referee for suggestions on the text improvement.

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