REMARKS ON FINITELY PRESENTED BANDS

by JOSEPH LAU

Abstract. For each variety of bands V, we give a formula for ϕ_{V}(m, k), which is the largest integer such that for every band B in V generated by m generators and k relations, there is a subset of the generators of size ϕ_{V}(m, k) which generates a (relatively) free sub-band of B as a basis. We also determine the semilattice structure of a finitely presented band and the complete structure of a finitely presented normal band.

1. Introduction and Preliminaries. Let V be a variety of algebras. For integers m > k, we define ϕ_{V}(m, k) to be the largest integer such that for every algebra B in V presented by m generators a_{1}, …, a_{m} and k relations, it is possible to choose ϕ_{V}(m, k) elements from among a_{1}, …, a_{m} which generate a (relatively) free subalgebra of B as a basis.

The question of finding ϕ_{V} for the variety of groups was posed by Lyndon ([7, Question 5.29]). Romanovskii [8] showed that ϕ_{V}(m, k) = m − k in this case, but the proof is not constructive. For the variety of semigroups Shneerson [10] gave a constructive proof, based on linear algebra, that ϕ_{V}(m, k) = m − k. Standard linear algebra methods, applied to abelian groups (see [5, Chapter 10]), shows that ϕ_{V}(m, k) = m − k for the variety of abelian groups also. This implies, by an easy exercise, that ϕ_{V}(m, k) = m−k for the variety of commutative semigroups also.

Let V be a variety of algebras. Let A be an alphabet and R a set of relations. Then ⟨A | R⟩_{V} will denote the algebra in V generated by A, subject to the relations of R. The subscript V is sometimes dropped when it is clear from context. We record the following properties of ϕ_{V}.

Lemma 1.1. For any variety of algebras V,

(i) ϕ_{V}(m_{1} + m_{2}, k_{1} + k_{2}) ≤ ϕ_{V}(m_{1}, k_{1}) + ϕ_{V}(m_{2}, k_{2}), and (ii) ϕ_{V}(m, k) ≤ ϕ_{V}(m − 1, k − 1).

Proof. If S_{1} = ⟨A_{1} | R_{1}⟩_{V} and S_{2} = ⟨A_{2} | R_{2}⟩_{V} are two algebras defined with disjoint alphabets, then A_{0} ⊆ A_{1} ∪A_{2} generates a relatively free subalgebra as a basis in ⟨A_{1} ∪ A_{2} | R_{1} ∪ R_{2}⟩_{V} only if A_{0} ∩ A_{i} does so in S_{i} for i = 1, 2. Hence the first inequality.

Suppose S = ⟨A | R⟩_{V}. Choose a ∈ A and a letter b not in A. We prove (ii) by considering ⟨A, b | R, a = b⟩_{V}, using similar reasoning to above. □

In this paper, we give a formula for ϕ_{V}(m, k) for each variety of bands V. Section 2 proves a combinatorial lemma which is the main ingredient of the proof of the formula. Section 3 gives an upper bound on ϕ_{V}(m, k) in the case of semilattices. The formula is proved in Section 4 by first elucidating the semilattice structure of a finitely presented band. Finally, the results in Section 4 are improved in Section 5 to give complete information on the structure of a finitely presented normal band.

We shall use the following notation: the free semigroup over a set A is denoted by F_{A}. If w ∈ F_{A}^{1} we define its content C(w) to be the set of elements of A appearing in w. If w ∈ F_{A}, denote its initial and terminal letters by i(w) and t(w) respectively. Let P_{A} be the semilattice of finite subsets of A under union.

2. A Combinatorial Lemma.

Lemma 2.1. Let A be a set of size m. Let A_{1}, A_{2}, …, A_{k} be subsets of A such that |A_{i}| ≥ l for each i. Then there exists a subset A′ of A such that

(i) |A′| ≤ min\{⌊\frac{m+(l−1)k}{2l−1} ⌋, k\}, (ii) A′ ∩ A_{i} ≠ ∅ for each i.

Proof. Let n = lk − m. We shall deal with the case when n ≤ 0 first, and then prove the lemma for n > 0 by induction.

If n ≤ 0, then m ≥ lk and ⌊\frac{m+(l−1)k}{2l−1} ⌋ ≥ k. For each i choose a_{i} ∈ A_{i}. The lemma is proved by letting A′ = {a_{1}, a_{2}, …, a_{k}}.

Now suppose n > 0. Since |A_{i}| ≥ l for each i and lk > m, some a_{0} ∈ A is in (at least) two of the A_{i}’s. In other words,

k′ = |{i | a_{0} ∉ A_{i}}| ≤ k − 2.

Since lk′ − (m − 1) ≤ lk − m − (2l − 1) < n, by induction there is a subset B of A \ {a_{0}} such that

|B| ≤ min\{⌊\frac{(m − 1) + (l − 1)k′}{2l − 1} ⌋, k′\}

≤ min\{⌊\frac{(m + (l − 1)k − (2l − 1)}{2l − 1}⌋, k − 2\}

≤ min\{⌊\frac{m + (l − 1)k}{2l − 1} ⌋, k\} − 1,

and B ∩ A_{i} ≠ ∅ if a_{0} ∉ A_{i}. The lemma is proved by letting A′ = {a_{0}} ∪ B. □

Obviously, the bound on |A′| can be attained when m ≥ lk, and when m = l, the bound can be improved to |A′| = 1. I do not know the necessary and sufficient conditions for the bound in Lemma 2.1 to be sharp.

Using Lemma 2.1 in the case l = 2 allows us to prove the following combinatorial lemma, on which the proof of the formula in Section 4 depends.

Lemma 2.2. Let A be a set of size m. Suppose A_{1}, A_{2}, …, A_{k} and B_{1}, B_{2}, …, B_{k} are subsets of A such that for each i, either

(a) |A_{i}| ≥ 2 and |B_{i}| ≥ 2, or

(b) |A_{i}| = 1 and A_{i} ⊆ B_{i}.

Then there exists a subset A′ of A such that

(i) |A′| ≤ k + min\{⌊\frac{m−k}{3} ⌋, k\},

(ii) A′ ∩ A_{i} ≠ ∅ and A′ ∩ B_{i} ≠ ∅ for each i.

Proof. By reordering we can assume |A_{i}| = 1 if and only if i ∈ {1, 2, …, k_{0}} for some k_{0} ≤ k. Let C = A_{1} ∪ A_{2} ∪ ⋯ ∪ A_{k_{0}}. Then by condition (b),

k′ = |{i | C ∩ A_{i} = ∅}| + |{i | C ∩ B_{i} = ∅}| ≤ 2(k − k_{0}) ≤ 2(k − |C|).

Applying Lemma 2.1 with l = 2, we obtain a subset D of A \ C so that

|D| ≤ min\{⌊\frac{(m − |C|) + k′}{3} ⌋, k′\}

≤ min\{⌊\frac{(m − |C|) + 2(k − |C|)}{3}⌋, 2(k − |C|)\}

= (k − |C|) + min\{⌊\frac{(m − |C|) − (k − |C|)}{3}⌋, k − |C|\}

≤ k + min\{⌊\frac{m − k}{3} ⌋, k\} − |C|,

D ∩ A_{i} ≠ ∅ if C ∩ A_{i} = ∅, and D ∩ B_{i} ≠ ∅ if C ∩ B_{i} = ∅. The lemma is proved by letting A′ = C ∪ D. □

Note that the proofs of the lemmas provide an algorithm for finding the required A′. The algorithm guarantees an upper bound on the size of A′, but may not give the smallest possible A′.

3. Semilattices. In this section let V denote the variety of semilattices. Let

S = ⟨A | u_{i} = v_{i}, i ∈ I⟩_{V}

be a finitely presented semilattice. Let α be the congruence on P_{A} generated by the relation

ρ = {(C(u_{i}), C(v_{i})) | i ∈ I}.

Then S = P_{A}/α.

Lemma 3.1. Each α-class Bα has a unique maximal element, denoted by m(Bα).

Proof. We give an algorithm for finding m(Bα), given a subset B of A. First we let M = B. For each i ∈ I, if C(u_{i}) ⊆ M or C(v_{i}) ⊆ M (but not both), then replace M by M ∪ C(u_{i}) ∪ C(v_{i}). (This corresponds to the elementary ρ-transition M → M ∪ C(u_{i}) or M → M ∪ C(v_{i}).) Repeat this procedure until for each i, either C(u_{i}) ∪ C(v_{i}) ⊆ M or C(u_{i}) ⊈ M and C(v_{i}) ⊈ M.

By the construction, it is clear that M is not α-equivalent to another subset N if N ⊈ M. Therefore M is maximal and unique. □

The word problem can be easily solved by the procedure given in the proof: BαC if and only if m(Bα) = m(Cα).

Lemma 3.2. Let V denote the variety of semilattices. Then for m > k,

ϕ_{V}(m, k) ≤ m − (k + min\{⌊\frac{m − k}{3} ⌋, k\}).

Proof. We first prove a special case, then use it to prove the lemma in general.

Let E = ⟨a, b, c, d | ab = cd⟩_{V}. If B ⊂ {a, b, c, d} has 3 elements, then either {a, b} ⊂ B or {c, d} ⊂ B. Suppose without loss of generality that {a, b} ⊂ B. Since m({a, b}α) = {a, b, c, d} = m(Bα), B does not freely generate a sub-semilattice of E. Therefore ϕ_{V}(4, 1) ≤ 2.

Case 1 Suppose ⌊\frac{m−k}{3} ⌋ ≥ k, that is, m = 4k + l for some l ≥ 0. Then by Lemma 1.1 (i),

ϕ_{V}(m, k) ≤ ϕ_{V}(l, 0) + ∑_{i=1}^{k} ϕ_{V}(4, 1) ≤ l + 2k = m − 2k.

Case 2 If ⌊\frac{m−k}{3} ⌋ ≤ k, then, writing m′ = m − (k − ⌊\frac{m−k}{3} ⌋) and k′ = k − (k − ⌊\frac{m−k}{3} ⌋),

ϕ_{V}(m, k) ≤ ϕ_{V}(m′, k′) ≤ m′ − 2k′ = m − (k + ⌊\frac{m − k}{3} ⌋),

where the first inequality follows from Lemma 1.1 (ii), and the second inequality follows from case 1, because ⌊\frac{m′−k′}{3} ⌋ = ⌊\frac{m−k}{3} ⌋ = k′. □

4. Varieties of Bands. In this section we determine a formula of ϕ_{V}(m, k) for each variety of bands V. Along the way we determine the semilattice structure of bands in V defined by a presentation. We assume familiarity with the basic results from the theory of semigroups and bands, which can be found in [6].

Theorem 4.1. In any band B, J equals the minimum semilattice congruence on B, and B is a semilattice of rectangular bands.

Corollary 4.2. Let T be a sub-band of a band B. Then for all x, y in T, xJ y in T if and only if xJ y in B.

Proof. If xJ y in a band, then x and y lie in the same rectangular band. Therefore xyx = x and yxy = y. □

Let A be a fixed alphabet. For a variety of bands V, let R_{V} be the fully invariant congruence on F_{A} generated by the defining identities for V. Thus F_{A}/R_{V} is the free band in V over A.

The lattice LB of varieties of bands has been determined by Birjukov [1], Fennemore [2] and Gerhard [3] (see also [4]). Only the following fact will be needed in this paper ([9], Remark 3.11):

Let LB_{0} be the lattice obtained from LB by taking away the four varieties of trivial, right zero, left zero, and rectangular bands. If V ∈ LB_{0}, then C(w) = C(z) if (w, z) ∈ R_{V}.

We first treat the case of the subvarieties of rectangular bands.

Theorem 4.3. ϕ_{V}(m, k) = m − k for the variety of left [right] zero semigroups. ϕ_{V}(m, k) = max{m−2k, 1} for the variety of rectangular bands.

Proof. Let LZ and RZ denote the varieties of left and right zero semigroups respectively. If V = LZ, then

⟨A | u_{j} = v_{j}, j ∈ J⟩_{V} ≃ ⟨A | i(u_{j}) = i(v_{j}), j ∈ J⟩_{V},

so the result is clear. The case of right zero semigroups is dual.

If V is the variety of rectangular bands, then it is easy to see that

\begin{aligned}⟨A | u_{j} & = v_{j}, j ∈ J⟩_{V} ≃ ⟨A | i(u_{j})t(u_{j}) = i(v_{j})t(v_{j}), j ∈ J⟩_{V} \\ & ≃ ⟨A | i(u_{j}) = i(v_{j}), j ∈ J⟩_{LZ} × ⟨A | t(u_{j}) = t(v_{j}), j ∈ J⟩_{RZ}, \end{aligned}

and the result follows. □

For the rest of the section let V be a variety in LB_{0}.

Let B be a band in the variety V defined by the presentation

B = ⟨A | u_{i} = v_{i}, i ∈ I⟩_{V}

where I is some index set and u_{i}, v_{i} ∈ F_{A} for each i ∈ I. Let β be the congruence on F_{A} generated by the relation

R = {(u_{i}, v_{i}) | i ∈ I} ∪ R_{V}.

Then B = F_{A}/β. Let α be the congruence on P_{A} generated by the relation

ρ = {(C(u_{i}), C(v_{i})) | i ∈ I}.

Now any two distinct β-equivalent elements of F_{A} are connected by a finite sequence of elementary R-transitions. If the transition is of the form

puq → pvq (p, q ∈ F_{A}^{1}, (u, v) ∈ R_{V}),

then C(puq) = C(pvq). On the other hand, if the transition is of the form

pu_{i}q → pv_{i}q or pv_{i}q → pu_{i}q (p, q ∈ F_{A}^{1})

for some i ∈ I, then there is a corresponding elementary ρ-transition

C(p)C(u_{i})C(q) → C(p)C(v_{i})C(q) or C(p)C(v_{i})C(q) → C(p)C(u_{i})C(q).

Therefore

wβ = zβ ⇒ C(w)α = C(z)α.

Suppose now that w, z ∈ F_{A} are such that wβJ zβ in B. Then there exist x, y ∈ F_{A}^{1} such that (xβ)(wβ)(yβ) = zβ. Therefore

C(x)C(w)C(y)α = C(z)α

by the above, and C(w)C(z)α = C(z)α. Hence C(z)α ≤ C(w)α in the semilattice P_{A}/α. Similarly C(w)α ≤ C(z)α, and we have shown that

wβJ zβ ⇒ C(w)α = C(z)α.

The converse is also true, and we prove it in two steps. Firstly, since B is a band, a^{2} = a and abJ ba for all a, b in B, so that, clearly, if w ∈ F_{A} and C(w) = {a_{1}, a_{2}, …, a_{n}}, then wβ J (a_{1}a_{2} … a_{n})β. Secondly note that if {a_{1}, a_{2}, …, a_{n}}, {b_{1}, b_{2}, …, b_{m}} ∈ P_{A} are connected by an elementary ρtransition, say

\begin{matrix}\{a_{1}, a_{2}, …, a_{n}\} = & \{c_{1}, …, c_{k}\} ∪ C(u_{i}) \\ → & \{c_{1}, …, c_{k}\} ∪ C(v_{i}) = \{b_{1}, b_{2}, …, b_{m}\}, \end{matrix}

then

a_{1}a_{2} … a_{n}β J c_{1} … c_{k}u_{i}β  (by the first step)
= c_{1} … c_{k}v_{i}β
J b_{1}b_{2} … b_{m}β  (by the first step).

We summarize the result in the following

Theorem 4.4. Let V ∈ LB_{0}. Let w, z ∈ F_{A}. Then wβJ zβ if and only if C(w)α = C(z)α. Thus the semilattice of the J -classes of B is isomorphic to the semilattice defined by the presentation SL(B) = ⟨A | u_{i} = v_{i}, i ∈ I⟩.

In particular, if B is the free band in V, then wβJ zβ if and only if C(w) = C(z). Before proving the formula for ϕ_{V} we give two corollaries.

Corollary 4.5. Suppose that a subset A_{0} of A generates a (relatively) free sub-band of B. Then A_{0} generates a free sub-semilattice of SL(B).

Proof. Let A_{1} and A_{2} be nonempty subsets of A_{0} such that A_{1}α = A_{2}α. We need to show that A_{1} = A_{2}. Choose w, z ∈ F_{A} such that C(w) = A_{1} and C(z) = A_{2}. Because A_{1}α = A_{2}α, wβJ zβ in B. By Corollary 4.2, wβJ zβ in the sub-band generated by A_{0}. Therefore C(w) = C(z) by hypothesis. □

Corollary 4.6. Let w ∈ F_{A}. Then there exists z ∈ F_{A} such that wβ = zβ and C(z) = m(C(w)α).

Proof. Choose y ∈ F_{A} such that C(y) = m(C(w)α). Then by Theorem 4.4, wβJ yβ. Hence we can find u, v ∈ F_{A}^{1} such that wβ = (uβ)(yβ)(vβ). The corollary is proved with z = uyv, since

m(C(w)α) = C(y) ⊆ C(uyv) ⊆ m(C(uyv)α) = m(C(w)α). = 0
□

Theorem 4.7. Suppose V ∈ LB_{0}. Then for m > k,

ϕ_{V}(m, k) = m − (k + min\{⌊\frac{m − k}{3} ⌋, k\}).

Proof. Let A be an alphabet of size m and let

S = ⟨A | u_{1} = v_{1}, u_{2} = v_{2}, …, u_{k} = v_{k}⟩_{V}.

We need to show that there exists a subset A_{0} of A of the correct size which generates a (relatively) free sub-band of S as a basis.

Case 1 Suppose that for each i, either

(a) |C(u_{i})| ≥ 2 and |C(v_{i})| ≥ 2, or

(b) |C(u_{i})| = 1 and C(u_{i}) ⊆ C(v_{i}). By Lemma 2.2, we can obtain a subset A′ of A such that |A′| ≤ k +

min\{⌊\frac{m−k}{3} ⌋, k\}, and if A_{0} = A \ A′, then u_{i}, v_{i} ∉ F_{A_{0}} for any i. Therefore, as none of the relations can be applied, A_{0} generates a (relatively) free sub-band of S as a basis. (Note that the hypothesis that V ∈ LB_{0} is used here.)

Case 2 Suppose that one of the relations, say u_{k} = v_{k}, is of the form a_{0} = w, where a_{0} ∈ A \ C(w). Let u′_{i} [v′_{i}] be the word in F_{A\{a_{0}}} obtained by substituting each occurence of a_{0} in u_{i} [v_{i}] by w. Then it is easy to check that

S ≃ ⟨A | u′_{1} = v′_{1}, u′_{2} = v′_{2}, …, u′_{k−1} = v′_{k−1}, a_{0} = w⟩_{V} ≃ S′,

where we put

S′ = ⟨A \ {a_{0}} | u′_{1} = v′_{1}, u′_{2} = v′_{2}, …, u′_{k−1} = v′_{k−1}⟩_{V},

and A_{0} ⊆ A \ {a_{0}} generates a (relatively) free sub-band of S as a basis if it generates a (relatively) free sub-band of S′ as a basis. Since

\begin{matrix}(m − 1) − ((k − 1)+ & min\{⌊\frac{(m − 1) − (k − 1)}{3}⌋, k − 1\}) \\ ≥ m − (k+ & min\{⌊\frac{m − k}{3} ⌋, k\}), \end{matrix}

the problem can be reduced to the first case.

Hence we have shown that

ϕ_{V}(m, k) ≥ m − (k + min\{⌊\frac{m − k}{3} ⌋, k\}).

Lemma 3.2 and Corollary 4.5 show that the reverse inequality also holds. So the proof is complete. □

Note that the proof of the theorem, combined with the procedure in Section 2, gives a procedure to obtain a subset of A of size ϕ_{V}(m, k) which generates a relatively free sub-band.

5. Structure of Finitely Presented Normal Bands. In this section we apply results in Section 4 to obtain complete information on the structure of a finitely presented normal band.

Let B be a normal band defined by the finite presentation

B = ⟨A | u_{i} = v_{i}, i ∈ I⟩.

Let the congruences β and α be as defined in Section 4.

For each element τ of P_{A}/α, denote the J -class corresponding to τ by J_{τ}. By general theory, J_{τ} is a rectangular band. Let RB(τ) be the rectangular band defined by

RB(τ) = ⟨m(τ) | u_{i} = v_{i}, i ∈ I′⟩,

where i ∈ I′ if and only if u_{i}, v_{i} ∈ F_{m(τ)}.

Theorem 5.1. For each α-class τ, J_{τ} ≃ RB(τ).

Proof. Let m(τ) = {a_{1}, …, a_{n}}. By Corollary 4.6, we can assume the elements of J_{τ} are wβ where C(w) = m(τ). Using the identity abcd = acbd for normal bands, we can see that J_{τ} is generated by b_{k} = a_{k}a_{1} ⋯ a_{n}a_{k}β, k = 1, …, n.

For each j ∈ I′, let u′_{j} [v′_{j}] be the element in J_{τ} obtained by substituting each occurence of a_{k} in u_{j} [v_{j}] by b_{k}. Then

\begin{aligned}u′_{j} = i(u_{j})a_{1} ⋯ a_{n}t(u_{j})β & = u_{j}a_{1} ⋯ a_{n}u_{j}β \\ & = v_{j}a_{1} ⋯ a_{n}v_{j}β = i(v_{j})a_{1} ⋯ a_{n}t(v_{j})β = v′_{j}. \end{aligned}

Hence, by the universal property of RB(τ), there is a homomorphism φ from RB(τ) onto J_{τ}, mapping a_{k} to b_{k} for k = 1, …, n.

On the other hand, the map wβ ↦ w defines a homomorphism from J_{τ} to RB(τ) which is clearly inverse to φ. For this map to be well-defined we need to check that if w, z ∈ F_{m(τ)} and wβ = zβ, then w = z in RB(τ). This is true because the identities for normal bands are satisfied by rectangular bands, and the only relations u_{i} = v_{i} applicable here are when i ∈ I′. □

If τ_{1}, τ_{2} are two α-classes, τ_{1} ≥ τ_{2}, then m(τ_{1}) ⊆ m(τ_{2}), and there is a homomorphism φ_{τ_{1},τ_{2}} : RB(τ_{1}) → RB(τ_{2}) defined by a ↦ a for each a ∈ m(τ_{1}).

Thus we have expressed B as a strong semilattice of finitely presented rectangular bands. Corollary 4.5 can be improved in the case of normal bands.

Proposition 5.2. A subset A_{0} of A generates a free normal sub-band of B if and only if A_{0} generates a free sub-semilattice of SL(B), and A_{0} generates a free sub-rectangular band of RB(A_{0}α).

Proof. We shall only prove the ‘if’ part, the other direction being obvious.

If A_{0} generates a free sub-semilattice in SL(B), then Theorem 4.4 shows that for w, z ∈ F_{A_{0}}, wβ = zβ only if C(w) = C(z).

It remains to show that if C(w) = C(z) = A′ ⊆ A_{0}, then wβ = zβ if and only if the words w and z are equal in the free rectangular band over A′. Since A′ ⊆ A_{0}, m(A′α) ⊆ m(A_{0}α). The homomorphism φ_{A′α,A_{0}α} maps a to a for each a ∈ A′. The hypothesis implies that A′ generates a free rectangular band in RB(A_{0}α), and hence in RB(A′α). □

The results in this section are valid for left [right] normal bands by replacing ‘rectangular bands’ with ‘left [right] zero semigroups’.

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School of Mathematics and Statistics, University of Sydney, N.S.W. 2006, Australia.