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In this section we describe subsemigroups that have elements both above and

below the diagonal. Let P be a subsemigroup of B with PflC/ / 0 and SO U 0.

Without loss of generality we can assume that q = t(^S) < «(P) = p observing

that the other case is dual to this by using the anti-isomorphism

We now state our main result of this section:

T heorem 7.10 Let S be a subsemigroup o /B such that 5'flf/y^0, S'nC/y^0

and q = l{S) < k{S) = p. There exist d G N, Fd C D n Lq, F Ç Tq^p, I Ç {q,... ,p — I}, P C {0,..., d — 1} with 0 G P such that

S = FdU F U Ai,p,d U

To prove this theorem we need the following elementary result from number

theory, the proof of which we include for completeness:

Lem m a 7.11 Let oi,..., &i,..., 7;, ro G No be arbitrary with ai > 0, 6i > 0

and let

d — gcd((%4,... 5 u/j, b\,,.. , b[),

Then there exist numbers Oi,..., —/3i,..., ~Pi G Nq such that:

(i) aiüi -b ... 4- 4- P\bi 4-... 4- Pibi = d;

(ii) O'!,..., Ofc, —/3i,..., —Pz > Tq.

Proof. We start by assuming, without loss of generality, that , . . . , ajç, ^ 4 , * . ' J bi 0.

Since d = gcd(ai,..., a^, 6i,..., 6^), we can write

k I

d = Y + E

CHAPTER 7. SUBSEMIGROUPS OF THE BICYCLIC MONOID B 102 for some integers « i , ... , A- Let H be any positive integer and let

P — H k la i.. .akbi.. .bi, Q ~ P/k, R = P/l. We can then write d = j2 = E + p - p + Y P ' f i i i= l j —i i= l k I k I

= Y

i = l

+ 0) + E - ^) = E (“f +

j = l 4=1

+ E

i= l k I = ^ ^ T ^ y i = l j = l

It is clear that when H increases all numbers a i, —/3i, •.., — A increase as

well and so the result holds. N

Pr o o f o f T heorem 7.10. Let Fo = S n D n L g and S' ~ S \Fd' We have S' = S n (Md n (Rgp U Ep)) where d = gcd(A(5')) and so S' is a subsemigroup. We observe that the elements dP G Fd act as identities in S '. Let x G S' CU

and G S' OU such that 0(æ) = i{S) — q and 4/(y) = k,{S) = p. Let Y Ç S' he

a finite set such that:

(i) 37, p G y ;

(ii) Ai n 5' n 5 'p 0 = > A^ n E y^ 0 for z G {g..., p - 1} {Y contains at least

one representative for each line in the strip with elements in 5");

(iii) {(z — p) mod d : A* Pi E 0 Ep y^ 0} = {(z - p) mod d : A,; (1 6" H Ep y^ 0} (E contains at least one representative for each class of lines in the square having a representative in S');

(iv) gcd(A(E)) = d.

Such E can be obtained by choosing a finite set Ei (with at most p—g+d elements)

q

P p +d

P P+d

Figure 7.4: Moving using and letting Y = U Eg- Let

E n (D U Z7) = ...,

where x = d^hi\ g = 4 < 2g < ... < 4 , 4 > 4 , 4 > 4 , • • • , 4 > 4 and let y n R =

where y — p = 4 < 4 < • • • < 4 and ki > li,... ,ks > 4 -

We are going to show that

Before proving this we will make an observation showing the importance of these

two elements. This observation is illustrated in Figure 7.4.

Let d P be an element in Md fl {Sq^p U Ep). We have d P d lf^ ^ — dN'^^ which

means intuitively that we can move d positions to the right in the grid using the

element If z > p then we also have d’'^^PdN = d~^^d which means that

we can move d positions down. If 7 > p + d then we have dPc^'^^c^ = dP~^ which

means that we can move left. Finally, if z > p + d then we have dP~^^dN — d~^N and so we can move up.

CHAPTER 7. SUBSEMIGROUPS OF THE BICYCLIC MONOID B 104 by (ivj. Since 4 — 4 < 0 and ki — k > 0, Lemma 7.11 can be applied and we can write

d = ai(4 “ 4 ) "b ' • ■ + Ckr(4 “ jr) + — 4) + • • • + “ 4) (7.1)

with the constants

, . . . , 0!j-, /3x J..., /3g ^ max "(4 ,...,2^,4j-**5 4}-

We can now consider the product which is equal to

( ^ n Û ^ Û + a i O ' l “ Û ) ) ( g î 2 ^ i 2 + a ; 2 L 2 “ î 2 ) ^ . , , ( c b ^ i r + U r O V -ir ) j

Since ai > max{4,..., 4} and 4 ~ 4 > 1 we have 4 + «1(4 “ 4) > 4, t = 1,..., r

and therefore, we can compute the above product working from the left hand side

to obtain

^iifjii+ai{ji-ii)+a2ij2-i2)+..>+ar{jr-ir) Çj 2)

We now consider the product (c^'6^')'^' ... which is equal to

^gis+/3s(fcs —ls)iJ's ^ +p2(k‘i—l2) ^ {ki —li) y,i ^

Since A > max{4, • • •, 4} and — 4 > 1 we have

4 + A(^i “ 4) > 4, t = 1,..., s

and we can compute this product from the right to obtain

^l+Pl{kl—ll)-\-g2{k2 — l2)-\'.--+gs{ks—la)y-l

Multiplying the elements (7.2) and (7.3) of S' we obtain

g U y i ’l + O c i i j l — u ) + 0 ! 2 (i 2 — t 2 ) + . . . + C > : j ’ ( j 'r — n ) + / 3 2 ( / î 2 “ ‘ Î 2 ) + ” . + ^ a ( f c a ~ ^ s ) ^ n

since q — A < l i ~ p and using equation (7.1). So G S '.

Since d | (4 - 4) we can write 4 - i\ = td for some t G N. Since p > 4 we have p + M > 4 and therefore

= cû-ji+P+M()P ^

We conclude that d P G S' as well. We now take the constants Q!i, . . ., a^j /?!J •. •, Ps ^ max'(4J ■ • •, 4’, 4, • • •, 4}

to be such that

d = a i( 4 — 4) + • • • + 0)r{jr ~ 4) + Pi{h — ^i) + ... + A (4 ~ Ls) (7.4)

and we consider the following element of S':

ii)+a2(j2—■i2)+...+Q;r(>—ïr)gO+/3l(Aîl—6)+^2(fc2—^2) + ...+/3s(fca— ^

Since 4 = g < P = 4 this element can be written as

^PJjP+<^l(jl—il)+a2{j2—i2) +—+<Xr {jr—ir) QP+Pl{kl—h)+g2{k2~l2) + —+Ps{ks~l3)yP

and it is equal to by equation (7.4). Therefore we have G S' as we wanted to show.

We are now going to prove that

S' n Ep = Ep,d,p

where

P = {(z — p) mod d : Li n E n Ep / 0}.

We will first show that Ep,d,p Ç S'. Let cP+r+«d^+r+w ^ By definition of E there is dP G E n Ep such that (z — p) mod d — r. Therefore, since Y Ç S' C Md, w e have dP = c^'^r+u'djjp+r+v'd^ As we have seen we can move

from dV to using the elements and which means

CHAPTER 7. SUBSEMICROUPS OF THE BICyCLIC MONOID B 106

elements d '^^d ,d P G S'. We will now show that S' D Ep Ç Ep,d,p- Let

d P G S' n Ep. By definition of P and by (iii) in the definition of Y we have

(z - p) mod d = r G P. Since S' Ç Mj we have cW = for some UyV > 0 and so dP G Ep^d+- We conclude that S' fl Ep — Ep^d.p-

We now prove that

s n Sq^p = A l ,p , d

where

I = {i : q < i < p — 1\dP G S' for some j}.

In fact, from any element d P G S' 11 S^,p we can move left and right using the elements dd"^^ and d'^'^d in order to obtain the whole line Ai,p,d. Since S' Ç Mj

it follows that S' D Sg,p = A/,p,d- We conclude that

S — F U Ep^djP O A i,p ,d

where F = S (1 T^^p is a finite set, and this implies S = Fd U F U E p ,d ,p U A i,p ,d

as required. H

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