CC BY
10URAL MATHEMATICAL JOURNAL, Vol. 3, No. 1, 2017
FINITE NILSEMIGROUPS WITH MODULAR CONGRUENCE LATTICES
Alexander L. Popovich
Ural Federal University, Ekaterinburg, Russia, [email protected]
Abstract: This paper continues the joint work [2] of the author with P. Jones. We describe all finitely generated nilsemigroups with modular congruence lattices: there are 91 countable series of such semigroups. For finitely generated nilsemigroups a simple algorithmic test to the congruence modularity is obtained.
Key words: Semigroup, Nilsemigroup, Congruence lattice.
Introduction
In [2] the characterization of nilsemigroups with distributive and modular congruence lattices had been obtained. The basic notion in that result was the width of a semigroup, considered as a poset under division. Recall that the width of a poset is the maximal integer n such that the poset contains an antichain of n elements. It was proved in [2] that the congruence lattice of a nilsemigroup is distributive [modular and not distributive] if and only if it has the width 1 [the width 2].
A poset of the width 1 is a chain. Semigroups, whose congruence lattice form a chain, were investigated in the works [1, 3, 4]. There is no complete classification for such semigroups, in the same time some important cases (finite semigroups, commutative semigroups, permutative semigroups) were considered. It is known that finitely generated nilsemigroups whose congruence lattices form a chain are cyclic nilsemigroups. Thus we have a description of finitely generated nilsemigroups with distributive congruence lattices.
In this paper we describe all finitely generated nilsemigroups with the modular congruence lattice up to isomorphism or dual isomorphism. The set of all such semigroups has been splited into series (almost all of them are infinite), each of them has 4 or less natural parameters. The list of all series is given in the table below.
We prove the following theorem:
Theorem 1. Let S be a finitely generated nilsemigroup. Then the following are equivalent:
a) Con S is modular and not distributive;
b) S is (generated by two elements a and b and the poset {a2,ab,ba,b2} under division has the width 2;
c) S is isomorphic or dually isomorphic to a suitable semigroup in the following table:
N Name Presentation Restrictions
1 A(n) a2 = ab = ba = b2, an = 0 n ^ 2
2 Bi(n) a2 = ab = b2, an = 0 n ^ 3
3 B2.i(m,n) a2 = b2, ab = ba, am = ban = 0 m ^ 3, n ^ 2, |m — n| = 1
4 B2.2(m,n) a2 = b2, ab = ba, am = bam-l, an = 0 n ^ m ^ 3
5 B3.i(m,n) a2 = ab = ba, am = bn n > m ^ 3
6 B3.2(m,n) a2 = ab = ba, am = bn = 0 n,m ^ 3, n ^ m — 1, n = m
7 B3.3(m,k) a2 = ab = ba, am = bm, ak = 0 k ^ m ^ 3
8 B4.l(m, n) a2 = ab, b2 = ba, am = bn = 0 lm — nl = 1; m,n ^ 3
9 B4.2(m,n) a2 = ab, b2 = ba, am = bm, ak = 0 k ^ m ^ 3
10 Ci a2 = ab, b2 = ba = 0
11 C2 a2 = b2 = ab, ba = 0
12 C3 a2 = b2 = ab = 0
13 C4 a2 = b2, ab = ba = 0
14 C5 a2 = ab = ba, b2 = 0
15 Co ab = ba, a2 = b2 = 0
16 C7.1 (n) a2 = ab = ba = bn n ^ 3
17 C7.2 (n) a2 = ab = ba = bn = 0 n ^ 3
18 .Di.i(m,n) b2 = ba = am = anb m ^ 3, m — 1 ^ n ^ 2
19 Di.2(m,n) b2 = ba = am, anb = 0 m ^ n ^ 2, m ^ 3
20 D2.i(m, n, k) ab = ba = am, an = bk = 0 n > m ^ 3, k ^ 3, n ^ k(m —1) + 1
21 D2.2(m, n, k) ab = ba = am, an = bk = 0, an-:L = bk- m,k ^ 3, m ^ n — 2 ^ (k — 1)(m — 1), n = (m — 1)(k — 1) + 1
22 D2.3(m,n,q) ab = ba = am, a(m-i)q = bq, an = 0 m ^ 3,q ^ 2, n ^ (m — 1)q + 1
23 D3.i(m,n,k) ab = ba, b2 = am, an = akb = 0 m ^ 3, k ^ 2, k + m ^ n ^ k, n ^ m + 1
24 D3.2(n,k,q) ab = ba, b2 = a2(n-k), an = akb = 0, aq+n-k = aqb n ^ 3, k ^ 2, n ^ k, n ^ 2n — 2k + 1, k > q ^ 2
25 D3.3(m,n,k,q) ab = ba, b2 = am, an-k+q = aqb, an = ak b = 0 m ^ 3, k ^ 2, k + m ^ n ^ k, k ^ min(n—k+q, q+m), k > q ^ 2
26 D4(n) ba = bn, a2 = ab n ^ 3
27 Ei.i(m, n, k) b2 = ba = amb, an = akb = 0 m ^ 2, 2m ^ k ^ m, n ^ k, n ^ 3
28 Ei.2(m,n,k) b2 = ba = amb, an = akb m ^ 2, 2m ^ k ^ m, n > k, n ^ 3
29 E2.i(m) a2 = b2 = (ab) f = (ba) = 0 m ^ 3
30 E2.2(m) a2 = b2 = (ab) t = 0 m ^ 3
31 E2.3(m) a2 = b2 = (ab) , (ba) t = 0 m ^ 3
32 E2.4(m) a2 = b2 = (ab) t = (ba) t , (ba) = 0 m ^ 3
33 E2.5(m) a2 = b2 = (ab) "2, (ba) "2 = 0 m ^ 3, m is odd
34 E2.o(m) a2 = b2 = (ab) "2, (ab)= x , X " + 1 (ba)~ = 0 m ^ 3
N Name Presentation Restrictions
35 #3.i(n,m) ab = ba = an = bm n,m ^ 3
36 £3.2 (n,m) ab = ba = an = bm = 0 n, m ^ 3
37 E4 a2 = b2, ba = 0
38 E5A(m,n,k) ab = ba, b2 = amb, an = akb = 0 n ^ k ^ m ^ 2
39 E5.2(m,n,k,q) ab = ba, b2 = amb, an-k+q = aqb, n ^ k ^ m ^ 2, n ^ 3, n — k = m,
ak = an = 0 k ^ min(n — k + q,q + m), q ^ 2
40 E5.3(m,n,q) ab = a9 b ba, b2 = amb, an = 0, am+q = n ^ m ^ 2, n ^ 3, q ^ 2
41 E6.1 a2 = ab, b2 = ba2
42 E6.2 a2 = ab, b2 = 0
43 Ez(n) a2 = ab, ba = 0, bn = 0 n ^ 3
44 G(m, n) ab = am, ba = bn m, n ^ 3
45 H1.1(m, n, k) ba = amb, b2 = an, an+1 = ak b = 0 n ^ k > m ^ 2
46 H12(m,n,k) ba = an+1 = amb, b2 = an = ak-1b, = ak b = 0 n ^ k > m ^ 2
47 H2.1(m, n, k, l) b2 = bam, ab = an, ak = ba' = 0 m ^ 2, k ^ l > k > n > m + 1, m m + n ^
48 H2.2(m,n,k,l) b2 = bam, ab = an, ak-1 = ba'-1, m ^ 2, k > n > m + 1, m + n ^
ak = ba' = 0 k ^ l > m, k = l + n — 1
49 H2.3(m,n,k,q) b2 = bam, ab = an, aq+n-1 = baq, m ^ 2, n > m + 1, m + n ^ k ^
ak = ba' = 0 q + n — 1
50 H2.4(m, k, l) b2 = bam, ab = am+1, ak = ba' = 0 m ^ 2, k ^ l ^ m + 1
51 H2.5(m,k,l) b2 = bam, ab = am+1, ak-1 = m ^ 2 k > m + 1, k ^ l > m,
ba1 1, ak = ba1 = 0 k = l + m
52 H2.6(m,k,q) b2 = ak = bam, ab = am+1, aq+m = baq, 0 m ^ 2, n> m, k ^ q + n — 1, q > 2
53 In(n) a2 = (ab) , b2 = (ba) "2 n ^ 2
54 11.2 (n) a2 = (ba) b2 = (ab) 222ti n ^ 1
55 I1.3(n) a2 = (ba) b2 = (ab) 2n+2 2n+2 (ab) 2 = (ba) 2 = 0 n ^ 1
56 I1.4(n, m, k) a2 = (ab) 2s2ti, b2 = (ba) n . . k 2 = (ba) 2 = 0 n ^ 1, k,m > 2n + 1, \k - m\ ^ 1
(ab)
57 J1.1(n,m) ab = ba = am, b2 = an m ^ 3, 2m — 2 > n > m
58 J1.2(n,m) ab = ba = am, b2 = an = 0 m ^ 3, 2m — 2 > n > m
59 J1.3(m,k) ab = ba = am, b2 = a2m-2, ak = 0 m ^ 3, k > 2m — 2
60 J2(n) b2 = an, ab = ba = an+1 = 0 n ^ 3
61 J3(n) b2 = ab = an, ba = an+1 = 0 n ^ 3
62 J4(n) ab = an, b2 = ba = an+1 = 0 n ^ 3
63 L1(n, m) b2 = am, ban = 0, ab = 0 m ^ 3, m + 1^ n^ 2
64 L2.1(n,m) ba = am, bn = ab = 0 m ^ 3, n ^ 3
65 L2.2(n,m) ba = am = bn-1, am+1 = bn = 0 m ^ 3, n ^ 4
66 L3.1(m,k,n) ab = am, b2 = ak, ban+1 = 0 k > m k — m ^ 3, k = 2m — 2, k ^ n ^
67 L3.2(m,k,q) ab = am, b2 = ak = baq k>m ^ 3, k = 2m — 2 k > q ^
k — m + 1
N Name Presentation Restrictions
68 L3.3(m,k,q) ab = am, b2 = ak, aq+m-1 = baq k > m ^ 3, k = 2m—2, k — m+1 > q ^ k — 2m + 2
69 L3A(m,n,l) ab = am, b2 = a2m-2, bal = an = 0 m ^ 3, l ^ m — 1, l + m ^ n ^ 2m — 1
70 L3.5(m,n,l) ab = am, b2 = a2m-2, bal = an = 0, an-1 = bal-1 m ^ 3, l ^ m — 1, l + m ^ n ^ 2m — 1
71 L3.6(m,n,q) ab = am, b2 = ba2m-2, aq+m-1 = baq, an = 0 m ^ 2, q ^ 2, n ^ q + m
72 L3.7(m, l, k, n) ab = am, b2 = bal, an = bak = 0 2m — 1 ^ n ^ k > l ^ m ^ 3
73 L3.8(m, l, k, n) ab = am, b2 = bal, an = bak = 0, an-1 = bal-1 2m — 1 ^ n ^ k > l ^ m ^ 3
74 L3.g(m,l,q,n) ab = am, b2 = bal, an = 0, baq+m-1 = baq 2m — 1 ^ n > l ^ m ^ 3, q ^ 2
75 L3.io(m,n, k) ab = am, b2 = 0, an = bak = 0 n> m ^ 3, m + k ^ n ^ k ^ 2
76 L3.ii(m,n,k) ab = am, b2 = 0, an = bak = 0, an-1 = bak-1 n> m ^ 3, m + k ^ n ^ k ^ 2
77 L3.i2(m,n,q) ab = am, b2 = 0, an = 0, aq+m-1 = baq n> m ^ 3, q ^ 2
78 N1.1(m, l, n, k) b2 = amb, ba = alb, an = akb = 0 n ^ k ^ l > m ^ 2, m + l ^ k, 2m > l
79 N1.2(m, l, n, k) b2 = amb, ba = alb, an = akb = 0, an-1 = ak-1b n ^ k ^ l > m ^ 2, m + l ^ k, 2m > l
80 N2.1(m, l, n, k) ba = amb, b2 = alb, an = akb = 0 n ^ k ^ l > m ^ 2, m +1 ^ k
81 N2.2(m,l,n,k) ba = amb, b2 = akb, an = alb = 0, an-1 = ak-1b n ^ k ^ l > m ^ 2, m +1 ^ k
82 N3.1 (m,k) a2 = (ab) 2 , b2 = (ab) 2 k > 2m + 1, m > 1
83 N3.2 (m,k) a2 = (ab) 2 , b2 = (ab) 2, b2a = ab2 = b3 = 0 k > 2m, m > 1
84 N3.3 (m,k) a2 = (ab) 2 , b2 = (ab) 2 = (ba) 2 k > 2m, m > 1
85 N3.4 (m,k) a2 = (ab) 2 , b2 = (ab) 2, ,, \ k (ba)2 = 0 k > 2m, m > 1
86 N3.5(m,k,n,l) a2 = (ab) 2 , b2 = (ba) 2 n 1 (ab) 2 = (ba) 2 = 0 k > m, m > 1, n, l ^ k, |n — l| ^ 1
87 N3.e(m,n,l) a2 = (ab) 2 , b2 = (ab) 2 = (ba)2 = 0 m > 1, n, l ^ m, |n — l| ^ 1
88 N3.7 (m) a2 = (ba) m, b2 = (ba) ^f1 m ^ 3
89 N3.8 (m) a2 = (ab)m, b2 = 0 m ^ 3
90 N3.9 (m) a2 = (ab)m = (ba)m, b2 = 0 n ^ 3
91 N4(m, n) ab = an = bm, ba = 0 m, n ^ 3
We show later that every row in this table, with some constants fixed, gives us exactly one semigroup up to isomorphism or dually isomorphism. Some rows have no parameters, which means that such rows defines only one finite semigroup.
Let us note that any two semigroups in this table are not isomorphic and are not dually isomorphic. Indeed, every nilsemigroup has exactly one basis, i.e. a minimal set of generators. Every generator is a maximal element under division order Conversely, every maximal element of (S, is an element of any basis. So, the set of maximal elements is the unique basis of S. Then
every automorphism of S maps the basis onto itself, which means that it preserves the presentation of S. All semigroups in the table have distinct presentations, that can be revised by a careful check.
It is easy to check that all semigroups in this table have a width 2. It gives us the implication from c) to a). The implication from a) to b) is proved in [2]. The rest of the paper is directed to prove that b) leads c).
Theorem 1 provides a simple test to determine whether the congruence lattice of a finite nilsemi-group is modular by checking the condition (b) or by searching the corresponding semigroup in the Table.
Theorem 1 has an important corollary for the class of nilpotent semigroups. Every nilpotent semigroup S satisfy the ascending chain condition under Then S has a basis, which consists of maximal elements of S under This basis form an antichain, so by result of [2], it has 1 or 2 elements. From Theorem 1 we have the following corollary:
Corollary 1. Every nilpotent semigroup with modular congruence lattice is finite. It is isomorphic or dually isomorphic to a suitable semigroup in the Table.
1. Preliminaries
We consider the division relation ^ on a semigroup S defined as a ^ b iff there exist s,t £ S1 such that b = sat. Since every nilsemigroup is J-trivial, the relation ^ is an order relation on a nilsemigroup.
Our starting point is the following statements that was proved in [1] as Corollary 2.
Proposition 1. Let S be a nilsemigroup such that Con S is modular. If S is finitely generated, then it is finite. If S is not cyclic, then it is (generated by two elements a, b and the poset {a2,ab,ba,b2} has width at most two.
We assume further in the paper that S is a finite nilsemigroup generated by two distinct elements a and b.
We say that an element x £ S is an atom, if x covers 0, i.e. x > 0 and, for every z £ S, the condition 0 < z ^ x implies z = x. Put
x > y iff there exist s,t £ S1 such that y = sxt and st = 1.
The relation > on S is antisymmetric and transitive. It is easy to see that, for x,y £ S, x > y implies x > y, and x > y implies x ^ y (the converse is false, since 0 > 0, but 0 > 0).
Lemma 1. 1) An element x £ S is equal to zero if and only if x > x.
2) For every s £ S either s = s'a or s = s'b for some s' £ S1.
3) For every t £ S either t = at' or t = bt' for some t' £ S1.
4) If x £ S satisfies xa = ax = xb = bx = 0, then x is an atom or a zero.
The proof is obvious.
P
Let u be a word of n letters. Define un for 0 ^ p ^ n — 1 as a p-element prefix of u. For an
P r j -i p mod n
arbitrary positive integer p, put un = u[p/n]u n .
Lemma 2. Let c,d be letters and let p be a positive integer. Then:
1) (cd) 2 = c(dc) t .
2) (cd) 2 = (cd) c, if p is odd.
3) (cd) 2 = (cd) d, if p is even.
4) (cd) 2 (dc) 2 = (cd) V', if p is odd.
5) (cd) 2 (cd) 2 = (cd) ^, if p is even.
The proof is obvious.
Lemma 3. 1) If a2 < ab, then either a2 < b2 or a2 < ba.
2) If ba < ab, then either ba < a2 or ba < b2.
3) If a2 > b2 > ab, ba > b2 and ab = 0, then ab < ba.
4) If a2 > ab > b2, ba > ab and b2 = 0, then b2 < ba.
5) If a2 > ab > ba and ba = 0, then b2 > ba.
Proof. 1) Let a2 < ab. Then a2 = sabt for some s,t G S1. If s = s'a for some s' G S1, then a2 < a2, which implies a2 = 0 < b2. If s = s'b for some s' G S1, then a2 = s'babt and a2 < ba. Let s = 1 and a2 = abt. If t = at' for some t' G S1, then a2 < ba. If t = bt' for some t' G S1, then a2 < b2.
2) The proof is similar to 1).
3) If a2 > b2, then b2 = sa2t for some s,t G S1. If s = s'b for some s' G S1, then b2 < ba, a contradiction. If t = bt' for some t' G S1, then b2 < ab, a contradiction. So b2 = ak for some k ^ 3. Then ab <b2 = ak, so ab = uakv for some u,v G S1. If u = u'a or v = av' for some u',v' G S1, then ab < ak+1 = ab2, i.e. ab = 0, a contradiction. If v = bv' for some v' G S1, then ab < ab and ab = 0. If u = u'b for some u' G S1 , then ba > ab.
4) If a2 > ab, then ab = sa2t for some s,t G S1. If s = s'b for some s' G S1, then ab < ba, a contradiction. If t = bt' for some t' G S1, then ab = 0, contrary to ab > b2. So ab = ak for some k ^ 3. Then b2 < ab = ak, so b2 = uakv for some u,v G S1. If u = u'a or v = av' for some u',v' G S1, then ab < ak+1 = aba, i.e. ab < b2, a contradiction. If v = bv' for some v' G S1, then b2 < akb = ab2 and b2 = 0, a contradiction. If u = u'b for some u' G S1, then ba> b2.
5) If a2 > ab, then ab = sa2t for some s,t G S1. If s = s'b for some s' G S1, then ab < ba, a contradiction. If t = bt' for some t' G S1, then ab = 0, contrary to ab > ba. So ab = ak for some k ^ 3. Then ba < ab = ak, so ba = uakv for some u,v G S1. If u = u'a or v = av' for some u',v' G S1, then ba ^ ak+1 = aba < ba, i.e. ba = 0, a contradiction. If v = bv' for some v' G S1, then ba ^ akb = ab2 < b2. If u = u'b for some u' G S1, then ba < ba, which means ba = 0, a contradiction. □
2. Finite nilsemigroups of width 2
The elements a2, ab, ba, b2 form a subposet of S. This subposet has no more than 4 elements and has no antichains with 3 elements. We enumerate all such posets in the following list.
B
1 I - V A II M N
D
E
F
G
H
K
AV
M
N
i >
A
O
Q
For each poset A-Q we examine all possibilities of mapping the set {a2,b2,ab,ba} onto the poset. We consider two cases be equal if one of them can be obtained from another either by replacing a to b and vice versa or by replacing ab to ba and vice versa. Indeed, these cases give us isomorphic or dually isomorphic semigroups. Some cases are forbidden by Lemma 3, we don't mention them.
I
Let us note that in cases B, D, F, G, H, I, K, M, O the elements a2,b2,ab,ba are not equal to zero, since every element of a nilsemigroup divides zero.
Series A. a2 = b2 = ab = ba. Then every element of S, except b, can be written as ap for some positive p. Let n be the least positive integer such that an = 0. Then S = A(n).
Series B. The following cases are possible:
a2 = b2 = ab ba a2 = b2 ab = ba
• • • •
B1 B2
a2 = ab = ba b2
a2 = ab b2 = ba
B3 B4
Case B1. Let x be an element of S. If a or b2 is a left divisor for x, then x = ap for some p. If ba2 is a left divisor for x, then x = ap for some p, since ba2 = b3. So, every element of S, except b and ba, can be written as ap for some p. Let n be the least positive integer such that an = 0. We obtain the semigroup B1(n).
Case B2. It is easy to show that every element can be written as ap or bap for some p ^ 0. Let n and l be the least positive integers such that an = ba1 = 0. Then In — ll ^ 1 and n ^ 3, l ^ 2. If In — ll = 1, then S = B2.1 (n, l).
Let am = bam-1 for some m ^ 3. Then ap = bap-1 for all p ^ m. Let n be the least positive integer such that an = 0. We obtain the semigroup B2.2(m,n).
Case B3. In this case every element can be written as abp-1 or bp for some p ^ 1. Let m be the least positive integer such that abm-1 = bn for some n ^ 3, m ^ 3 and n ^ m — 1. The following cases are possible:
Case B3.1 m = n. Then abm = a(abm-1) = abn, so abm = 0. The element abm-1 = bn is a single atom or a zero. Then S = B3.1(m,n) or S = B3.2(m,n) respectively.
Case B3.2. m = n. Then abp-1 = bp for all p ^ m. Let k be the least positive integer such that bk = 0. We have that S = B3.3(m, k).
Case B4. In this case every element can be written as abp-1 = ap or bp for some p ^ 0. Let m and n be the least positive integers such that am = bn. If m < n then am < bam = bm+1 ^ bn, so am = bn = 0 and n = m + 1. If m > n, then bn < abn = an+1 ^ am, so am = bn = 0 and m = n + 1. We got lm — nl = 1 and S = B4.1. If m = n, then ap = bp for all p ^ n. Let k be the least positive integer such that ak = 0. We deduce that S = B4.2(m, k).
Series C. The following cases are possible:
a2 = ab
b2 = ba C1
a2 = b2 = ab
ba C2
ba
a2 = b2 = ab
C3
"a2 = b2 ta2 = ab = ba fab = ba tb2
"ab = ba *b2 *a2 = b2 ^a2 = ab = ba
C4 C5 C6 C7
Case C1. Since b2 < a2, we have b2 = sa2t for some s,t £ S1. If s = s'a for some s' £ S1, then ba = b2 = s'a3t = s'abat, which means that ba > ba, so b2 = 0. Cases s = s'b and t = at' for some s',t' £ S1 are similar. If t = bt' for some t' £ S1, then ba = sa2bt' = sa3t' = sabat', which implies ba > ba. So, b2 = ba = 0.
An element a2 is a single atom. Indeed, a2b = a3 = aba = 0 and ba2 = 0. Then S = C1.
Case C2. Since ba < a2, then ba = sa2t for some s,t £ SIf s = s'a for some s' £ S\ then ba = s'a3t = s'abat, which impies ba = 0. Cases s = s'b, t = bt' and t = at' for some s', t' £ S1 are similar. So, ba = 0.
An element a2 is a single atom. Indeed, a2b = a3 = aba = 0 and ba2 = 0. Then S = C2.
Case C3. By the same arguments as before, we have a2 = b2 = ab = 0. The element ba is a single atom. Then S = C3.
Case C4. Using arguments of case C1, we have ab = ba = 0. The element a2 = b2 is a single atom. Then S = C4.
Case C5. Using arguments of case C2, we have b2 = 0. The element a2 = ab = ba is a single atom. Then S = C5.
Case C6. Using arguments of case C2, we have a2 = b2 = 0. The element ab = ba is a single atom. Then S = C6.
Case C7. We have a2 = ab = ba = bn for some n ^ 3. The element a2 is an atom or a zero, since ab2 = ba2 = a2b = a3 = abk < ab2. If a2 is an atom, then S = C7.1(n). If a2 is a zero, then S = C7.2 (n).
Series D. The following cases are possible:
• ab
ta
.b2
tb2
b2 = ba ab = ba fb2 • ab = ba ba •a2
D1 D2 D3 D4
ab = ba fb2 = ba b = ba fa2 = b2
a2 -b2 ab •a2 a2 • ab ab »ba
D5 D6 D7 D8
Case D1. We have b2 < a2, so b2 = ba = sa2t for some s,t £ S1. If s = s'b or t = bt' for some s', t' £ S1, then ba < ab or ba < ba, a contradiction. So, b2 = ba = am for some m ^ 3. The element am+1 = ba2 = b2a = bam = a2m-1 with m = 2 divides itself, which implies that it is a zero. The elements bab = b3 = b2a and aba = am+1 are also equal to zero, which means that ba is an atom. The element am-1b is an atom or a zero.
Every element of S can be written as ap or ap-1 b for some p ^ m. Let q ^ 1 and 1 < n < m be the least positive integers such that aq = anb. If q < m, then am = aqam-q = anbam-q = anamam-q-1 < am, so am = ba = 0, a contradiction. If q = m, then an+1b = am+1 = 0, so am is a single atom and S = D1.1(m,n). If q > m, then anb = 0 and S = D1.2(m,n).
Case D2. We have ab < a2, so ab = ba = sa2t for some s,t £ S1. If s = s'b or t = bt' for some s',t' £ S1, then ba < ab, a contradiction. So, ab = ba = am for some m ^ 3. Then every element of S can be written as ap or bp for some p. Let n and k be the least positive integers such that an = 0 and bk = 0. Then n ^ k(m - 1) + 1 and k ^ 3.
If ap = bq implies ap = 0, then S = D2.1(m,n, k). Let p,q be the least positive integers such that ap = bq = 0. Then ap+1 = bqa = a(m-1)q+1. If p = (m - 1)q, then ap+1 = 0 and p + 1 = n, q + 1 = k. In this case S = D2.2(m,n,k). If p = (m — 1)q, then ar(m-1) = br for all r ^ p, so k = [n/(m — 1)] and S = D233(m,n, q).
Case D3. We have b2 = am for some m ^ 3. Every element of S can be written in the form ap or apb for some p. Let n be the least positive integer such that an = 0 and let k be the least positive integer such that akb = 0. Since akb2 = ak+m, we have k + m ^ n ^ k.
2
2
a
a
If ap = aqb for some p,q implies ap = 0, then S = D31(m,n,k). Let p,q be the least positive integers such that ap = aqb = 0. Then ap+r = aq+rb for all r ^ 0, which implies n — p = k — q, so p = n — k + q. We have apb = aqb2 = aq+m, so either q + m — p = p — q or apb = 0. In the former case p = q + m/2 and m — 2(n — k), whence S = D32(n, k, q). In the latter case k ^ min(p, q + m) and S = D3.3(m, n, k, q).
Case D4. We have ba = bn for some n ^ 3. Then bba = bn+1 = bab = baa = bna = b2n_1 <bn+1, so bba = bab = baa = 0. Also a3 = aba = abn = an+1 < a3, so aba = 0. We got that ba is an atom. The element a2 is also an atom, because ba2 = 0, a3 = a2b = 0. Elements of S are equal to a, or to a2, or to b for i = 1... n. We got a semigroup D4(n).
Case D5. We have a2 < ab = ba, so a2 = sabt for some s,t G S1. If s = s'a or t = at' for some s',t' G S1, then a2 < a2 and a2 = 0 < b2, a contradiction. If s = s'b or t = bi' for some s',t' G S1, then a2 < b2, a contradiction.
Case D6. We have ab < a2 = b2, so ab = sa2t for some s, t G S1. If s = s'a or t = bt' for some s', t' G S1, then ab < ab and ab = 0 < ba, a contradiction. If s = s'b or t = at' for some s', t' G S1, then ab < ba, a contradiction.
Case D7. We have a2 < b2, so a2 = sb2t for some s,t G S1. If s = s'a for some s' G S1, then a2 < ab, a contradiction. If s = s'b for some s' G S1, then a2 = s'b3t = s'babt < ab, a contradiction. Let s = 1. If t = at' or t = bt' for some t' G S1, then a2 = b2at' = b3t' = babt' < ab, a contradiction.
Case D8. We have ab < a2, so ab = sa2t for some s,t G S1. If s = s'a or t = at' for some s',t' G S1, then ab ^ a3 = ab2 < ab, a contradiction. If s = s'b or t = bt' for some s',t' G S1, then ab ^ b3 = a2b < ab, a contradiction.
Series E. The following cases are possible:
»ab
b2 = ba
ab
El
ba
a2 = b2
E2
b2
hb = ba E3
ab
= b2
2
ab = ba ba
= ab b2
= ab
E7
E4 E5 E6
Case E1. We have b2 < ab, so b2 = sabt for some s,t £ S1. If s = s'b or t = at' or t = bt' for some s',t' £ S1, then b2 < ba = b2, which implies b2 = 0. Anyway, there exists m ^ 2 such that b2 = ba = amb. Then every element of the semigroup S can be written as ap or apb for some p.
Let k and n be the minimal numbers such that akb = an = 0. Obviously, n ^ k and k ^ m. We have a2mb = amb2 = b3 = b2a = bamb = b2am-1b. Since m = 1, the element b2a divides itself, which means a2mb = 0 and k ^ 2m.
If aq = arb implies aq = 0, then S = ^1.1(m, n, k). Let aq = arb = 0 for some q > r > 0. Then ar+1b = aq+1 = arba = ar+mb < ar+1b, so aq+1 = ar+1b = 0, which means q = n — 1 and r = k — 1. Then S = E1.2(m,n,k).
Case E2. We have a2 < ab and a2 < ba, so a2 = b2 = (ab)m/2 or a2 = b2 = (ba)m/2 for some m ^ 3. Without loss of generality we suppose that a2 = b2 = (ab)m/2. Then every element of S can be written in the form (ab)p/2 or in the form (ba)p/2 for some p. The following cases are possible:
Case E2.1. m = 2n + 1 for some n ^ 1, so a2 = b2 = (ab) 2+ . Then a3 = (ab) 2+ a = (ab)~a2 = (ab)~b2 = (ab)^^b3 = (ab)^^a3b, so a3 = 0. Therefore a2b = b3 = ba2, which means
2n+2 2n+2 2n+3 2n+2 2n+2 7 \ 2n 2 , , ,
(ab) 2 = (ba) 2 . Then (ab) 2 = (ab) 2 a = (ba) 2 a = b(ab) 2 a2 = 0 and, analogously,
2
2
a
a
2
2
2
a
a
a
2
2
(ba) n2+ = 0. So, a2b is an atom or a zero. If a2 = a2b = (ba) ™ = 0, then S = E2.1(2n + 1). If a2 = 0 and (ba)h = 0, then S = E2^(2n + 1). If a2 = 0 and (ba)h = 0, then S = E2,3(2n + 1). If a2 = (ba)h = 0 and a2b = 0, then S = E2A(2n + 1). If a2 = (ba)h and a2b = 0, then S = E2.5(2n + 1). If a2 = 0, (ba)f =0, a2 = (ba)f and a2b = 0, then S = E2£(2n + 1).
Case E2.2. m = 2n for some n ^ 2, so a2 = b2 = (ab)Then a3 = a(ab)~ = a2(ba) 2 =
2n_2 2n_3 2n
b3(ab) = ba3(ba) , so a3 = 0. From here we obtain ba2 = b3 = a2b = (ab) ~ b =
, . 2n — 2 2 , n 2n — 2 3 , , i2 r 2 /i \ h
(ab) 2 ab2 = (ab) 2 a3 = 0, which means that a2 is an atom or a zero. If a2 = (ba) 2 = 0, then S = E2.i(2n). If a2 =0 and (ba)h = 0, then S = E2.2(2n). If a2 = 0 and (ba)h = 0, then S = E2.3(2n). If a2 = (ba)h = 0, then S = E2A(2n). If a2 = 0, (ba)h = 0 and a2 = (ba)h, then S = E2.6(2n).
Case E3. We have ab < a2, whence ab = sa2t for some s, t £ S1. If s = s'b or t = bt' for some s', t' £ S1, then ab < ab or ba < ba, which means ab = ba = 0. Anyway, ab = am for some m ^ 3. Analogously, ab = bn for some n ^ 3. Every element of the semigroup S can be written in the form ap or bp for some p. Let n be the least positive integer such that ab = ba = an and m be the least positive integer such that ab = ba = bm. Then aba = an+1 = abm = anbm-1 = a2n-1bm-2 < an+1, so aba = 0. By the same arguments, abb = 0, which means that ab is either an atom or a zero. If ab is an atom, S = E3.1(n, m). If ab is a zero, S = E3.2(n, m).
Case E4. We have ba < a2 = b2, so ba = sa2t for some s,t £ S1. If s = s'a or t = at' for some s', t' £ S1, then ba ^ a3 = b2a < ba. If s = s'b or t = bt' for some s', t' £ S1, then ba ^ b3 = ba2 < ba, which implies ba = 0. Now we have ab2 = a3 = b2a = 0, a2b = b3 = ba2 = 0, aba = bab = 0, so the semigroup S consists only of five elements and S = E4.
Case E5. We have b2 < ab = ba, so b2 = sabt for some s,t £ S1. If s = s'b or t = bt' for some s',t' £ S1, then b2 < b2, which means b2 = 0. Anyway, there exists m such that b2 = amb. Let m be the minimal integer with such a property.
Every element of S can be written as ap or apb for a suitable p. Let n and k be minimal positive integers such that an = 0 and akb = 0.
If ap = aqb implies ap = 0 for some p, q, then S = E5.1(m, n, k).
Let p, q be the least positive integers such that ap = aqb = 0. Then p < n, q < k and ap+r = aq+rb for all r ^ 0, so n — p = k — q, which means p = n — k + q. We have an-k+q = aqb, so an-k+qb = aq+mb. If n — k = m, then an-k+qb = aq+mb = 0, so k ^ min(n — k + q,q + m). In this case S = E5.2(m, n,k,q). If n — k = m, then S = E5.3(m, n, q).
Case E6. Since a2 = ab, every element of S can be written as ap or bqap for some p. Then b2 = 0 or b2 = ban for some n. If n ^ 3, then b2 < a3 = ab2 < b2, which implies b2 = 0. So,
b2 = ba2 = 0 or b2 = 0.
Let b2 = ba2 = 0. Then a3 = ab2 = aba2 = a4, so a3 = 0. Thus, b2a = ba3 = 0, b3 = ba2b = ba3 = 0 and ab2 = a3 = 0, so b2 is an atom. Then S = E6.1. If b2 = 0, then a3 = 0. Hence a2 and ba are atoms and S = E6.2.
Case E7. Using the same arguments as in case E6, for some m ^ 2, we have ba = amb < a3 = aba < ba, so ba = 0. Let n be the index of b. Then S consist of elements a, a2, b,b2,..., bn-1,0 and is isomorphic to E7(n).
Series F. The following cases are possible:
.b2 = ba
• ab
ab
a2 = b2
ba
Fl
F2
ab = ba b2
F3
ab = ba ba
a2 = ab
F4 F5
Case Fl. We have a2 < b2 = ba, but a2 < ab. So, a2 = 0 and a2 = bka for some k ^ 2. These arguments are true for ab, so ab = bla for some l ^ 2. Then a2 ^ ab or a2 ^ ab, a contradiction.
Case F2. ab < a2, so ab = sa2t for some s,t G S1. If s = s'b for s' G S1, then ab < ba. If t = bt' for t' G S1, then ab < ab. If s = s'a or t = at' for some s', t' G S1, then ab ^ a3 = ab2 < ab2. All the possibilities lead to a contradiction. Cases F3-F5 are analogous to Fl or F2.
Series G. Only one case is possible:
a2 ab
b2 ba
G1
Case G1. We have ab < a2, so ab = am for some m ^ 3. Similarly, ba = bn for some n ^ 3. Then am+1 = a2b = aba = abn = ambn-1 = a2m-1bn-2 < am+1, so am+1 = amb = bam = 0. Similarly, bn+1 = bab = b2a = 0. So, ab and ba are atoms.
Every element of the semigroup S can be written as ap or bp for a suitable p. Let aq = br for some q < m and r ^ n. Then aq+1 = bra = 0 and am = 0, a contradiction. If q = m and r = n, we have ab = ba, a contradiction. We obtain that S = G1(m,n).
Series H. The following cases are possible:
a b2
ab ba
a2 ab
ba b2
a2 ab
b2 ba
Hl
H2
H3
Case Hl. We have b2 < a2, so b2 = an for n ^ 3. Since ba < ab and ba < b2, the equality ba = amb holds for some n > m ^ 2. Every element of the semigroup S can be written as ap or apb for some p. Now an+1 = b2a = bamb = am2 b2 = am2+n < an+1, so b2 = an+1 = 0, a contradiction. Therefore b2 = ban = amn = 0 and b2 is an atom. Let k be the least positive integer such that akb = 0. We have anb = b3 = 0, so m < k ^ n.
If the equality ap = aqb for some p ^ n and q ^ k implies ap = 0, then S = H1.1(m, n, k). Let ap = aqb for some p ^ n and q ^ k. Then aq+1b = ap+1 = aqba = aq+mb, so ap+1 = 0. If p < n, then b2 = 0, a contradiction. Let p = n and q > m. We have aq+1b = 0, so q = k — 1. We deduce S = H1.2(m,n,k).
Case H2. We have ab = an for n ^ 3 and b2 = bam for m ^ 2. Since b2 > ab, then n ^ m + 1. Every element of the semigroup S can be written as ap or bap for some p.
Let n > m+1. Then an+m = abam = ab2 = anb = a2n_1 < an+m, which implies an+m = 0. Let k and l be the minimal integers such that ak = 0 and bal = 0. Therefore n < k and m < l ^ k ^ m+n.
If ap = baq implies ap = 0, then S = H2.1(m,n,k,l). Let ap = baq = 0. Since ab < ba, then p > n. Therefore baq+1 = ap+1 = abaq = aq+n. If p + 1 = q + n, then ap+1 = 0. This implies
2
2
a
a
2
2
b
2
a
p + 1 = k and q + 1 = l, so S = H2.2(m, n,k,l). Let p + 1 = q + n. Then aq+n-l+r = baq+r for every r ^ 0, so l = k — n + 1 and k > q + n — 1. We obtain S = H2.3(m, n, k, q).
Let n = m + 1. Let k and l be the minimal integers such that ak = 0 and bal = 0. It is obvious that m < k — 1, m < l and k ^ l.
If ap = baq implies ap = 0, then S = H24(m, k, l). Let ap = baq = 0. Since ab < ba, we have p> m + 1. Therefore baq+1 = ap+1 = abaq = aq+m+1. If p + 1 = q + m + 1, then ap+1 = 0. This means p + 1 = k and q + 1 = l, so S = H2.5(m, k, l). Let p + 1 = q + n. Then aq+m+r = baq+r for every r ^ 0, so l = k — m and k > q + m, whence we get S = H2.6(m, k, q).
Case H3. We have ab = am for m ^ 3 and ba = an for n ^ 3. Then ab > ba or ba > ab, a contradiction.
Series I. The following cases are possible:
ab\y\ba a2 KN.b2 I1
a2 * * ab
b2 *+ba I2
a2 Mb2
ab **ba I3
a2 * * ba ab KN.b2 I4
Case I1. We have a2 < ab, a2 < ab, but a2 < b2, which means that a2 = (ab)l/2 or a2 = (ba)l/2 for some l ^ 3. We suppose without loss of generality that a2 = (ab)l/2. Then b2 = (ba)l/2. Every element can be written in the form (ab)p or (ba)p for some p. Two cases are possible:
Case I1.1:
aa2 = a(ab) 2 = a2(ba) 2 = (ab) 2 (ba) 2 = (ab) 2 b2(ab) 2 = (ab) 2 (ba) 2 (ab) 2 =
2n ) 2
(ab) ^^ (ab) ~~ < (ab) ^^, so a3 = 0. Analogously, b3 = 0. Then a2b = (ab) ^b = (ab) ^^b2 = (ab) 2 (ba)~ = (ab) 2 < a3, so a2b = 0. Similarly, ba2 = ab2 = b2a = 0, which means that a2 and b2 are atoms. Hence 5 = I\.i(n).
= (ab) 2 and b2 = (ba) 2 for some n ^ 2. Then (ab) ^ =
2n- 1 , , , 2n , 2n- 1 " "
2n+ 1
aa
2n
a
2n-2
2n + 1
Case I1.2:
2n + 1 Q , . 2H+ 1 2W + 2 Q Q
= (oa) 2 and o2 = (ab) 2 for some n ^ 1. Then (ba) 2 = a2a = aa2 =
2n+2 2n+2 2n+2
(ab) 2 . It is easy to see that (ba) 2 = (ab) 2 is either an atom or a zero. If it is an atom, then S = Ii.2(n). If it is a zero, then S = /1.3.
Case I1.3: a2 = (ab)and b2 = (ba)~nr~ for some n ^ 1. Let m be the least positive integer such that (ab)t = 0 and k be the least positive integer such that (ba) 2 = 0. Then \m — k\ ^ 1 and m > 2n + 1, k > 2n + 1. Hence S = I1.4(n, m, k).
Case I2. We have b2 < ab and b2 < a2, so b2 = akb for some k ^ 3. By the same arguments ba = alb for some l ^ 3. Then b2 and ba are comparable, a contradiction. Cases I3 and I4 lead to a contradiction in a similar way.
Series J. The following cases are possible:
2n+1
2n+2
a2
ab = ba
b2 J1
a2
ab
b2 = ba J4
a b2
ab = ba J2
a2 = ab
b2
ba J5
a2
b2 = ab
ba J32
a2 = ab
ba
b2 J6
Case J1. We have ab = ba = am for some m ^ 3. Then b2 = an for m < n ^ 2m — 2. Hence an+i = ab2 = amb = a2m_1. If n < 2m — 2, then ab2 = 0 and b3 = 0. So, b2 is an atom or zero, which means that S = J1.1(n, m) or S = J1.2(n,m) respectively.
2
a
Let n = 2m — 2 and let k be the index of a. Then S = J3(m, k).
Case J2. We have b2 = an for some n ^ 3. Also, ab < b2, so ab = sb2t for some s,t £ SIf s = s'a or t = at' for s', t' £ Sthen ab ^ an+1 = ab2 < ab. Cases s = s'b or t = at' for s', t' £ S1 are similar. So, ab = ba = an+1 = 0 and S = J2(n).
Case J3. In this case 0 < b2 = ab < a2 implies that b2 = ab = an for some n ^ 3. Then an+1 = b2a < ba and anb = b3 = ban < ba. So, ba = an+1 = 0 and S = J3(n).
Case J4. We have ab = an for some n ^ 3. Then an+1 = aba < ba and anb = ab2 < b2 = ba. But ba < ab, so ba = 0 and S = J4(n).
Case J5. The inequality b2 < a2 implies that b2 = sa2t for some s,t £ S1. If s = s'b for s' £ S1, then b2 < ba, a contradiction. In other cases we have b2 < a3 = ab2, which implies b2 = 0 and b2 < ba, a contradiction.
Case J6. As in the previous case, ba ^ a3 = aba, whence ba = 0 and ba < b2, a contradiction.
Series K. All cases from this series are impossible by Lemma 3.
Series L.
L1 L2 L3
Case L1. We have b2 = am for some m ^ 3. If ab = ap for p > m, then ab = ap = ap-mb2 < ab. If ab = baq for q ^ m, then ab = baq = b3aq-m = ambaq-m < ab. Anyway, ab = 0. Every element of the semigroup S can be written in the form ap or bap for some p.
Since am+1 = ab2 = 0 and bam = b3 = amb = 0, the element b2 is an atom. Let n be the minimal integer such that ban = 0. Then ban-1 is an atom and n ^ m + 1. Every element of the semigroup S can be written in the form ap or bap for some p.
Let ap = baq for p,q ^ n. Then b2 = an = an-pap = an-pbaq = 0, a contradiction. So, S = L1(m,n).
Case L2. We have ba = am for some m ^ 3. Then am+1 = aba<ab, amb<ab, bam = a2m-1 <ab, but ab < ba, so ab = 0 and ba = am is an atom. Every element of the semigroup S can be written in the form ap or bp for some p. Let n be the least positive integer such that bn = 0.
If ap = bq for some p, q leads to ap = 0, then S = L2.1(m, n). Let ap = bq for p ^ m and q < k. Then ap+1 = abq = 0 = am+1, so p = m and q = n — 1. Then S = L2.2(m,n).
Case L3. ab = am for some m ^ 3. Since b2 < ba and b2 < ab = am, either b2 = an = 0 for some n ^ m + 1 or b2 = bal for some l ^ m or b2 = 0.
Case L3.1. b2 = ak = 0 and k = 2m — 2. Then b2a = afc+1 = ab2 = amb = a2m-1. Since k + 1 = 2m — 1, ak+1=0. Then b2 = 0, so b2 is an atom. Every element of the semigroup can be written in the form ap or bap for some p. Note that bak = b3 = akb = am+k-1 = 0. Let n be the least positive integer such that ban+1 = 0. Then k — m ^ n ^ k, since aba1 = al+m for all l.
If ap = baq implies ap = 0 for all p, q, then S = L3.1(m,k,n). Suppose that ap = baq = 0 for some p, q and let p, q be the least positive integers with such a property. Then k ^ p > q. We have ap+1 = abaq = aq+m, so either p = k or p = q + m — 1. If p = k, then ak = baq and ak+1 = aq+m, so q ^ k — m + 1 and S = L3.2(m, k,q). If p = q + m — 1, then ar+m—1 = bar for all r ^ q. But baq+m-1 = b2aq = 0, so aq+2m-1 = 0, which implies q ^ k — 2m + 2. We got S = L3.3(m, k, q).
Case L3.2. b2 = a2m-2 = 0. Every element of the semigroup can be written in the form ap or bap for some p. Then ba2m-2 = b3 = a2m-2b = a3m-3 and ba2m-2+r = a3m-3+r for all r ^ 1. Let
n be the least positive integer such that an = 0. Then n ^ 2m — 1, since b2 = 0. Let l be the least positive integer such that bal = 0. Then l ^ m — 1 and n ^ l + m.
If ap = baq implies ap = 0 for all p, q, then S = L34(m, n, l). Let ap = baq = 0 for some p, q and let p, q be the least positive integers with such a property. Then ap+1 = abaq = aq+m, so either ap+1 = aq+m = 0or p = q + m — 1. Let ap+1 = aq+m = 0. Then p + 1= n, q + 1 = l and q + m ^ n. Therefore S = L3.5(m, n, l).
Let p = q + m — 1. Then q ^ 2, n ^ q + m and we have S = L36(m,n,q).
Case L3.3. b2 = bal = 0 for some l. Since ab > b2, l ^ m. Every element of the semigroup S can be written either in the form ap or in the form bap for some p. Then a2m-1 = amb = ab2 = aba1 = al+m. Since l ^ m, a2m—1 = am+l = 0. Let n be the least positive integer such that an = 0 and k be the least positive integer such that bak = 0. Then l < k ^ n ^ 2m — 1.
If ap = baq for some p,q implies ap = baq = 0, then S = L37(m,l,k,n). Let p,q be the least positive integers such that ap = baq = 0. Obviously, q ^ 2. Then ap+1 = abaq = aq+m, so either ap+1 = aq+m = 0 or p = q + m — 1. In the former case we have p = n — 1 and q = l — 1, so S = L38(m, l, k, n). In the latter case we have ar+m-1 = bar for all r ^ q, then k = n — m + 1 and S = L39(m, l, q, n).
Case L3.4. b2 = 0. Every element of the semigroup can be written in the form ap or bap for some p. Let n be the least positive integer such that an = 0 and k be the least positive integer such that bak = 0. Then k ^ n ^ k + m.
Let p, q be the least positive integers such that ap = baq. Then one of the following possibilities holds:
1) ap = bq = 0;
2) ap+1 = aq+m = 0;
3) p = q + m — 1 and n > p + 1.
In the first case we have p = n and q = k, so S = L310(m,n,k). In the second case we have p + 1 = n and q + 1 = k, whence S = L3.11(m, n, k). In the third case we have ar+m-1 = bar for all r ^ q, so k = n — m + 1 and S = L3.12(m, n, q).
Series M. By Lemma 3, all cases lead to a contradiction.
Series N.
a\ sab a2\ sab ab \ sba a2 \ sb2
ba
a b2
■a2
2
ab ba
N1 N2 N3 N4
Case N1. We have b2 < ab, so b2 = sabt for some s,t G S1. If s = s'b or t = at' for s', t' G S1, then b2 < ba, a contradiction. If t = bt' for some t' G S1, then b2 < b2, so b2 = 0 < ba, a contradiction. Therefore b2 = amb for some m ^ 2. By the same arguments either ba = 0 or ba = alb = 0 for some l > m.
Case N1.1. ba = alb = 0 for some l > m. Every element of the semigroup can be written in the form ap or apb for some p. Note that a2mb = amb2 = b3 = bamb < ba, so l < 2m. Let n be the least positive integer such that an = 0 and let k be the least positive integer such that bak = 0. Obviously, l < k ^ n. Since am+lb = amba = b2a = balb = a12b2 = al2+mb, we obtain k ^ m + l.
Let p, q be the least positive integers such that ap = aqb. Then aq+1b = ap+1 = aqba = aq+lb = 0, so either p = n, q = k and S = N1.1 (m,l,n,k) or p = n — 1 and q = k — 1. This means that S = Ni.2(m,l,n,k).
Case N1.2. ba = 0. Every element of the semigroup S can be written either in the form ap or in the form apb for some p. Let n be the least positive integer such that an = 0 and let k be the least positive integer such that bak = 0. Trivially, k ^ n.
Let p, q be the least positive integers such that ap = aqb. Then ap+1 = aqba = 0, so either p = n, q = k and S = N1.1(m, l,n,l) or p = n — 1 and q = k — 1, i.e. S = N1.2(m, l, n, l).
Case N2. We have ba < ab, so ba = sabt for some s,t £ S1. If s = s'b or t = at' for some s',t' £ S1, then ba = 0 < b2, a contradiction. If t = bt' for some t' £ S1, then ba < b2, a contradiction. Therefore ba = amb for some m ^ 2. By the same arguments either b2 = 0 or b2 = a1 b = 0 for some l > m.
Case N2.1. b2 = alb = 0 for some l > m. Every element of the semigroup can written in the form ap or apb for some p. Let n be the least positive integer such that an = 0 and let k be the least positive integer such that akb = 0, then l < k ^ n. Since am+lb = alba = b2a = bamb = am +lb < am+lb, we have k ^ m + l.
Let p, q be the least positive integers such that ap = aqb. Then aq+1b = ap+1 = aqba = aq+mb, so either p = n, q = k and S = N2.1(m, l,n,k) or p = n — 1 and q = k — 1, i.e. S = N2.2(m, l, n, k).
Case N2.2. b2 = 0. Every element of the semigroup can be written in the form ap or apb for some p. Let n be the least positive integer such that an = 0 and let k be the least positive integer such that akb = 0. Then k ^ n.
Let p, q be the least positive integers such that ap = aqb. Then aq+1b = ap+1 = aqba = aq+mb = 0, so either p = n, q = k and S = N2.1(m, l,n,l) or p = n — 1 and q = k — 1, i.e. S = N2.2(m, l, n, l).
Case N3. We have a2 < ab and a2 < ba, but a2 < b2. So a2 = (ab) 2 or a2 = (ba) 2 for some p. Obviously, every element of the semigroup can be written in the same form.
2m |1 2k |1 2k | 2
Case N3.1. a2 = (ab) 2 , b2 = (ab) 2 = 0 for some k > m > 1. Then (ab) 2 = (ab) 2 b = b3 = b(ab) 2 = (ba) 2 . So, (ab) 2 = (ab) 2 a = (ba) 2 a = (ba) 2 a2 =
2k+1 2k+1 2k+3 2k+3 1/7 \ 2k+3
(ba) 2 (ab) 2 < (ab) 2 , whence (ab) 2 = 0. Analogously, (ba) 2 = 0.
2k+1 2k+1 2k+1 2k+1 2k+1 If (ab) 2 = (ba) 2 and (ab) 2 = 0, then S = N3.1(m, 2k + 1). If (ab^^ = (ba) 2 = 0 2k+1 2k+1 2k+1 and (ab)— = 0, then S = N3.2(m, 2k + 1). If (ab)~ = (ba)~, then S = N3.3(m, 2k + 1). If 2k+1 2k+1 (ab)— = 0 and (ba)~ = 0, then S = N3.4(m, 2k + 1).
2m |1 2k | 2 2k |3
Case N3.2. a2 = (ab) 2 , b2 = (ab) 2 = 0 for some k ^ m > 1. Then (ba) 2 = b(ab) ^^ = b3 = (ab) ^^ b = (ab) ^^ b2 = (ab) ^^ (ab) ^^ =
/ 2 n 2k|1 , T.2:k, , , 2m|^ 2k|1 4k|2m|^ 2k|2 2k|3 3
(ab) 2 a2(ba) 2 = (ab) 2 (ab) 2 (ba) 2 = (ab) 2 < (ba) 2 , so (ba) 2 = b3 = 0. Therefore b2ab < b3 and b2ab = 0.
If b2a = 0, then S = N3.1(m, 2k + 2). If b2a = 0 and (ba)^ = 0, then S = N3.2(m, 2k + 2). 2k+2 2k+2 2k+2 2k+2 If (ab)— = (ba)~ = 0, then S = N3.3(m, 2k + 2). If (ab)~ = 0 and (ba)~ = 0, then
S = N3.4(m, 2k + 2).
2m+1 2k+1
Case N3.3. a2 = (ab)^^, b2 = (ba)^^ for some m,k ^ 2 and k > m. Let n and l be the
n 1
least positive integers such that (ab)2 = (ba)2 = 0. Clearly, n,l ^ 2k + 1 and \n — l\ ^ 1. Then S = N3.5(m,k,n,l).
Case N3.4. a2 = (ab)^2^, b2 = 0. Let n and l be the least positive integers such that
n 1
(ab) 2 = 0 and (ab) 2 = 0. Obviously, \n — l\ < 1. Then S = N3.a(m,n,l).
2 / \ 2m 2m|1 3 , x 2m 2 / \ 2m — 1
Case N3.5. a2 = (ab) 2 for some m > 1. Then (ab) 2 = a3 = a(ab) 2 = a2(ba) 2 =
/ , 2m —1 2m — 1 2 2m — 2 . 2m —1 2 2m —2 2m|1
(ab) 2 (ba) 2 = (ab) 2 b2(ab) 2 . It is easy to see that (ab) 2 b2(ab) 2 < (ab) 2 , so
2m|1 , , 2m|2 , 0 , 2m|1 0
(ab) 2 = 0. Therefore (ba) 2 = 0. Then b2 = (ba) 2 or b2 = 0.
If b2 = (ba)^ = 0, then S = N3.r(2m). If (ba)^ = 0 and b2 = 0, then S = N3.8(2m). If b2 = (ba) ^ = 0, then S = N3.9(2m).
O , . 2m+1 2m+2 , 2m + 2 2m + 3 2m + 3
Case N3.6. a2 = (6a) 2 . Then (ab) 2 = a3 = (6a) 2 , so (ab) 2 = (ba) 2 = 0. Therefore b2 = (ab) or b2 = 0.
If b2 = (ab)^ =0, then 5 = N3,7(2m + 1). If(ab)^ = 0 and b2 = 0, then 5 = N3.g(2m+1). If b2 = (ab) ^ = 0, then 5 = N3.9(2m + 1).
Case N4. We have ab = an = bm for some n,m ^ 3. Then bab < ba, a2b = an+1 = aba < ba, ab2 = bm+1 = bab < ba, but ba < ab. So, ba = 0 and ab is an atom. If ap = bq for 1 < p < n and 1 < q < m, then ap+1 = 0, which means ab = an = 0 = ba, a contradiction. Then 5 = N4(n, m).
Series O, P, Q leads to a contradiction by Lemma 3 or by arguments from series F. Theorem 1 is now proved. □
REFERENCES
1. Nagy A., Jones P.R. Permutative semigroups whose congruences form a chain. Semigroup Forum, 2004. Vol. 69, no. 3. P. 446-456. DOI: 10.1007/s00233-004-0131-3
2. Popovich A.L., Jones P.R. On congruence lattices of nilsemigroups. Semigroup Forum, 2016. P. 1-7. DOI: 10.1007/s00233-016-9837-2
3. Schein B.M. Commutative semigroups where congruences form a chain. Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys., 1969, Vol. 17, P. 523-527.
4. Tamura T. Commutative semigroups whose lattice of congruences is a chain. Bull. Soc. Math. France, 1969, Vol. 97, P. 369-380.
