Abstract
The group of fractions of a semigroup S, if exists, can be written as G = SS−1. If S is abelian, then G must be abelian. We say that a semigroup identity is transferable if being satisfied in S it must be satisfied in G = SS−1. One of problems posed by G.Bergman in 1981 asks whether the group G must satisfy every semigroup identity which is satisfied in S, that is whether every semigroup identity is transferable. The first non-transferable identities were constructed in 2005 by S.V.Ivanov and A.M. Storozhev.
A group G is called Hopfian if each epimorphizm G → G is the automorphism. The residually finite groups are Hopfian, however there are many problems concerning the Hopfian property e.g. of infinite Burnside groups, of finitely generated relatively free groups [11, Problem 15]. We prove here that if G = SS−1 is an n-generator group of fractions of a relatively free semigroup S, satisfying m-variable (m < n) non-transferable identity, then G is the non-Hopfian group.
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−1S.
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−1S = SS−1.
A group G satisfies an identity u(x1,…,xm)≡ v(x1,…,xm) if for every elements g1,…,gm in G the equality u(g1,…, gm) = v(g1,…,gm) holds.
An identity in a semigroup has a form u(x1,…,xn)≡ v(x1,…, xn) where the words 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
Proposition 2.1
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−1N.
Proof
An n-variable identity a(x1,…,xn)≡ b(x1,…,xn) in S implies a 2-variable identity if we replace the i-th variable by xyi. In view of the cancellation property, it can be written as 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−1t = s′t′−1, we obtain G = SS−1. Since by assumption G = F/N, we have F = ℱℱ−1 N. □
Remark
Note, that if we use the last letters on both sides of the identity, we get right Ore condition and the equality G = S−1S.
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].
Definition 2.2
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 xn yn = yn xn are transferable [8].
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 : = Fn be a free group and ℱ be a free semigroup with the unity, both freely generated by the set X = {x1, x2, x3,…,xn}.
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 ℱφ = ℱ/ρ, that is
Each relation a = b in S is defined by the pair (a, b)∊ρ. We consider the set
which consists of words corresponding to relations a = b in S.
Proposition 3.1
Let S be a generating semigroup in a group G = F/N, A = N ∩ ℱℱ−1 and AF be the normal closure of the set A in F. Then if S satisfies an identity, the following equality holds
Proof
There is the natural homomorphism F → F/AF, such that ℱ→ ℱ/μ, where the congruence μ consists of pairs (a, b) for ab−1∊ AF ∩ ℱℱ−1
Since by [10, Corollary 12.8] AF ∩ ℱℱ−1 = A, we have μ = ρ. So S is the generating semigroup in the group F/AF. Then, if S satisfies an identity, we have by Proposition 2.1 F/AF = SS−1, which implies F = ℱℱ−1 AF. □
Lemma 3.2
Let S be a generating semigroup in a group G = F/N, and A = N ∩ ℱℱ−1. If S satisfies an identity, then N = AF.
Proof
Since F = ℱℱ−1 AF. and AF⊆ 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.
Lemma 4.1
Let G = SS−1 = F/N. The normal subgroup N is positively invariant if and only if S is a relatively free semigroup.
Proof
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 = AF 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 wEnd 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 wEnd+ 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.
Corollary 4.2
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:
A semigroup identity a ≡ b is transferable if for every N◃ F:
Corollary 4.3
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 G0 := F/N0, where
Corollary 4.4
For every semigroup S satisfying an identity a ≡ b its group of fractions G = F/N is a quotient group of G0 = F/N0.
5 GB-counterexamples and non-Hopfian groups
Definition 5.1
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.
Definition 5.2
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.
Remark
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 G0 = F/N0, 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(x1,…,xm)≡ b(x1,…,xm). We show that if 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”.
Lemma 5.3
Let a generating semigroup S of a group G satisfy an identity. Then for all g1, g2,…,gm in G there are s1, s2,…,sm, and r in S such that gi = sir−1, i = 1, 2,…,m.
Proof
In view of Proposition 2.1, G = SS−1. So for m = 1 the statement is clear. Let gi = tiq−1 for i ≤ m−1 and gm = ab−1. By left Ore condition there exist q′, b′ ∊ S such that qq′ = bb′. We denote r : = qq′ = bb′, si : = tiq′ and sm := ab′. Then
as required. □
Theorem 5.4
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(x1,…,xm)≡ b(x1,…,xm), n > m. Then G is the non-Hopfian group.
Proof
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,
where a : = a(x1,…,xm), and b:= b(x1,…,xm) are the words in ℱ.
Let α be an automorphism in the free group F, which maps x1→ x1 and xi → xix1−1 for i > 1. Then
Note that α−1 maps xi → xi x1, i > 1, hence α−1 is the positive endomorphism. Since N is the positively invariant subgroup in F, we have Nα−1 ⊆ N, and hence
To show that the inclusion (2) is proper, assume the contrary, that N = Nα. By assumption the word w := ab−1, where w := w(x1, x2, …, xm)∊ N defines the identity a ≡ b in S but not in G, so the condition (1) holds. It means that for some words g2, g3,…,gm+1 in F, the value of w is not in N:
Together with the word w(x1, x2, …, xm), N must contain the word w(x2, x3,…,xm+1). Since we assumed that (2) is: Nα = N, we have
By Lemma 5.3 for every g2, g3,…,gm+1 in G there are s2, s3,…,sm+1, and r in S such that gi = sir−1. The map x1→ r and xi → si is positive. So since N is positively invariant, we get w(g2, g3,… gm+1)∊ N, which contradicts (3).
So the inclusion (2) is proper N⊊ Nα. Then
that is F/N is isomorphic to its quotient, which proves that G = F/N is the non-Hopfian group. □
Corollary 5.5
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 AF = 〈A〉
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〉 = AF. Our next Theorem describes the conditition for this equality to hold.
Theorem 6.1
Let G = F/N where N is positively invariant, A = N ∩ ℱℱ−1. The equality 〈A〉 = AF holds if and only if G is a relatively free group of finite exponent.
Proof
If 〈A〉 = AF 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−1a =
We can write
Conversely, let xn belong to N. To show that AF = 〈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 (xn−1a)(xn−1b)−1∊ A. Now, since x± n ∊ N ∩ ℱℱ−1 =: A, we get x−1(ab−1)x = x−n(xn−1a)(xn−1b)−1xn∊〈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〉 = AF does not hold because otherwise G would have a finite exponent, which is not so.
Acknowledgement
The author is grateful to Referee for suggestions on the text improvement.
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