SU B SE MI G R OU PS OF G R OU PS: PR E SE N TATI ON S, MA L C E V PR E SE N TATI ON S, A N D A U TOMATI C STR U C TU R E S Alan J. Cain, Edmund F. Robertson, Nik Ruˇskuc School of Mathematics and Statistics University of St Andrews North Haugh St Andrews Fife KY16 9SS United Kingdom Email: alanc, edmund,
[email protected] Tel: +44 (0)1334 46 3228 Fax: +44 (0)1334 46 3748
abstract All finitely generated subsemigroups of virtually nilpotent groups admit finite Malcev presentations. (A Malcev presentation is a presentation of a special type for a semigroup that can be embedded in a group.) All automatic or asynchronous automatic semigroups embeddable into groups admit finite Malcev presentations. Finitely generated subsemigroups of virtually free groups are automatic. A finitely generated subsemigroup of the free product of a free group and an abelian group that fails to have a finite Malcev presentation is exhibited. Therefore the class of groups all of whose finitely generated subsemigroups admit finite Malcev presentations is properly contained in the class of coherent groups. Finitely generated subsemigroups of the free product of a free monoid and an abelian group are asynchronously automatic and therefore have finite Malcev presentations. 1
introduction
Malcev presentations are a particular species of semigroup presentation that can be used to define any semigroup embeddable into a group. The theory of Malcev presentations has perhaps been neglected: until recently, the only works on the subject were due to Spehner, who introduced the concept of a Malcev presentation [24] and proved that all finitely generated subsemigroups of free semigroups have finite Malcev presentations [25]. Recent work by the authors [3] has shown that all finitely generated subsemigroups of virtually free groups admit finite Malcev presentations. (A virtually free group is a group with a free subgroup of finite index.) The present paper continues this investigation into Malcev presentations. Sections 2 and 3 review the necessary preliminaries on Malcev presentations and automatic structures for semigroups. Section 4 proves a technical theorem which is then applied to deduce the following result:
1
Theorem 1. Every finitely generated subsemigroup of a virtually nilpotent group admits a finite Malcev presentation. There is an important link between the theory of Malcev presentations and the new and active field of automatic semigroups. Whilst automatic groups [7] are always finitely presented, automatic semigroups [4] need not be. However, Section 5 shows that: Theorem 2. Every automatic semigroup that can be embedded in a group admits a finite Malcev presentation. As with groups, this result extends from synchronous to asynchronous automatism: Theorem 3. Every asynchronous automatic semigroup that can be embedded in a group admits a finite Malcev presentation. All finitely generated subsemigroups of a free semigroup are automatic [4, Theorem 8.2]. Section 6 generalizes this: Theorem 4. Finitely generated subsemigroups of virtually free groups are automatic. Section 7 contains an example of a semigroup that does not admit a finite Malcev presentation, but which arises as a subsemigroup of the free product of a free group and an abelian group. The Kurosh Subgroup Theorem implies that this free product is coherent. (Recall that a group is coherent if all of its finitely generated subgroups are finitely presented.) Therefore: Theorem 5. The class of groups all of whose finitely generated subsemigroups have finite Malcev presentations is properly contained in the class of coherent groups. Theorem 2 and the example just mentioned show that a positive subsemigroup of an automatic group need not be automatic. Contrast the example with the following result from Section 8: Theorem 6. Let M be the [monoid] free product of a free monoid and an abelian group. Every finitely generated subsemigroup of M is asynchronously automatic and so has a finite Malcev presentation. 2
preliminaries
This section summarizes the necessary definitions and results regarding Malcev presentations and Cayley graphs, and also introduces notation used throughout the paper. 2.1
Malcev presentations
The theory of Malcev presentations was introduced by Spehner [24], though they are based on Malcev’s necessary and sufficient condition for the embeddability of a semigroup in a group [15, 16]. (Details of the embeddability condition can also be found in [6, Chapter 12].) Although this section contains the definitions and results about Malcev presentations required for the rest of the paper, the reader is assumed to be familiar with the basic theory of [ordinary] semigroup presentations. 2
Definition 2.1. Let S be any semigroup. A congruence σ on S is a Malcev congruence if S/σ is embeddable in a group. T If {σi : i ∈ I} is a set of Malcev congruences on S, then σ = i∈I σi is also a Malcev congruence on S. ThisQis true because S/σi embeds in a group Gi Q for each i ∈ I, so S/σ embeds in i∈I S/σi , which in turn embeds in i∈I Gi . The following definition therefore makes sense.
Definition 2.2. Let A+ be a free semigroup; let ρ ⊆ A+ × A+ be any binary relation on A+ . Denote by ρM the smallest Malcev congruence containing ρ — namely, \ ρM = σ : σ ⊇ ρ, σ is a Malcev congruence on A+ .
Then SgMhA | ρi is a Malcev presentation for [any semigroup isomorphic to] A+/ρM . If both A and ρ are finite, the the Malcev presentation SgMhA | ρi is said to be finite.
Let SgMhA | Ri be a Malcev presentation for a semigroup S. If S has a finite Malcev presentation, then it admits one of the form SgMhA | Qi, where Q is a finite subset of R. The notation SgMhA | ρi distinguishes the Malcev presentation with generators A and defining relations ρ from the ordinary semigroup presentation SghA | ρi, which defines A+/ρ# . (Recall that ρ# denotes the smallest congruence containing ρ.) Similarly, GphA | ρi denotes the group presentation with the same set of generators and defining relations. Let X be a subset of a group G. Denote by SghXi the subsemigroup generated by X and by GphXi the subgroup generated by X. Fix A+ and ρ as in the Definition 2.2 and let S = A+/ρM . Let AL , AR be two sets in bijection with A under the mappings a 7→ aL , a 7→ aR , respectively, with A, AL , AR being pairwise disjoint. Extend the mappings a 7→ aL , a 7→ aR to anti-isomorphisms from A∗ to (AL )∗ and (AR )∗ , respectively. Let τ = ρ ∪ (bbR a, a), (abbR , a), (bL ba, a), (abL b, a) : a ∈ A ∪ AL ∪ AR , b ∈ A .
Let G = Sg A ∪ AL ∪ AR | τ . The semigroup G is actually the universal group of S, with G ≃ GphA | ρi. It can be shown that ρM = τ# ∩ (A+ × A+ ), or equivalently that ρM = {(u, v) : u, v ∈ A+ represent the same element of GphA | ρi}.
(2.1)
Therefore, two words u, v ∈ A+ represent the same element of S if and only if there is a sequence u = u0 → u1 → . . . → un = v, with n > 0, where, for each i ∈ {0, . . . , n − 1}, there exist pi , qi , qi′ , ri ∈ (A ∪ AL ∪ AR )∗ such that ui = pi qi ri , ui+1 = pi qi′ ri , and (qi , qi′ ) ∈ τ or (qi′ , qi ) ∈ τ. In fact, it can be shown that u, v ∈ A+ represent the same element of S if any only if there exists such a sequence with pi ∈ (A ∪ AL )∗ and ri ∈ (A ∪ AR )∗ . This restriction on the letters that can appear in pi and ri simply means that no changes can occur to the left of an aL or to the right of an aR . Such a sequence is called a Malcev ρ-chain (or simply a Malcev chain) from u to v. If (u, v) ∈ ρM , then (u, v) is said to be a Malcev consequence of ρ. 3
2.2
Words, prefixes, and suffixes
Following [7], the notation used in this paper distinguishes a word from the element of the semigroup or group it represents. Let A be an alphabet representing a set of generators for a semigroup S. For any word w ∈ A+ , denote by w the element of S represented by w. Similarly, if A represents a set of generators for a group G, let w be the element of G represented by w ∈ (A∪A−1 )∗ . In both cases, for any set of words W, W is the set of all elements represented by at least one word in W. Denote the identity of A∗ — the empty word — by ε. Denote the length of u ∈ A∗ by |u|. Let u = u1 · · · un , where ui ∈ A. For t ∈ N ∪ {0}, let if t = 0, ε u(t) = u1 · · · ut if 0 < t 6 n, u1 · · · un if n < t, and let
ut+1 · · · un u[t] = ε
if 0 6 t < n, if n 6 t.
So u(t) is the prefix of u up to and including the t-th letter; u[t] is the suffix of u after and not including the t-th letter. Observe that for all t ∈ N ∪ {0}, u = u(t)u[t], and that if one formally assumes that ut = ε for t > n, then u(t + 1) = u(t)ut+1 and u[t] = ut+1 u[t + 1]. 2.3
Cayley graphs
Definition 2.3. Let S be a semigroup and A an alphabet representing a set of generators for S. The Cayley graph Γ (S, A) of S with respect to A is the directed graph with vertex set S and, for each pair of vertices s, t, an edge from s to t labelled by a ∈ A if and only if sa = t. Let S be a semigroup that embeds in a group. If S is not already a monoid, then a new vertex representing a two-sided identity can be adjoined to Γ (S, A). Throughout this paper, assume that Γ (S, A) has a vertex representing a twosided identity as a basepoint. b be the walk in Γ (S, A) starting at the basepoint Given a word u ∈ A+ , let u b ends at the vertex u. and labelled by u. The walk u Definition 2.4. Let s, t be two vertices in Γ (S, A). The [undirected] distance from s to t, denoted d(s, t), is defined to be the infimum of the lengths of the undirected paths from s to t. The open ball of radius r centred on s ∈ S is the set Br (s) = {t ∈ S : d(s, t) < r}. 3
automatic semigroups
The concept of an automatic structure has been generalized from groups [7] to semigroups [4]. Asynchronous automatic structures have also been generalized from groups [7, Chapter 7] to monoids [9]. Although few properties 4
of asynchronous automatic groups carry over to general semigroups, many of them do survive if attention is restricted to semigroups embeddable into groups. This section gathers the necessary results for both synchronous and asynchronous automatic semigroups that embed in groups. Although some of the results appear here for the first time, the given proofs are only sketches: generally, the proofs are minor variations on the group case. Definition 3.1. Let S be a semigroup. A rational structure for S is a pair (A, L), where A is a finite alphabet representing a set of generators for S, and L is a regular language over A such that L = S. Let (A, L) be a rational structure for a semigroup S. For each a ∈ A ∪ {ε}, define La = {(u, v) : u, v ∈ L, ua = v}. (3.1) Let $ be a new symbol not in A. Let A(2, $) = {(a, b) : a, b ∈ A ∪ {$}} − {($, $)}. Define the mapping δA : A+ × A+ → A(2, $)+ by if m = n, (u1 , v1 ) · · · (um , vn ) (u1 · · · um , v1 · · · vn ) 7→ (u1 , v1 ) · · · (un , vn )(un+1 , $) · · · (um , $) if m > n, (u1 , v1 ) · · · (um , vm )($, vm+1 ) · · · ($, vn ) if m < n,
where ui , vi ∈ A. The symbol $ is usually called the padding symbol.
Definition 3.2. Let S be a semigroup. A synchronous automatic structure for S is a rational structure (A, L) such that, for each a ∈ A∪{ε}, the set La δA is a regular language over A(2, $). A synchronous automatic semigroup is a semigroup that admits a synchronous automatic structure. By default, for the purposes of this paper, ‘automatic’ means ‘synchronously automatic’. Definition 3.3. Let S be a semigroup. An asynchronous automatic structure for S is a rational structure (A, L) such that, for each a ∈ A ∪ {ε}, the relation La is rational. An asynchronous automatic semigroup is a semigroup that admits an asynchronous automatic structure. Recall that a rational relation is recognized by an asynchronous two-tape automaton [20, Theorem 1.1]. The deterministic rational relations are those recognized by deterministic asynchronous two-tape automata. The deterministic rational relations form a strictly smaller class than the rational relations [20, Example 2.1]. However, any asynchronous automatic semigroup embeddable into a group admits an asynchronous automatic structure all of whose relations (3.1) are deterministic. (Shapiro [23] shows this for groups; the corresponding theorem for semigroups embeddable into groups follows using the results in this section.) Therefore, for brevity, the remainder of the paper assumes all rational relations to be deterministic. Chapter 7 of [7] contains all the definitions and results about asynchronous two-tape automata required hereafter. Every [synchronous] automatic semigroup is asynchronously automatic: define the two-tape automaton to simulate a deterministic automaton by reading from alternating tapes [7, Lemma 7.1.2]. 5
Definition 3.4. An asynchronous two-tape automaton is called boundedly asynchronous if there exists k ∈ N such that the automaton never reads more than k letters in a row from one tape. A boundedly asynchronous automatic structure is one in which all the relations La can be recognized by boundedly asynchronous automata. Theorem 3.5. Let S be a semigroup that embeds in a group. Let (A, L) be an asynchronous automatic structure for S. Then there exists a language L ′ ⊆ L such that (A, L ′ ) is a boundedly asynchronous automatic structure for S. Proof of 3.5. The proof is analogous to that in [7, Theorem 7.2.4]. At those points where inverses are required or the identity, work in the universal group of S. 3.5 Definition 3.6. Let S be a semigroup, and (A, L) a rational structure for S. A departure function for S is a function D : R → R such that, if w ∈ L, r, s > 0, t > D(r), and s + t 6 |w|, then the distance between w(s) and w(s + t) in the Cayley graph of S exceeds r. The existence of a departure function means that a word in L labels a walk that eventually departs from every finite neighbourhood. Definition 3.7. Let S be a semigroup and A an alphabet representing a set of generators for S. Let u, v ∈ A+ . The Hausdorff distance between the two walks b, b u v in Γ (S, A) is
[ [ Br (u(t)) . h = inf r : u ⊆ Br (v(t)) and v ⊆
b is at most h from some point on b In other words, every point on u v, and vice versa. The following lemma will be needed in Section 8.
Lemma 3.8. Suppose (A, L) is a rational structure for S, where S is a semigroup that embeds in a group. Suppose that L maps finite-to-one onto S. For all r ∈ N, there exist only finitely many words y ∈ A+ such that xyz ∈ L for some x, z ∈ A+ , and d(x, xy) < r. Proof of 3.8. The proof of [2, Lemma 7.4] applies unchanged.
3.8
Theorem 3.9. Let S be a semigroup that embeds in a group. Let (A, L) be a rational structure for S. Then (A, L) is a boundedly asynchronous automatic structure for S if and only if it satisfies the following conditions: 1. There exists a departure function D for S. 2. There exists a constant λ ∈ N such that, for u, v ∈ L with ua = v for a ∈ b and b A ∪ {ε}, the walks u v in the Cayley graph of S are at most a Hausdorff distance λ from one another. Proof of 3.9. The proof is analogous to that in [7, Theorem 7.2.8]. There is one difference: rather than using an open ball in the Cayley graph of S, take the ball to be in the Cayley graph of the universal group of S. 3.9
6
Corollary 3.10. Let S be a semigroup that embeds in a group. Let (A, L) be an asynchronous automatic structure for S. Let w ∈ A+ . Then, for u, v ∈ L with b and b uw = v, the Hausdorff distance between u v is bounded by a quantity dependent only on w. 3.10 The proof of Theorem 3.9 actually shows that a slightly stronger result holds. Proposition 3.11. Let S be a semigroup with asynchronous automatic structure (A, L). Let a ∈ A ∪ {ε} and let (u, v) ∈ La . Let A be an asynchronous twotape automaton recognizing La . When the automaton has read t letters in total, let sL (t) be the number of letters read from the left-hand tape and sR (t) the number read from the right-hand tape, so that t = sL (t) + sR (t). There exists a constant λ ∈ N, dependent only on (A, L), such that for all t ∈ N ∪ {0}, there exist pt , qt ∈ A∗ such 3.11 that u(sL (t))pt = v(sR (t))qt with |pt | + |qt | < λ. In the synchronous case, this result is simpler: Proposition 3.12. Let S be a semigroup with automatic structure (A, L). Let a ∈ A ∪ {ε} and let (u, v) ∈ La . There exists a constant λ ∈ N, dependent only on (A, L), such that for all t ∈ N ∪ {0}, there exist pt , qt ∈ A∗ such that u(t)pt = v(t)qt with 3.12 |pt | + |qt | < λ. The next result applies to semigroups generally, not just to those embeddable into groups. Theorem 3.13 ([4, Theorem 7.2]). Let S be a semigroup, and let SI denote the semigroup formed by adjoining a two-sided identity to S. Then SI is automatic if and 3.13 only if S is automatic. Two further results about [synchronous] automatic semigroups embeddable into groups will be needed. Theorem 3.14. Let S be a semigroup that embeds in a group. Let (A, L) be a rational structure for S. Then (A, L) is an automatic structure for S if and only if there exists a constant λ ∈ N such that for all a ∈ A ∪ {ε}, if u, v ∈ L are such that ua = v, then for all t the distance from u(t) to v(t) is less than λ. Proof of 3.14. Again, the proof is analogous to that in [7, Theorem 2.3.5], with one difference: rather than using an open ball in the Cayley graph of S, take 3.14 the ball to be in the Cayley graph of the universal group of S. All finitely generated abelian groups are automatic [7, Chapter 4]. There exist finitely generated commutative semigroups that are not automatic [10]. However, finitely generated subsemigroups of abelian groups — which are precisely the finitely generated commutative cancellative semigroups — are automatic: Proposition 3.15. Let S be a subsemigroup of an abelian group, and let A = {a1 , . . . , an } be a finite alphabet representing a set of generators for S. Let ≺ be the ShortLex ordering on A+ based on a1 ≺ . . . ≺ an . Let K = {w ∈ a∗1 · · · a∗n − {ε} : w is the ≺-minimal representative for w}. Then (A, K) is an automatic structure for S. 7
Proof of 3.15. One can represent elements of S using tuples: the tuple (α1 , . . . , αn ) αn + 1 represents the same element as aα 1 · · · an . The ShortLex ordering on A becomes (α1 , . . . , αn ) ≪ (β1 , . . . , βn ) ⇐⇒
n X
αi
1, then put all variables in u, v equal to one another and use the cancellative property to deduce that S is again a group (of finite exponent). If |u|, |v| > 1, write u = xu ′ and v = yv ′ , so that |u| = 1 + |u ′ | and |v| = 1 + |v ′ |. Observe that if x and y were the same variable, cancellation would show u ′ = v ′ to be a law, which would contradict the minimality of |u|. Hence every two elements of S have a common right multiple. Now invoke Ore’s Theorem 4.1 to show that the group of right quotients SS−1 exists. The symmetry of the hypotheses 4.2 shows that S−1 S exists as well. Theorem 4.3. Let S be a finitely generated semigroup that embeds in its group of right (respectively, left) quotients SS−1 (respectively, S−1 S). Suppose that this group is finitely presented. Then S admits a finite Malcev presentation. Proof of 4.3. Suppose the semigroup S embeds in its group of right quotients: the left quotient case follows in a similar manner. Let X be a finite generating set for S. Let H = SS−1 . The group H has a finite group presentation GphA | Ri, where the alphabet A represents the set of generators X, and where the righthand side of each defining relation in R is the empty word. Let w be a reduced word on A ∪ A−1 . Suppose that −1 −1 −1 w = s−1 1 t1 s2 t2 · · · sk tk sk+1 ,
where s1 , sk+1 ∈ A∗ and si , tj ∈ A+ for 2 6 i 6 k and 1 6 j 6 k. Since H = SS−1 , there exist ui , vi ∈ A+ for 1 6 i 6 k such that −1 −1 −1 −1 s−1 for 2 6 i 6 k. 1 t1 = u1 v1 and vi−1 si ti = ui vi
The relations in the finite set −1 −1 −1 −1 Sw = {(s−1 1 t1 , u1 v1 )} ∪ {(vi−1 si ti , ui vi ) : 2 6 i 6 k}
therefore hold in H. Observe that, using the relations in Sw , one can rewrite the word w to obtain one of the form pq−1 , where p, q ∈ A∗ , that represents the same element of H: −1 −1 −1 −1 w = s−1 1 t1 s2 t2 s3 t3 · · · sk tk sk+1 −1 −1 −1 −1 → u1 v−1 1 s2 t2 s3 t3 · · · sk tk sk+1 −1 −1 −1 → u1 u2 v−1 2 s3 t3 · · · sk tk sk+1 .. . −1 −1 → u1 u2 u3 · · · v−1 k−1 sk tk sk+1 −1 → u1 u2 u3 · · · uk v−1 k sk+1 .
Let S be the union of the Sw as w ranges over all left-hand sides of defining relations in R. The set S is a finite subset of (A−1 )∗ A∗ × A+ (A−1 )+ . The group H is finitely presented by GphA | R ∪ Si. Replace the set R by a set Q ⊆ A+ (A−1 )+ × {ε} by rewriting left-hand sides in R using S, thus obtaining a new presentation GphA | Q ∪ Si for H. 9
Replace each pair (pq−1 , ε) ∈ Q by (p, q) and each pair (pq−1 , s−1 t) ∈ S by (sp, tq) to obtain a new presentation GphA | Pi for H, where P ⊆ A∗ × A∗ . This is a finite presentation for H all of whose defining relations are between positive or empty words. If any defining relation in P has one side empty, pre- or post- multiply both sides of it by some letter a ∈ A. This ensures that P ⊆ A+ × A+ . Let U be the set of all valid relations in S, or equivalently the set of all valid relations in H that lie in A+ × A+ . It is clear that U contains P, and so H = GphA | Pi = GphA | Ui . Therefore, H is the universal group of S = SghA | Ui. Therefore, by (2.1), 4.3 PM = U and so SgMhA | Pi is a finite Malcev presentation for S. Corollary 4.4. Let G be a group. Let S be a subsemigroup of G. Suppose that S satisfies a non-tautological semigroup law. Then the subgroup of G generated by S is isomorphic to the universal group of S. Proof of 4.4. Repeat the reasoning in the proof of Theorem 4.3 without the assumption of finite generation or finite presentability of H = GphSi. The presentation hA | Pi is a group presentation for H and a Malcev presentation for S. Thus H is [isomorphic to] the universal group of S. 4.4 Theorem 1. Every finitely generated subsemigroup of a virtually nilpotent group admits a finite Malcev presentation. Proof of 1. Let G be a virtually nilpotent group. Let K be a nilpotent subgroup of G of finite index. By passing to the core of K in G, assume without loss that K is normal. By [19, Theorem 1], K satisfies a non-tautological law u = v. As observed in [12, Section IV], G must satisfy a non-tautological law obtained by replacing each variable x of u and v by xn , where n is the exponent of the finite group G/K. Let X be a finite subset of G. Let S = SghXi and H = GphXi. Since G satisfies a non-tautological law, S does also. By Lemma 4.2, the group H and the group of right quotients SS−1 coincide. The group H is also virtually nilpotent. It is finitely generated and so finitely presented. The conditions of Theorem 4.3 are thus fulfilled and so S admits a finite Malcev presentation. 1
Nilpotent groups can contain finitely generated subsemigroups that do not admit finite ordinary presentations: Example 4.5. Let H be the Heisenberg group: H = Gphx, y, z | (yx, xyz), (zy, yz), (xz, zx)i . The group H is nilpotent — indeed, it is the free nilpotent group of class 2 and rank 2. Every element of H can be represented by a unique word of the form xα yβ zγ , where α, β, γ ∈ Z; see, for example, [11]. Clearly, any word in {x, y, z}+ can be transformed to one in this normal form using only the relations (yx, xyz), (zy, yz), and (xz, zx). The subsemigroup T of H consisting of elements represented by words in {x, y, z}+ is therefore presented by T = Sghx, y, z | (yx, xyz), (zy, yz), (xz, zx)i .
(4.1) 10
Identify H — and so T — with words in normal form. Let A = {a, b, c} be an alphabet representing elements of T in the following way: a = y2 , b = xz, c = x2 z. Let S be the subsemigroup of T generated by A. Observe that for all n ∈ N, ca2n b = x2 zy4n xz = x3 y4n z4n+2 = xzy2n x2 zy2n = ban can . Therefore (ca2n b, ban can ) is a valid relation in S for all n ∈ N. An easy argument shows that for any proper subword w of ca2n b, the element w can be factorized into elements of A in only one way. For example, a subword of the form cak represents x2 y2k z. Since the relations in (4.1) do not change the numbers of letters x or y, the occurrence of x2 must arise from one letter c or two letters b. The presence of the single z precludes the latter possibility, since moving to normal form cannot decrease the number of letters z. Similarly, the y2k requires k letters a. Therefore, any word representing x2 y2k z must be a rearrangement of cak . However, any letters a to the left of the c would lead to more than one z being present in the normal form. Similar arguments apply to subwords of the form ak b and ak . Therefore any presentation for S on this set of generators must have ca2n b as one side of a defining relation for each n ∈ N. Thus S is not finitely presented. 5
automatism and malcev presentations
Every automatic or asynchronous automatic group has a finite presentation [7, Theorems 2.3.12 and 7.3.4]. On the other hand, automatic semigroups need not be finitely presented [4, Examples 3.9 and 4.4]. However, restricting attention to subsemigroups of groups yields the following result: Theorem 2. Every automatic semigroup that can be embedded in a group admits a finite Malcev presentation. Proof of 2. The proof begins along similar lines to [7, Theorem 2.3.12]. Let S be an automatic semigroup that can be embedded in a group; let (A, L) be an automatic structure for S. Suppose the undirected fellow traveller constant for (A, L) is λ. For a ∈ A, let γa ∈ L represent the same element as a. Let T = {(a, γa ) : a ∈ A}. The relations in T are valid in S. Let (u, v) ∈ A+ ×A+ be a relation that holds in S. Then u and v label walks in the Cayley graph Γ (S, A) from the basepoint to the same vertex. Suppose u = u1 · · · uk and v = v1 · · · vl , where ui , vi ∈ A. Let αi be a representative in L of u(i) for 0 6 i 6 k and similarly βj a representative of v(j) for 0 6 j 6 l (see Figure 1). Assume that α1 = γu1 and β1 = γv1 . The relations (u1 , α1 ), (v1 , β1 ), (αi ui+1 , αi+1 ), (βj vj+1 , βj+1 ), and (αk , βl )
(5.1)
hold in S for i = 0, . . . , k − 1 and j = 0, . . . , l − 1. Lemma 5.1. The relation (u, v) is a consequence of the relations (5.1).
11
u(t + 2)
u=v
u(t + 1) u(t)
αt+1
αt+2
αt
u
v
Basepoint
Figure 1: Walks in Γ (S, A) labelled by valid relations. s
µ[t + 1℄
r 2 ( ζ)
ν[t + 1℄ a 1 ( ζ)
pt+1 qt+1 µt+1
νt+1 pt
x
r4 ( ζ) a2 (ζ)
ζ r 1 ( ζ)
r3 ( ζ)
y
qt
µ(t)
ν(t)
w ( ζ)
w ′ ( ζ)
Basepoint
(a)
Basepoint
(b) Figure 2: Loops in Γ (S, A).
Proof of 5.1. The sequence u = u1 · · · uk → α1 u2 · · · uk → α2 u3 · · · uk → . . . → αk−1 uk → αk → βl → βl−1 vl → . . . → β1 v2 · · · vl → v1 · · · vl = v shows that (u, v) is a consequence of the given relations.
5.1
Choose and fix one of the relations (5.1) other than (u1 , α1 ) and (v1 , β1 ). To simplify notation, write it as (µs, ν), where µ, ν ∈ L and s ∈ A ∪ {ε}. Suppose µ = µ1 · · · µ|µ| and ν = ν1 · · · ν|ν| , where µi , νi ∈ A. Since (µ, ν) ∈ Ls , Proposition 3.12 asserts that for t ∈ N ∪ {0}, there exist pt and qt in A∗ such that µ(t)pt = ν(t)qt , with |pt | + |qt | being bounded by the constant λ. Assume that p0 , q0 = ε and that pm = s and qm = ε, where m = max{|µ|, |ν|}. Geometb and rically, these pt and qt give m − 1 undirected loops between the walks µ b, the total length of each loop being bounded by 2λ + 2 (see Figure 2(a)). ν Consider the set of all loops ζ in Γ (S, A) of the form shown in Figure 2(b), where |a1 (ζ)|, |a2 (ζ)| 6 1 and |ri (ζ)| 6 λ/2 for i = 1, . . . , 4. There is no require12
ment that the vertices x and y be distinct: this covers the case when t = 0 in Figure 2(a). There are only a finite number of possible labels for such loops. Let Z be a set of such loops in Γ (S, A) containing exactly one with each possible label. Consider some loop ζ ∈ Z and retain the notation from Figure 2(b). [ w \ ′ (ζ) from the basepoint of Γ (S, A) to the vertices x and Choose walks w(ζ), y, respectively. The two relations (w(ζ)r1 (ζ), w ′ (ζ)r3 (ζ)) and (w(ζ)a1 (ζ)r2 (ζ), w ′ (ζ)a2 (ζ)r4 (ζ))
(5.2)
both hold in S. Let Q be the set of relations (5.2) thus obtained as ζ ranges over Z. Since Z is a finite set, so is Q. Lemma 5.2. The relation (µs, ν) is a Malcev consequence of Q. Proof of 5.2. Let t ∈ {0, . . . , m − 1}. The key is to show that there is a Malcev R Q-chain that leads from µ(t + 1)pt+1 qR t+1 ν[t + 1] to µ(t)pt qt ν[t]. Let ζ be the loop in Z with the same label as that in Figure 2(a). In this case, a1 (ζ) = µt+1 , a2 (ζ) = νt+1 , r1 (ζ) = pt , r2 (ζ) = pt+1 , r3 (ζ) = qt , r4 (ζ) = qt+1 ; and the corresponding relations (5.2) become (w(ζ)pt , w ′ (ζ)qt ) and (w(ζ)µt+1 pt+1 , w ′ (ζ)νt+1 qt+1 ).
(5.3)
The following is the desired Malcev chain: R µ(t + 1)pt+1 qR t+1 ν[t + 1] = µ(t)µt+1 pt+1 qt+1 ν[t + 1]
→ µ(t)w(ζ)L w(ζ)µt+1 pt+1 qR t+1 ν[t + 1] → µ(t)w(ζ)L w ′ (ζ)νt+1 qt+1 qR t+1 ν[t + 1] → µ(t)w(ζ)L w ′ (ζ)νt+1 ν[t + 1] = µ(t)w(ζ)L w ′ (ζ)ν[t] → µ(t)w(ζ)L w ′ (ζ)qt qR t ν[t] → µ(t)w(ζ)L w(ζ)pt qR t ν[t] → µ(t)pt qR t ν[t]. Concatenating all such chains yields a Malcev chain from µs to ν, and thus 5.2 (µs, ν) is a Malcev consequence of Q. By Lemma 5.2, each pair in (5.1) is a Malcev consequence of Q. Lemma 5.1 states that (u, v) is a consequence of (5.1) and T. Therefore, since neither T nor Q depend on (u, v), each valid relation in S is a Malcev consequence of T ∪ Q, 2 and so SgMhA | T ∪ Qi is a finite Malcev presentation for S. Theorem 2 extends to the case of asynchronous automatism. Theorem 3. Every asynchronous automatic semigroup that can be embedded in a group admits a finite Malcev presentation.
13
Proof of 3. This proof is similar to that of Theorem 2. First, by Theorem 3.5, assume without loss of generality that a boundedly asynchronous automatic structure exists. Following the proof for the synchronous case, show that b, b every valid relation is a consequence of those in (5.1). The walks u v, where (u, v) ∈ Ls are then linked by loops in a manner similar to Figure 2(a), but by appealing to Proposition 3.11 rather than Proposition 3.12. One must also use the definition of a boundedly asynchronous automaton to ensure that such a loop whose ‘left side’ is of length 1 cannot have a ‘right side’ exceeding length 2k, where the automaton does not read more than k letters in a row from either tape. Equipped with these loops of length at most 2k + 2λ + 1, one obtains a finite set of relations of the form (5.2). Reasoning as in Lemma 5.2 shows that the valid relation (µs, ν) is a Malcev consequence of the relations arising from 3 these loops. 6
automatism of subsemigroups of virtually free groups
Recall that a virtually free group is a group with a free subgroup of finite index. Muller and Schupp [18, Lemma 3] proved that a finitely generated virtually free group has context-free word problem. The proof of this result relies on the obvious fact that a finitely generated virtually free group has a free normal subgroup of finite index and therefore has a presentation
(6.1) Gp Y, D | di yh d−1 i = ui,h , di dj = zi,j dk ,
where h ranges over 1, . . . , m and i, j over 1, . . . , n; and where Y is the set of yh , D is the set of di , and ui,h and zi,j are words on Y ∪ Y −1 . The set Y is a basis for a free normal subgroup N of finite index n, and
Gp d1 , . . . , dn | di dj = dk
is the ‘multiplication table’ presentation for the finite group F/N. Muller and Schupp observe that any word can be rewritten using the relations in (6.1) to a unique normal form wd, where w is a reduced word on Y ∪ Y −1 and d ∈ D.
Theorem 4. Finitely generated subsemigroups of virtually free groups are automatic. Proof of 4. Let G be a virtually free group presented by (6.1). Identify elements of G with their normal form representatives. Thus the length |g| of an element g ∈ G is the length of its normal form representative. Use · to denote concatenation of reduced words on Y ∪ Y −1 ∪ D. Let A be a finite alphabet representing a subset of G, and let S be the subsemigroup of G generated by A. The first step is to prove: Proposition 6.1. The semigroup S — viewed as a set of normal form representatives — is a regular language over Y ∪ Y −1 ∪ D. The proof proceeds by constructing a finite state automaton over Y∪Y −1 ∪D that recognizes S. To immediately define the automaton would be to sacrifice clarity to brevity: some motivational remarks will aid understanding of the ideas behind the proof. View the process of multiplying elements of G in the following way. For elements w1 d1 , w2 d2 of G, with w1 , w2 being reduced words on Y ∪ Y −1 and 14
d1 , d2 ∈ D: concatenate w1 d2 and w1 d2 to obtain w1 d1 ·w2 d2 ; move d1 to the right using the defining relations of the presentation (6.1) to obtain w1 · w3 d3 , where w3 is a reduced word on Y ∪ Y −1 and d3 ∈ D; and freely reduce w1 · w3 to obtain a reduced word w4 . Thus w1 d1 w2 d2 = w4 d3 . Suppose xy is a two-letter subword of w4 d3 , with w4 d3 = w · xy · w ′ , and suppose that w·x = wx is a prefix of w1 that is unaffected by the free reduction of w1 · w3 . There are two ways in which the letter y can end up being next to the letter x: either w · x · y = wxy is also a prefix of w1 unaffected by free reduction, or y is a letter of w3 d3 such that the suffix of w1 after x and the prefix of w3 d3 before y reduce to the empty word. These two alternatives are the key to defining the automaton A below. Suppose the input to the automaton is wd ∈ G. If A has successfully read a prefix w ′ x of its input, is has [non-deterministically] ‘guessed’ a word u ∈ A+ such that w ′ x is a prefix of u ∈ S. The automaton keeps track of the letter x just read and the suffix s of u after w ′ x. The automaton can only read a letter y if either y is the first letter of s, or if there is a word v ∈ A+ such that forming the product u v brings x and y together in the manner described above. In this case, the ‘guessed’ word becomes uv. Observe that the automaton does not have to store this ‘guessed’ word, merely the difference between it and the input read thus far. Moreover, the number of possible differences can be assumed to be finite, by assuming that no proper prefix of the word v would serve to bring x and y together. Consider a simple example from a free group: forget about the letters d, so that multiplication is simply concatenation and free reduction. Let X = {xy2 , y−1 z, z−1 y−1 zx, x−1 y, . . .} be a subset of the free group on x, y, z. Now, xyz is in the semigroup T generated by X, since xy2 · y−1 z · z−1 y−1 zx · x−1 y = xzy. Analyze this factorization as follows: =ε
=ε
}| { z }| { z −1 −1 −1 x yy · y z · z y z x · x−1 y = xzy.
The automaton can immediately read x, since there are words in T starting with this letter. If the automaton ‘guesses’ that this word has a factorization starting with xy2 (as indeed it does, although there may be other factorizations), it moves to a state storing x (the letter just read) and y2 (the suffix of the first factor). It can then read z, since the suffix y2 , the factor y−1 z, and the prefix z−1 y−1 cancel, and z and x are not mutually inverse. The automaton moves to the state storing z and the suffix x. At this stage, the automaton does not need any information about factorization to the right of this z, so it does not need to store any more information. The letter x may be affected by cancellation, but the definition of the automaton requires that it will not ‘guess’ any factorization that would cause cancellation to affect the letter z. The automaton must be nondeterministic to deal with many possible factorizations. The definitions below are slightly more complicated than one would infer from this example, because of the letters d and the more complex multiplication. 15
Proof of 6.1. Let T = {w : w is a suffix of the normal form of da ∈ G, d ∈ D, a ∈ A}, and let Q = (T × (Y ∪ Y −1 ∪ {ε})) ∪ {q∞ }.
Let A = (Q, Y ∪ Y −1 ∪ D, η, (ε, ε), {q∞ }) be a finite state automaton, where the transition function η : Q × (Y ∪ Y −1 ∪ D) → Q consists of the following transitions and is elsewhere undefined: 1. ((s, y), z) 7→ (t, z) for z ∈ Y ∪ Y −1 if there exist a ∈ A, w ∈ A∗ such that swa = z · t, and y−1 6= z. 2. ((s, y), z) 7→ (t, z) for z ∈ Y ∪ Y −1 if s = z · t, |t| > 1, and y−1 6= z. 3. ((s, y), d) 7→ q∞ for d ∈ D if there exist a ∈ A, w ∈ A∗ such that swa = d.
4. ((s, y), d) 7→ q∞ for d ∈ D if s = d.
Lemma 6.2. S ⊆ L(A).
Proof of 6.2. Let f ∈ S and let u ∈ A+ with u = u1 · · · un , ui ∈ A, and u = f. Let 0 = i0 < i1 6 i2 6 · · · 6 i|f| = n be such that u(ij ) is the shortest prefix of u such that f(j) is a prefix of u(ij ) that remains unaffected by subsequent right-multiplication by uij+1 , . . . , un in forming u. (Observe that i|f| = n since any part of u after i|f| would affect at least the element of D at the end of u(i|f| ). Even if the letter after u(i|f| ) represented the identity, the d at the end would be involved in rewriting using the defining relations in (6.1), although it would simply be rewritten back to d.) Assume the automaton has read the prefix f(j), ending in y, and has reached a state (s, y), where s is the suffix of u(ij ) after f(j). There are now four possibilities: 1. ij+1 > ij and z ∈ Y ∪ Y −1 . A transition of type 1 exists, with w and a in its definition being such that u(ij )wa = u(ij+1 ). The state reached by this transition is (t, z) with t being such that u(ij+1 ) = f(j + 1) · t. 2. ij+1 = ij and z ∈ Y ∪ Y −1 . A type 2 transition exists and leads to a state (t, z) with s = z · t and so t is such that u(ij+1 ) = f(j + 1) · t. 3. ij+1 > ij and z ∈ D. A type 3 transition exists, with w and a in its definition being such that u(ij )wa = u(ij+1 ). The state reached by this transition is q∞ . 4. ij+1 = ij and z ∈ D. A transition of type 4 exists and leads to q∞ .
Induction on j shows that A accepts f, since entering state q∞ occurs only on reading z ∈ D, which can only happen at the end of f. 6.2 Lemma 6.3. L(A) ⊆ S.
16
Proof of 6.3. Let f ∈ L(A). Only transitions labelled by D reach q∞ , and the definitions of transitions labelled by Y ∪ Y −1 mean that f contains no adjacent letters that are mutually inverse. Therefore f is certainly in normal form. Choose and fix a walk γ in A that is labelled by f. Let γ = γ1 · · · γn , where each γi is a single transition. Let si be the first component of the state to which γi leads, with sn formally defined as ε. For transition γi of type 1 or 3, define γi ρ to be wa, where w ∈ A∗ and a ∈ A are as in the definition of this transition. If γi is of type 2 or 4, define γi ρ = ε. For some i, assume that γ1 ρ · · · γi ρ = f(i) · si .
(6.2)
Suppose γi is labelled by z, so that f(i + 1) = f(i) · z. If γi+1 is of type 2 or 4, then si = z · si+1 and γi+1 ρ = ε, and so: γ1 ρ · · · γi ργi+1 ρ = γ1 ρ · · · γi ρ = f(i) · si = f(i) · z · si+1 = f(i + 1) · si+1 . If, on the other hand, γi+1 is of type 1 or 3, then γi+1 ρ = wa, where w, a are such that si wa = z · si+1 , whence: γ1 ρ · · · γi ργi+1 ρ = γ1 ρ · · · γi ρwa = f(i) · si wa = f(i) · z · si+1 = f(i + 1) · si+1 . In either case, (6.2) holds for i + 1. Observe that (6.2) holds trivially for i = 0. Induction now shows that it holds for all i. In particular, it holds for i = n. Therefore γ1 ρ · · · γn ρ = f, and so f ∈ S. 6.3 Lemmas 6.2 and 6.3 together prove Propostion 6.1.
6.1
If S is not already a monoid, adjoin an identity to obtain a monoid M, otherwise let M = S. Let 1 be a new symbol not in A that represents the identity. The strategy is now to construct from A a generalized finite state automaton W that recognizes a language over A ∪ {1} that maps onto M. y Let p −→ q be a transition in A. Suppose it is of type 1 or 3. Let w, a be as in the definition of this transition, with w of minimal length. Define y (p, y, q)ρ = wa. For all other transitions p −→ q, define (p, y, q)ρ = ε. (This definition of ρ echoes that in the proof of Lemma 6.3.) Let m = max{| im ρ|}. The construction of W is as follows. Retain the state set and the start and y accept states from A. Replace the label on each transition p −→ q of the automaton A by (p, y, q)ρ1m−|(p,y,q)ρ| , so that each edge in W has a label of length m. The following two lemmata relate a word recognized by A to the word labelling the corresponding path in W. 17
Lemma 6.4. If u labels a walk in W from the start state to an accept state and s labels the corresponding walk in A, then u = s and m|s| = |u|. Proof of 6.4. This lemma follows from the proof of Lemma 6.3, noting the coincidence of the definition of ρ in that proof and in the construction of W. 6.4 Lemma 6.5. Let u ∈ L(W) and let t ∈ N ∪ {0} with m | t. Then the first t/m letters of u form a word wu over X ∪ X−1 ∪ D that label the walk in A corresponding to the walk labelled by u(t) in W. Furthermore, if su is the first component of the state to which this walk leads (in either A or W), then u(t) = wu · su . Proof of 6.5. Observe that, since m | t, upon reading the prefix u(t), the generalized finite state automaton A does indeed enter a state. Reasoning parallel to that in the proof of Lemma 6.3 shows that u(t) = wu · su . 6.5 Observe that Lemma 6.4 and the fact that A recognizes S together imply that L(W) maps onto S. The remainder of the proof consists of showing that if M = S, then (A ∪ {1}, L(W)) is an automatic structure for M; and that if M 6= S, then (A ∪ {1}, L(W) ∪ {1}) is an automatic structure for M. First case. Suppose M = S. Let u, v ∈ L(W) be such that ua = v for a ∈ A ∪ {1} ∪ {ε}. In order to invoke Theorem 3.14, it is necessary to show that for all t ∈ N ∪ {0}, u(t) and v(t) are within a bounded distance of one another. One may assume without loss that t 6 max{|u|, |v|}. Since cancellation can proceed at most l = max{|da| : d ∈ D, a ∈ A} leftwards from the end of an element of S, u and v must be equal except for suffixes of length at most l. Let t 6 max{|u|, |v|} and suppose that m | t. 1. If t is such that |u| − t/m is greater than l, apply Lemma 6.5 to u and v to obtain wu , su , wv and sv . The prefix wu , being of length t/m < |u| − l, is confined to the part of u that is unaffected by cancellation in forming −1
the product u a, and so wu and wv are identical. Therefore u(t) v(t) = s−1 u sv , and since there are only a finite number of choices for su and sv , u(t) and v(t) are within a bounded distance of one another. 2. If t is such that |u| − t/m does not exceed l, then |u| = m|u| 6 ml + t, so |u| − |u(t)| 6 ml. The distance between u(t) and v is at therefore at most ml + 1. Consider the maximum amount of cancellation that can occur in forming the product ua to see that |v| > |u| − l. So m|v| > m|u| − ml, whence |v| > |u| − ml by Lemma 6.4. So t 6 max{|u|, |v|} 6 |v| + ml, or |v| > t − ml. On the other hand, considering the case when no cancellation occurs in forming the product ua shows that |v| 6 |u| + l. So |v| = m|v| 6 m|u| + ml 6 2ml + t. Therefore −ml 6 |v| − t 6 2ml, or |v| − t 6 2ml. So the distance from v to v(t) is at most 2ml. So the total distance from u(t) to v(t) is at most 3ml + 1.
18
Clearly, if m ∤ t, then a distance at most 2m is added to these bounds. Therefore (A∪{1}, L(W)) possesses the fellow traveller property. The conditions of Theorem 3.14 are thus satisfied and so (A ∪ {1}, L(W)) is an automatic structure for M. Second case. Now suppose the other case holds: that M 6= S and a ∈ A ∪ {1} ∪ {ε}. The reasoning above remains valid except when u or v is 1. Suppose a = ε or a = 1. Clearly, u = 1 if and only if v = 1, since otherwise there would be a word in L(W) ⊆ A+ representing the identity of M, contradicting the definition of M. Trivially, therefore, u(t) and v(t) are within a bounded distance of one another for all t. If a ∈ A, then v cannot be 1, since otherwise ua would represent the identity, contradicting the definition of M. So assume u = 1, so that v = a. Then v can have at length at most m|a|, by the definition of W. Therefore, for all t, the distance between u(t) and v(t) is bounded. Therefore (A ∪ {1}, L(W) ∪ {1}) satisfies the conditions of Theorem 3.14 and so is an automatic structure for M. The monoid M is therefore automatic. 4 Now apply Theorem 3.13 to show that S is automatic. Theorems 2 and 4 together give a new proof that finitely generated subsemigroups of virtually free groups have finite Malcev presentations [3, Theorem 3]. However, the proof in [3] has the advantage of being an algorithm for finding a finite Malcev presentation. In contrast, Theorem 2 does not give a constructive method for finding the relations (5.2). 7
a semigroup with no finite malcev presentation
This section exhibits an example of a semigroup that fails to admit a finite Malcev presentation, but which embeds in a group all finitely generated subgroups of which are finitely presented. Denote by FG(Z) the free group on the alphabet Z. Let G be the free product FG(x, y, z, s, t) ∗ (Z × Z × Z). Identify elements of G with alternating products of elements of FG(x, y, z, s, t) (viewed as reduced words) and of (Z × Z × Z) (viewed as triples of integers). Let A = {a, b, c, d, e, f, g, h, i, j} be an alphabet, and let these letters represent a set of elements X in the following way: a = x2 y, b = y−1 (1, 0, 0)y, c = y−1 z, d = z−1 (0, 1, 0)z, e = z−1 x2 , f = x2 s, g = s−1 (1, 0, 1)s, h = s−1 t, i = t−1 (0, 1, −1)t, j = t−1 x2 . Let S be the subsemigroup of G that X generates. Lemma 7.1. The semigroup S is presented in terms of the generators X by SghA | Ri, where R = {(abα cdα e, fgα hiα j) : α ∈ N ∪ {0}} .
19
u1 \v1 b c d e f g a 1 2 1 1 1 3 1 1 1 1 b c 2 2 2 2 1 1 1 d 1 1 e f 1 g h i
h 2 2 2 2 2 2 2
i 1 1 2 1 1 1 1 2
j 1 1 2 1 1 1 1 2 1
Table 1: Possibilities for u1 and v1 . Proof of 7.1. Let α ∈ N ∪ {0}. Then abα cdα e = x2 y(y−1 (1, 0, 0)y)α y−1 z(z−1 (0, 1, 0)z)α z−1 x2 = x2 yy−1 (α, 0, 0)yy−1 zz−1 (0, α, 0)zz−1 x2 = x2 (α, 0, 0)(0, α, 0)x2 = x2 (α, α, 0)x2 = x2 (α, 0, α)(0, α, −α)x2 = x2 ss−1 (α, 0, α)ss−1 tt−1 (0, α, −α)tt−1 x2 = x2 s(s−1 (1, 0, 1)s)α s−1 t(t−1 (0, 1, −1)t)α t−1 x2 = fgα hiαj. So all of the relations in R hold in S. Define a set N of normal forms to be the set of all words in A+ that do not contain fgα hiα j for any α ∈ N ∪ {0}. Clearly, since there are no overlaps between these forbidden words and words abα cdα e, every element of S is represented by at least one element of N. Let u = u1 · · · up , v = v1 · · · vq ∈ N (ui , vi ∈ A for all i) be distinct words in normal form, and suppose they represent the same element of S. Without loss of generality, suppose that u1 6= v1 and that u precedes v in the lexicographic ordering based on a ≺ b ≺ c ≺ . . . ≺ j. Observe that x−1 appears in no word in X, nor does a negative number appear in the first or second components of any tuple. Letters x and tuples therefore cannot be cancelled. The possibilities marked 1 in Table 1 are immediately excluded for this reason. If u1 = c, the only way the occurrence of z can be cancelled from c is if u2 = d or u2 = e, which leaves the y−1 unaffected. Reasoning parallel to this, together with the observation in the last paragraph, excludes the possibilities marked 2 in Table 1. Suppose that u1 = b and v1 = c. Then the occurrence of y must be cancelled from b by a string of β letters b and one c. So u = y−1 (β + 1, 0, 0)zuβ+3 · · · up v = y−1 zv2 · · · vq . However, the occurrence of z must now be cancelled from v. Yet, using a string of letters d or e can only replace z with a tuple with 0 in the first component, 20
followed by x2 or an uncancelled z. This is a contradiction, which eliminates the possibility marked 3 from Table 1. Therefore u1 = a and v1 = f. Either u2 = b or v2 = g in order to cancel the occurrence of y or s. Suppose the former; the latter case is analogous. Suppose that u2 = . . . = uα+1 = b and uα+2 6= b for some α. Then u = x2 (α, 0, 0)yuα+2 · · · up . To match the first component of this tuple, v2 = . . . = vα+1 = g and vα+2 6= g. So v = x2 (α, 0, α)svα+2 · · · vq . Since uα+2 6= b and vα+2 6= g, either uα+2 = c or vα+2 = h. Once again, suppose the first case; the other follows in the same way. So u = x2 (α, 0, 0)zuα+3 · · · up . Repeating the reasoning above, with reference to the second component of the tuple rather than the first, shows that vα+2 = h and that uα+3 is the first of a string of β letters d and vα+3 a string of letters i. This shows that u = x2 (α, β, 0)z· · · up , v = x2 (α, β, α − β)t· · · vq . The only way the letters z and t can be cancelled is if the next letters of u and v are e and j, respectively. In order to match the third component of the tuple, therefore, α = β. So v begins fgα hiα j, which contradicts v’s membership of N. The set N is therefore a set of unique normal forms for S, and so SghA | Ri 7.1 does indeed present S. The following result will be needed shortly: Theorem 7.2 ([1]). Let J and L be finitely presented groups. Then their free product with amalgamated subgroup K, J ∗K L, is finitely presented if and only if K is finitely generated. 7.2 Proposition 7.3. The semigroup S does not admit a finite Malcev presentation. Proof of 7.3. By the preceding lemma, S is presented by SghA | Ri. Suppose S does admit a finite Malcev presentation. Then it admits one of the form SghA | Ni, where N = {(abα cdα e, fgα hiα j) : α ∈ {0, . . . , n}} , for some n ∈ N ∪ {0}. Since SgMhA | Ri ≃ SgMhA | Ni, GphA | Ri ≃ GphA | Ni. So H = GphA | Ri is finitely presented. Now, H = FG(a, b, c, d, e) ∗K FG(f, g, h, i, j) , where K ≃ Gphabα cdα e, α ∈ N ∪ {0}i ≃ Gphfgα hiα j, α ∈ N ∪ {0}i. If α > β, then abα cdα e(abβ cdβ e)−1 = abα cdα−β c−1 b−β a−1 , with α − β > 0. If α 6= β 6= γ, then at least the middle c remains uncancelled in the 21
product abα cdα e(abβ cdβ e)−1 abγ cdγ e. This shows that B = {abα cdα e : α ∈ N ∪ {0}} is Nielsen-reduced and so forms a basis for K (see, for example, [13, Section I.2]). The amalgamated subgroup K is therefore not finitely generated. By Theorem 7.2, H cannot be finitely presented. This is a contradiction, and 7.3 so S has no finite Malcev presentation. This example is particularly interesting because all finitely generated subgroups of G must, by the Kurosh Subgroup Theorem (see [13, 14]), be finitely presented. It is clear, therefore, that the coherence of a group is not sufficient — though it is of course necessary — to guarantee that all of its finitely generated subsemigroups admit finite Malcev presentations. (Observe that the universal group of S obviously does not coincide with its subgroup envelope in G.) In summary, therefore: Theorem 5. The class of groups all of whose finitely generated subsemigroups admit 5 finite Malcev presentations is strictly contained in the class of coherent groups. Applying Theorem 2 shows that the semigroup S is not automatic. Yet S is a subsemigroup of G, a free product of a free group and an abelian group. The Kurosh subgroup theorem and [7, Theorem 12.1.4] together show that the subgroup of G generated by S is automatic. Therefore S is an example of the positive subsemigroup of an automatic group that is not itself automatic. Question 7.4. Does the automatism of the universal group of a semigroup that embeds in a group imply the automatism of the semigroup itself? Although related to the question regarding positive subsemigroups just answered, one cannot hope to use an analogous argument to find an example of a non-automatic semigroup embeddable in a group whose universal group is automatic. Let T be a semigroup embeddable into a group. If the universal group of T were automatic, then it would be finitely presented. In particular, it would have a finite presentation GphA | ρi, where ρ ⊆ A+ × A+ , on a finite generating set for T , so SgMhA | ρi would be a finite Malcev presentation for T. 8
free products of free monoids and abelian groups
The following result contrasts the example from Section 7. Theorem 6. Let M be the [monoid] free product of a free monoid and an abelian group. Every finitely generated subsemigroup of M is asynchronously automatic and so has a finite Malcev presentation. Proof of 6. Suppose M = Y ∗ ∗ H, where Y is an alphabet and H is an abelian group. Identify M with alternating products of elements of Y ∗ and elements of H. Let A be a finite alphabet representing a subset of M. Partition A as A ′ ∪ A ′′ , where A ′′ = A ∩ H and A ′ = A − A ′′ . Let R be the subsemigroup of M generated by A and let R ′′ be the subsemigroup of H generated by A ′′ . Observe that any letters of A representing the identity of M are in A ′′ . Propostion 3.15 asserts the existence of an automatic structure (A ′′ , K) for ′′ R , where K maps bijectively onto R ′′ . If 1M ∈ R ′′ , let y be its unique representative in K. 22
Define the language of normal forms L as ∗ K (A ′ )+ K − {y} (A ′ )∗
if 1M ∈ R ′′ , and
(8.1)
∗ (K ∪ {ε}) (A ′ )+ K (A ′ )∗ − {ε}
otherwise. Observe that in both cases, a word in L alternates between nonempty words over A ′ and words from K. Assume that 1M ∈ R ′′ , so that L is given by (8.1). The other case is similar. Lemma 8.1. L = R. Proof of 8.1. Any w ∈ A+ is of the form w = u0 v1 u1 · · · vn−1 un−1 vn un , where ui ∈ (A ′′ )∗ and vi ∈ (A ′ )∗ , all except possibly u0 , un being nonempty. Furthermore, one can assume that no non-empty ui represents 1M . Replace each ui with the unique representative in K of ui . If u0 is empty, replace it by y. None of these replacements change w, and the resulting word is in L. 8.1 Hence L = A+ = R. Let w ∈ A+ . Then w can be uniquely written as an alternating product z0 s1 z1 · · · sn−1 zn−1 sn zn ,
(8.2)
where si ∈ Y + , zi ∈ H, and all except possibly z0 , zn not being 1M . For the purposes of this proof, call z0 and zn the H-prefix and H-suffix of w, respectively. Lemma 8.2. Let P = {z : z is the H-prefix of a, a ∈ A} ∪ {1H }, Q = {z : z is the H-suffix of a, a ∈ A} ∪ {1H } ∪ A ′′ . b, b Let p, r ∈ P, q, s ∈ Q, and let u, v ∈ K be such that puq = rvs. Then u v are walks a bounded Hausdorff distance apart in the Cayley graph of R ′′ , with the bound depending only on A. Proof of 8.2. Let T = {pq(rs)−1 : p, r ∈ P, q, s ∈ Q such that there exist w, w ′ ∈ K with pwq = rw ′ s}. Observe that T only depends on A. Suppose p, r ∈ P and q, s ∈ Q are such that pq(rs)−1 ∈ T . Then there exist w, w ′ ∈ K with pwq = rw ′ s. Since H is abelian, pq(rs)−1 = w ′ w−1 . For each t ∈ T , choose wt , wt′ ∈ (A ′′ )+ of minimal total length such that t = wt′ (wt )−1 . Now let p, q, r, s, u, v be as in the hypothesis, and let t = pq(rs)−1 . Then ut = v, and so uwt′ = vwt . Let k be the unique word in K representing uwt . Since (A ′′ , K) is an automatic structure for R ′′ , Corollary 3.10 shows that the b and from b walk b k is a bounded Hausdorff distance from u v, with the bound b, b v depending only on T , which in turn depends on A. Therefore the walks u are only a bounded Hausdorff distance apart, the bound depending on A alone. 8.2 23
Let u, v ∈ L. Suppose that u = v. Let t ∈ N. The strategy is to show that u(t) is only a bounded distance from some vertex on the walk b v. Let u(t) = ρu σu , where σu is the longest suffix of u(t) in (A ′′ )∗ . The immediate aim is to show that ρu is within a bounded distance of b v. v. Therefore If ρu = ε, then ρu is certainly within a bounded distance of b assume that ρu is nonempty. Write ρu in the form (8.2) as: ρu = z0 s1 z1 · · · sn−1 zn−1 sn zn . Since elements of Y ∗ cannot cancel, there is a shortest prefix ρv of v such that z0 s1 z1 · · · sn−1 zn−1 sn is a prefix of ρv . (Observe that the last letter of ρv must be in A ′ .) Then ρv = ρu z−1 n x, where x is a suffix of ρv . Since zn and x are both suffixes of elements of A ′ , of which there are a finite number, the distance between ρu and ρv is bounded by a quantity dependent only on A. If σu = ε, there is nothing further to prove. Therefore suppose σu is nonempty. Let τu be the longest prefix of u[t] in (A ′′ )∗ . The word σu τu is therefore the longest word over A ′′ ‘surrounding’ the t-th letter of u. By the definition of L, σu τu ∈ K − {y}. If τu does not stretch to the end of u, then the letter of u immediately following τu is in A ′ . Let p be the H-prefix of this letter, or let p = 1M if τu is indeed a suffix of u. Write ρv in the form (8.2) as: ′ ′ ′ ′ sm zm . zm−1 ρv = z0′ s1′ z1′ · · · sm−1
If m > n, then since ρv was defined as the shortest prefix of v such that ρv contains z0 s1 z1 · · · sn−1 zn−1 sn , the element represented by the last letter of ′ . It must ρv must contain a nonempty suffix of sn (perhaps all of it) and sm ′ ∗ therefore also contain the whole of zn . Since elements of Y cannot cancel, ′ = z σ τ p. Therefore σ τ = z−1 z ′ p−1 . Since σ τ ∈ K, it is the unique zn u u u u n u u n n ′ p−1 . Therefore, the number of possibilities for σ is representative of z−1 z u n n bounded by a quantity depending on A, and so u(t) is within a bounded distance of ρv . If m = n, then the situation is that shown in Figure 3. The suffix x of ′ of ρ . The elements u(t) and ρ differ by an ρv is actually the H-suffix zn v v element of H. Let τv be the longest subword of v following ρv that lies in (A ′′ )∗ . Let r be the H-prefix of the element represented by the letter of A ′ immediately following τv in v, or 1M if τv is a suffix of v. Since letters from Y ′ τ r. Lemma 8.2 applies to show that σ τ cannot be cancelled, zn σu τu p = zn v u u and τv label walks a bounded distance apart in the Cayley graph of R ′′ . So there exists a prefix χv of τv representing an element of R ′′ within a bounded distance of that represented by σu . (Indicated by the dashed line in Figure 3.) Since ρu and ρv are within a bounded distance of one another, so are ρu σu and ρv τv . Similar reasoning shows that v(t) is a bounded distance from a every verb for all t. Therefore the walks u b and b tex on u v are a bounded Hausdorff distance from one another. Now suppose that ua = v for some a ∈ A. If u does not end with a string from (A ′′ )+ , then it is clear that right-multiplication by a can only affect a ‘small’ part of the rightmost end of u. If u does have a suffix from (A ′′ )+ , then b and Lemma 8.2 applies. In either case, by reasoning parallel to that above, u b v are a bounded Hausdorff distance apart in Γ (R, A). 24
u(t)
u[t℄ u
ρu
σu
τu
p
zn z0 s1 z1 . . . sn
u=v
in H r
z′n χv
v ρv
τv
Figure 3: Words in L and corresponding elements of R ⊆ Y ∗ ∗ H. Lemma 8.3. There exists a departure function for R. Proof of 8.3. Let s ∈ R, and suppose, with the aim of obtaining a contradiction, that W ⊆ L is an infinite set of words representing s. Let α be the number of letters from Y appearing in s. Since every element of A ′ contains at least one letter from Y, any word w ∈ W can contain at most α letters from A ′ . Therefore, there is an infinite set W1 ⊆ W such that every element of W1 has the same projection onto (A ′ )+ . (‘Projection’ means ‘deletion of all letters not in A ′ ’.) Words in W1 differ only in subwords from (A ′′ )+ . However, each such subword must, by the definition of L, lie in K. Therefore there are words in K mapping to the same element of R ′′ , which contradicts the fact that K maps uniquely onto R ′′ . So L maps finite-to-one onto K. Therefore D(r) = max{|y| : x, y, z ∈ A+ , xyz ∈ L, d(x, xy) < r} is well-defined, since Lemma 3.8 implies that the sets on the right-hand side are finite. The function D is thus a departure function for R. 8.3 The rational structure (A, L) satisfies the hypotheses of Theorem 3.9 and so forms a boundedly asynchronous automatic structure for R. Theorem 3 shows 6 that R has a finite Malcev presentation. Section 7 showed that the free product of a free group and an abelian group contains finitely generated subsemigroups without finite Malcev presentations. Theorem 6 asserts that replacing the free group with a free monoid produces a monoid all of whose finitely generated subsemigroups have finite Malcev presentations. One may ask whether the other obvious restriction — replacing the abelian group with (N ∪ {0})n — also produces such a monoid. The answer to this question is no: taking the same alphabet A and the generators a = x2 y, b = y−1 (1, 1, 0, 0)y, c = y−1 z, d = z−1 (0, 0, 1, 1)z, e = z−1 x2 , f = x2 s, g = s−1 (1, 0, 1, 0)s, h = s−1 t, i = t−1 (0, 1, 0, 1)t, j = t−1 x2 in FG(x, y, z, s, t) ∗ (N ∪ {0})4 yields a subsemigroup isomorphic to the semigroup S of Section 7. 25
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