Left-cancellative monoids

The results in this subsection are uncited but not claimed as original. Definition. A monoidM is left-cancellative if, for allx, y, z∈M,xy =xz impliesy=z.

Definition. LetM be a monoid. For f, g, h∈M, if h=f g then we write f 4h and say that f is a left-divisor ofh, or f left-divides h. We say gis a

right-divisor of hand g right-divides h, writteng4Rh. For a subsetS⊆M:

DivL(S) denotes the set of all left-divisors of elements of S. DivR(S) denotes the set of all right-divisors of elements of S.

Definition. For a monoid M andf, g∈M, we say thatg isinvertible and f is an inverse ofg iff g=gf = 1.

Lemma 2.2.1. Let M be a left-cancellative monoid where 1 is the only invertible element. Then the left-divisibility relation4is a partial ordering. Proof. Reflexivity and transitivity of 4 are obvious. For antisymmetry, supposef, g ∈M,f 4g and g4f. Then there are h, h0 ∈M withf =gh and g = f h0. So g = ghh0 and left-cancelling g, we have hh0 = 1, which forcesh=h0 = 1. Thenf =g.

Definition. Let M be a monoid and suppose f, g, h ∈ M. If f 4 h and g 4 h then h is a common right-multiple of f and g. If h 4 h0 for any other common right-multipleh0 off and gthen we sayh is aleast common right-multiple, or right-lcm of f and g. We say h is a minimal common right-multiple of f and g if, whenever h0 is a common right-multiple of f and g, and h0 4h, we haveh0 =h.

More generally, for a subset S ⊆ M, we say that h ∈ M is a common right-multiple of S if s 4 h for all s ∈ S. Minimal and least common

right-multiples are defined analogously.

If h 4 f and h 4 g then h is a common left-divisor of f and g. If h0 4h for any other common left-divisorh0 of f and g then we sayh is a greatest common left-divisor, orleft-gcd off andg.

Note. Any right-lcm is a minimal common right-multiple.

Corollary 2.2.2. Suppose a monoidM is left-cancellative and 1is the only invertible element ofM. Then right-lcms and left-gcds are unique when they exist.

Proof. Let S ⊆ M and assume h, h0 are right-lcms for S. Then by the definition of right-lcm, we haveh 4h0 and h0 4h. Lemma 2.2.1 says that

4 is a partial ordering. Hence h = h0 by anti-symmetry. The proof for left-gcds is similar.

Notation. WheneverM is a monoid andf, g∈M have a unique right-lcm, it is denotedf ∨g.

Whenever S⊆M has a right-lcm, it is denotedW

S orW s∈Ss.

Whenever f, g∈M have a unique left-gcd it is denoted f∧g.

Note. The existence of a right-lcm for a collectionSof elements of a monoid M does not guarantee the existence of a right-lcm for every subset ofS. The following example illustrates this.

Example. Consider the monoid with presentation:

ha, b, c|abc=bca=cab, b2a=ab2i

ThenW

{a,b,c}=abc exists buta∨bdoes not.

Proof. Clearly abc∼ bca∼ cab by the defining relations, so a,b,c 4 abc. Again by the defining relations, the only way we can haveb₄g and c₄g for someg∈M is if abc4g. Hence W

It is easy to verify that ab2 and abc are minimal common right-multiples ofa and b, but neither divides the other. Hencea∨bdoes not exist. Lemma 2.2.3. Let M be a left-cancellative monoid where 1 is the only invertible element. Let a, f, g∈M. Then,

1. af∨ag exists if and only if f∨g does, and af∨ag=a(f∨g), 2. af∧ag exists if and only if f∧g does, and af∧ag=a(f∧g). Proof. We only show (1), as the proof of (2) is very similar.

Proof of ”if”. Supposef∨g exists. Clearlyaf, ag4a(f∨g). Now suppose af, ag4h for some h∈M. It remains to show thata(f∨g)4h. We have h=ah0 for someh0 ∈M. Left-cancellingagivesf, g4h0. Hencef∨g4h0, and thus a(f∨g)4ah0 =h.

Proof of ”only if”. Suppose af ∨ag exists. Then af ∨ag = ab for some b ∈M. Left-cancelling a, we have f 4b and g 4 b. Now suppose h ∈ M and f, g 4 h. Then af, ag 4 ah, so ab = af ∨ag 4 ah. left-cancelling a givesb4h, and b=f∨g, as required.