Semi Group Theory

8/13/2019 Semi Group Theory

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1. The Basic ConceptDenition : A semigroup is a pair (S, ) where S is a non-empty set and is an associativebinary operation on S . [i.e. is a function S × S → S with (a, b) → a b and for all a, b, c S we have a (b c) = ( a b) c].

We abbreviate “( S, )” by “S ” and often omit in “a b” and write “ ab”. By inductiona1a2 . . . an is unambiguous. Thus we write an for

a a . . . a

n times

.

Index Laws: We have that for all n, m N = {1, 2, . . . } the following index laws hold

an am = an + m

an m = anm .

Denition : A monoid M is a semigroup with an identity, i.e. there exists 1 M such

that 1 a = a = a1 for all a M .Putting a0 = 1 then the index laws hold for all n, m N 0.

Note. The identity of a monoid is unique.

Denition : A group G is a monoid such that for all a G there exists a b G withab = 1 = ba.

Example 1.1. Groups are monoids and monoids are semigroups. Thus we have

Groups Monoids Semigroups.

The one element trivial group {e} with multiplication tablee

e e

is also called the trivial semigroup or trivial monoid .

Example 1.2. A ring is a semigroup under × . A ring with identity is a monoid.1



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Example 1.3. N is a monoid under × . N forms a semigroup under + and N 0 is a monoidunder + and × .

Example 1.4. Let I, J be non-empty sets and set T = I × J with the binary operation

(i, j )(k, ℓ) = ( i, ℓ).

Then T is a semigroup called the rectangular band on I × J . To verify that T is a semigroupwe need to check that the associativity law holds. So

(i, j )(k, ℓ) (m, n ) = ( i, ℓ)(m, n ) = ( i, n ),

(i, j ) (k, ℓ)(m, n ) = ( i, j )(k, n) = ( i, n ),

for all (i, j ), (k, ℓ), (m, n ) T and hence multiplication is associative.

I

J

i

j

k

ℓ

(i, j )

(k, ℓ)

(i, ℓ)

Figure 1. The rectangular band.

Notice: (i, j )2 = ( i, j )( i, j ) = ( i, j ), i.e. every element is an idempotent. This can’t belooked at from ring theory because any ring where every element is an idempotent meansthat the ring is commutative. However, the rectangular band does not have to be commu-tative.

1.1. Adjoining an ElementLet S be a semigroup, which is not a monoid. Find a symbol not in S , call it “1”. We nowextend the denition of on S to S { 1} by



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a b = ab if a, b S,a 1 = a = 1 a for all a S,

1 1 = 1.

Then is associative (check this). Thus we have managed to extend multiplication in S to S { 1}. For an arbitrary semigroup S the monoid S 1 is dened by

S 1 =S if S is a monoid,S { 1} if S is not a monoid.

So, S 1 is “S with a 1 adjoined if necessary”.

Example 1.5. Let T be the rectangular band on {a}×{ b, c}. Then T 1 = {1, (a, b), (a, c)},which has multiplication table

1 (a, b) (a, c)1 1 (a, b) (a, c)

(a, b) (a, b) (a, b) (a, c)(a, c) (a, c) (a, b) (a, c)

Example 1.6 (The Bicyclic Semigroup / Monoid) . If A Z , such that |A| < ∞ , thenmax A is the biggest element in A. We also use this notation to represent

max{a, b} =a if a b,b if b a.

We note some further things about max:• max{a, 0} = a if a N 0,• max{a, b} = max {b, a},• max{a, a } = a,• max a, max{b, c} = max {a,b,c} = max max{a, b}, c .

Thus we have that ( Z , max) is a semigroup and ( N 0, max) is a monoid.

Note. The following identities hold for all a, b, c Z

( ) a + max {b, c} = max {a + b, a + c},max{b, c} = a + max {b− a, c − a}.

Put B = N 0 × N 0. Now, on B we dene a binary operation by

(a, b)(c, d) = ( a − b + t, d − c + t),where t = max {b, c}. We claim that B together with this operation forms a semi-group/monoid called the Bicyclic Semigroup/Monoid .



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Proof. With ( a, b), (c, d) B and t = max {b, c} we have t − b 0 and t − c 0. Thus wehave a − b + t a and d − c + t d. Therefore, in particular ( a − b + t, d − c + t) B somultiplication is closed. We have that (0 , 0) B and for any (a, b) B we have

(0, 0)(a, b) = (0 − 0 + max {0, a}, b − a + max {0, a}),= (0 − 0 + a, b − a + a),= ( a, b),= ( a, b)(0, 0).

Therefore (0 , 0) is the identity of B. We need to check associativity of max. Let ( a, b),(c, d), (e, f ) B. Then

(a, b)(c, d) (e, f ) = a − b + max {b, c}, d − c + max {b, c} (e, f ),

= a − b − d + c + max {d − c + max {b, c}, e}f − e + max {d − c + max {b, c}, e} .

(a, b) (c, d)(e, f ) = ( a, b) c − d + max {d, e}, f − e + max {d, e} ,= ( a − b + max {b, c − d + max {d, e}}

f − e − c + d + max {b, c − d + max {d, e}} .

Now we have to show that

a − b − d + c + max d − c + max {b, c}, e =

a − b + max b, c − d + max {d, e} ,

f − e + max d − c + max {b, c}, e =

f − e − c + d + max b, c − d + max {d, e} .

We can see that these equations are the same and so we only need to show

c − d + max d − c + max {b, c}, e = max b, c − d + max {d, e} .

Now, we have from ( ) that

max max{b, c}, c − d + e = max b, c − d + max {d, e} .

The RHS of this equation is

max b, c − d + max {d, e} = max b,max{c − d + d, c − d + e} ,= max b,max{c, c − d + e} ,= max {b,c,c− d + e},

= max max{b, c}, c − d + e .

Therefore multiplication is associative and hence B is a semigroup/monoid.



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Example 1.7. T X is a semigroup. See the examples class for proof.

1.2. Easy Facts / DenitionsA semigroup S is commutative if ab = ba for all a, b S . For example N with + iscommutative. B is not because

(0, 1)(1, 0) = (0 − 1 + 1 , 0 − 1 + 1) = (0 , 0),(1, 0)(0, 1) = (1 − 0 + 0 , 1 − 0 + 0) = (1 , 1).

Thus we have (0, 1)(1, 0) = (1 , 0)(0, 1). Notice that in B; (a, b)(b, c) = ( a, c). It is importantto see that in general we have NO cancellation in Semigroups. Thus we have that

ac = bc a = b,ca = cb a = b.

For example in the rectangular band on {1, 2} × { 1, 2} we have

(1, 1)(1, 2) = (1 , 2) = (1 , 2)(1, 2)but (1 , 1) = (1 , 2). A semigroup is cancellative if

ac = bc a = b,ca = cb a = b,

are true for all a, b, c S . For example, a group is cancellative (indeed, any subsemigroupof a group is cancellative). N 0 is a cancellative monoid, which is not a group.

1.3. ZerosA zero “0” of a semigroup S is an element such that, for all a S ,

0a = a = a0.

Adjoining a Zero

Let S be a semigroup, then pick a new symbol “0”. Let S 0 = S { 0}; dene a binaryoperation · on S 0 by

a · b = ab for all a S,0 · a = 0 = a · 0 for all a S,

0 · 0 = 0.



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Example 3.2. R = {cx | x X } is a subsemigroup of T X called the right zero semigroup.

Proof. cx cy = cy for all x, y X .

Example 3.3 (Bicyclic Monoid). Put E (B) = {(a, a ) | a N 0}, the set of all idempotentsin B. We claim that E (B) is a commutative submonoid of B. Clearly we have (0, 0) E (B)and for (a, a ), (b, b) E (B) we have

(a, a )(b, b) = ( a − a + t, b − b + t) where t = max {a, b},= ( t, t ),= ( b, b)(a, a ).

3.1. MorphismsDenition : Let S, T be semigroups then θ : S → T is a semigroup (homo)morphism if,for all a, b S ,

(ab)θ = aθbθ.If S, T are monoids then θ is a monoid (homo)morphism if θ is a semigroup morphism and1S θ = 1T .

Example 3.4. θ : B → Z given by (a, b)θ = a − b is a monoid morphism because

(a, b)(c, d) θ = ( a − b + t, d − c + t)θ t = max {b, c},= ( a − b) − (d − c),= ( a − b) + ( c − d),= ( a, b)θ + ( c, d)θ.

Furthermore (0 , 0)θ = 0 − 0 = 0.

Example 3.5. Let T = I × J be the rectangular band then dene α : T → T J by(i, j )α = c j . Then we have

(i, j )(k, ℓ) α = ( i, ℓ)α,= cℓ,

= c j cℓ,= ( i, j )α (k, ℓ)α.

So, α is a morphism.

Denition : A bijective morphism is an isomorphism .Isomorphisms preserve algebraic properties (e.g. commutativity). Suppose f : S → T is amorphism then with A S we have



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Af = {af | a A}.

If A = S then Sf is the image, Im f , of S . So, if A is a subsemigroup of S then Af willbe a subsemigroup of T . Now, conversely, if B T then we dene

Bf − 1 = {s S | sf B}.

If B is a subsemigroup of T then Bf − 1 is either or Bf − 1 is a subsemigroup of S , i.e.endomorphisms pull back subsemigroups.

Embeddings

Let α : S → T be a morphism. From the handout, Im α is a subsemigroup of T . If α is1:1 then α : S → Im α is an isomorphism and S is embedded in T .

Theorem 3.1 (The “Cayley Theorem” – for Semigroups) . Every semigroup is embedded in some T X

Proof. Let S be a semigroup and set X = S 1. We need a 1:1 morphism S → T X . Fors S , we dene ρs T X by xρs = xs. Now, dene α : S → T X by sα = ρs . First showthat α is 1:1. If sα = tα then ρs = ρt and so xρs = xρt for all x S 1; in particular1ρs = 1ρt and so 1s = 1t hence s = t and α is 1:1.

Let u, b S . For any x X we have

x(ρu ρv) = ( xρu )ρv) = ( xu)v = x(uv) = xρuv .

Hence ρu ρv = ρuv and so uαvα = ρu ρv = ρuv = ( uv)α. Therefore α is a morphism. Henceα : S → T X is an embedding.

Theorem 3.2 (The “Cayley Theorem” - for Monoids) . Let S be a monoid then there exists an embedding S → T X for some X .

Proof. We know from the proof of the Cayley Theorem for Semigroups that α : S → T X ,given by sα = ρs where xρs = xs, is a semigroup embedding. In the proof above we hadX = S 1 X = S because S is a monoid. We must check that α is a monoid morphism,i.e. we need to check 1α = I X . Now 1α = ρ1 and for all x X = S we have

xρ1 = x1 = x = xI X

and so 1α = ρ1 = I X .

Theorem 3.3 (The Cayley Theorem - for Groups) . Let S be a group. Then there exists an embedding S → S X for some X .

Proof. Exercise.



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3.2. IdempotentsDenition : e S is an idempotent if e2 = e. Also we dene the set of idempotents in S to be

E (S ) = {e S | e2 = e}.

Now, E (S ) may be empty, e.g. E (S ) = (N under +) but E (S ) may also be S . If S = I × J is a rectangular band then for any ( i, j ) S we have (i, j )2 = ( i, j )( i, j ) = ( i, j ) and soE (S ) = S .

Denition : If E (S ) = S , then S is a band .

For the bicyclic semigroup B we have E (B) = (a, a ) | a N 0 – from exercises 1. If S isa monoid then 1 E (S ). If S is a cancellative monoid, then 1 is the only idempotent: forif e2 = e then ee = e1 and so e = 1 by cancellation. In particular for S a group we haveE (S ) = {1}.

Lemma 3.1. Suppose ef = f e for all f, e E (S ). Then E (S ) = or E (S ) is a subsemi-group.

Proof. Let e, f E (S ). Then ( ef )2 = efef = eeff = ef and hence ef E (S ).

Denition : A commutative band is a semilattice .

From Lemma 3.1 if idempotents in S commute then E (S ) is empty or it’s a semilattice.

Example 3.6. E (B) = (a, a ) | a N 0 is a semilattice.

Example 3.7. A rectangular band I × J is not a semilattice (unless |I | = |J | = 1) since(i, j )(k, ℓ) = ( k, ℓ)( i, j ) i = k and j = ℓ.

Denition : Let a S . Then we dene a = {an | n N }, which is a commutativesubsemigroup of S . We call a the monogenic subsemigroup of S generated by a.

Proposition. a = (N , +) or a is nite.

Proof. If ai = a j for all i, j N with i = j then θ : a → N dened by ai θ = i is anisomorphism. Suppose that in the list of elements a, a 2, a3, . . . there is a repetition, i.e.a i = a j for some i < j . Let k be least such that ak = an for some n < k . Then k = n + rfor some n, r N – where n is the index of a, r is the period of a. Then the elementsa, a 2, a3, . . . , a n + r − 1 are all distinct and an = an + r .

We can express this pictorially by a map diagram.



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a a2

a3

a4

an

an +1

an + r − 1

an +2

Notice that

an +2 r = an + r + r = an + r ar = an ar = an + r = an

and hence an + rk = an for all k N 0. Let u N 0. Write u = qr + t with 0 t < r ,q, t N 0. Then an + u = an + qr + t = an + qr a t = an a t = an + t and so we have

a = {a, a 2, . . . , a n + r − 1} and | a | = n + r − 1.

Lemma 3.2 (The Idempotent Power Lemma) . If a is nite, then it contains an idem-potent.

Proof. Let n, r be the index and period of a. Choose s N 0 with s ≡ − n (mod r). Thens + n ≡ 0 (mod r) and so s + n = kr for k N . Then

(an + s )2 = an + n + s + s = an + kr + s = an + kr as = an as = an + s

and so an + s E (S ). In fact, {an , an +1 , . . . , a n + r − 1} is a group with identity an + s .

Corollary 3.1. Any nite semigroup contains an idempotent.

3.3. Idempotents in T X

We know cx cy = cy for all x, y X and hence cx cy = cx for all x X . Therefore cx E (T X )for all x X .

Example 3.8. Let us dene an element

α = 1 2 32 2 3 E (T X )

and a subset Z X . The “restriction of α to Z ”, denoted “α |Z ” is the map α Z : Z → Y such that Z (α Z ) = Zα .

Example 3.9. Let us dene an element

α = a b c d1 1 1 2 α |{c,d } = c d

1 2We can see that α is not one-to-one but α|{c,d } is.



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Let α T X (i.e. α : X → X ). Recall that Im α = {xα : x X } X .

Example 3.10. In T 3 we have Imc1 = {1}, Im I 3 = {1, 2, 3} then

Im 1 2 33 2 3 = {2, 3}.

Lemma 3.3 (The E (T X ) Lemma) . An element ε T X is idempotent ε| Im ε = I Im ε .

Proof. ε| Im ε means for all y Im ε we have yε = y. Then

ε E (T X ) ε2 = ε, xε2 = xε for all x X, (xε)ε = xε for all x X, yε = y for all y Im ε, ε| Im ε = I Im ε .

Example 3.11. Dene an element

α = 1 2 32 2 3 T 3,

this has image, Im α = {2, 3}. Now we can see that 2α = 2 and 3α = 3. Hence α E (T 3).

Example 3.12. We can similarly create another idempotent in T 5,

1 2 3 4 55 5 3 3 5 E (T 5).

Using Lemma 3.3 we can now list all the idempotents in T 3. We start with the constantmaps, i.e. ε E (T 3) such that | Im ε| = 1. These are

1 2 31 1 1 , 1 2 3

2 2 2 , 1 2 33 3 3 .

Now consider all elements ε E (T 3) such that | Im ε| = 2. These are

1 2 32 2 3 , 1 2 3

1 1 3 , 1 2 33 2 3 , 1 2 3

1 3 3 , 1 2 31 2 1 , 1 2 3

1 2 2 .

Now there is only one idempotent such that | Im ε| = 3, that is the identity map

1 2 31 2 3 .



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4. RelationsDenition : A (binary) relation ρ on A is a subset of A × A.

Convention: we may write “ aρb” for “(a, b) ρ”.

4.1. Partial OrdersWe dene a partial ordering on R .

a a for all a R ,a b and b c a c for all a, b, c R ,a b and b a a = b for all a, b R .

Notice that for any a, b R we have a b or b a. For X a set then P (X ) is the set of all subsets of X . Now is a partial order on P (X ). We have

A A for all A P (X )A B and B C A C for all A, B,C P (X )A B and B A A = B for all A, B P (X )

Notice that if |X | > 2 and x, y X with x = y then {x} { y} and {y} { x}. If ω = A × A is the universal relation on A, so xωy for all x, y A, then [x] = A for allx A. We have that

ι = (a, a ) | a Ais the equality relation and so xιy x = y and so [x] = {x} for all x A.

4.2. Algebra of RelationsIf ρ, λ are relations on A, then so is ρ ∩ λ. For all a, b A we have

a(ρ ∩ λ)b (a, b) (ρ ∩ λ) (a, b) ρ and (a, b) λ

aρb and aλb.We note that ρ λ means aρb aλb. Note ι ρ ρ is reexive and so ι ρ for anyequivalence relation ρ. We see that ι is the smallest equivalence relation on A and ω is thelargest equivalence relation on A. We note that

[a] = {b A | aρb}.If ρ is an equivalence relation then [ a] is the equivalence-class, a ρ-class, of a.



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Denition : An equivalence relation ρ on a semigroup S is a congruence if

(aρb and cρd) acρbd.

Lemma 4.2 (The Kernel Lemma) . Let θ : S → T be a semigroup morphism. Then ker θis a congruence on S .

Proof. We know ker θ is an equivalence relation on S . Suppose a,b,c,d S with

a ker θb and cker θd.Then aθ = bθ and cθ = dθ, so

(ac)θ = aθcθ = bθdθ = ( bd)θ.

Therefore ac ker θbd, so that ker θ is a congruence. Let ρ be a congruence on S . Then we dene

S/ρ = [a] | a S .Dene a binary relation on S/ρ by

[a][b] = [ab].We need to make sure that this is a well-dened relation. If [ a] = [a ′] and [b] = [b′] thenaρa ′ and bρb′ ; as ρ is a congruence we have abρa′b′ and hence [ab] = [a ′b′]. Hence ouroperation is well dened. Let [a], [b], [c] S/ρ then we have

[a] [b][c] = [a][bc],= a(bc) ,

= (ab)c ,= [ab][c],

= [a][b] [c].

If S is a monoid, then so is S/ρ because we have

[1][a] = [1a] = [a] = [a1] = [a][1]for any a S . Hence we conclude that S/ρ is a semigroup and if S is a monoid, then so isS/ρ . We call S/ρ the factor semigroup (or monoid) of S by ρ. Now, dene ν ρ : S → S/ρby

sν ρ = [s].Then we have



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sν ρ tν ρ = [s][t] denition of ν ρ ,= [st ] denition of multiplication in S/ρ,= ( st )ν ρ denition of ν ρ .

Hence ν ρ is a semigroup morphism. We now want to examine the kernel of ν ρ and so

s ker ν ρ t sν ρ = tν ρ denition of kerν ρ , [s] = [t] denition of ν ρ , sρt denition of ρ.

Therefore ρ = ker ν ρ and so every congruence is the kernel of a morphism.

Theorem 4.1 (The Fundamental Theorem of Morphisms for Semigroups) . Let θ : S → T

be a semigroup morphism. Then ker θ is a congruence on S , Im θ is a subsemigroup of T and S/ ker θ = Im θ.

Proof. Dene θ̄ : S/ ker θ → Im θ by [a]θ̄ = aθ. We have

[a] = [b] a ker θb aθ = bθ [a]θ̄ = [b]θ̄.

Hence θ̄ is well dened and one-to-one. For any x Im θ we have x = aθ = [a]θ̄ and so θ̄is onto. Finally,

[a][b] θ̄ = [ab]θ̄ = ( ab)θ = aθbθ = [a]θ̄[b]θ̄.Therefore θ̄ is an isomorphism and S/ ker θ = Im θ. Note that the analogous result holdsfor monoids.

5. IdealsNotation

If A, B S then we write

AB = {ab | a A, b B},A2 = AA = {ab | a, b A}.

Note. A is a subsemigroup if and only if A = and A2 A.

We write aB for {a}B = {ab | b B}. For example

AaB = {xay | x A, y B}.



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Facts:(1) A(BC ) = ( AB)C therefore P (S ) = {S | A S } is a semigroup – the power

semigroup of S .

(2) A B AC BC and CA CB for all A, B,C P (S ).(3) AC = BC A = B and CA = CB A = B, i.e. the power semigroup is notcancellative.

Denition : Let = I S then I is a right ideal if IS I (i.e. a I , s S as I ).Then I is a left ideal if SI I . Finally I is a (two sided) ideal if IS SI I .Any (left/right) ideal is a subsemigroup. If S is commutative, all 3 concepts coincide.

Example 5.1. Some examples of ideals.(1) Let i I then {i} × J is a right ideal in a rectangular band I × J .(2) We dene a right ideal {(x, y) | x m, y N 0} in the bicyclic semigroup B (m

xed).(3) Y X then we have {α T X | Im α Y } is a left ideal of T X .(4) For any n N we dene

S n = {a1a2 . . . an | a i S }.This is an ideal of S . If S is a monoid then S n = S for all n, since for any s S we can write

s = s 11 . . . 1

n − 1

S n .

(5) If S has a zero 0, then {0} (usually written 0), is an ideal.

Denition : For a semigroup S we have:(1) S is simple if S is the only ideal.(2) if S has a zero 0, then S is 0-simple if S and {0} are the only ideals and S 2 = 0.

Example 5.2. Let G be a group and I a left ideal. Let g G, a I then we have

g = ( ga− 1)a I and so G = I . Therefore G has no proper left/right ideals. Hence G is simple.

Exercise : G0 is 0-simple

Example 5.3. We have ( N , +) is a semigroup. Now dene I n (N , +) to be

I n = {n, n + 1 , n + 2 , . . . },which is an ideal. Hence N is not simple.

Note. {2, 4, 6, . . . } is a subsemigroup but not an ideal.

Example 5.4. The bicyclic semigroup B is simple.



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Proof. Let I B be an ideal, say (m, n ) I . Then (0, n) = (0 , m)(m, n ) I . Thus(0, 0) = (0 , n)(n, 0) I . Let (a, b) B . Then

(a, b) = ( a, b)(0, 0) I and hence B = I B is simple.

5.1. Principle IdealsWe make note of how the S 1 notation can be used. For example

S 1A = {sa | s S 1, a A},= {sa | s S { 1}, a A},

= {sa | s S, a A} { 1a | a A},= SA A.

In particular, if A = {a} then S 1a = Sa { a}. So,

S 1a = Sa a Sa, a = ta

for some t S . We have S 1a = Sa for a S if:• S is a monoid (then a = 1a).

• a E (S ) (then a = aa ).• a is regular , i.e. there exists x S with a = axa (then a = ( ax)a).But in ( N , +) we have 1 1 + N . Dually,

aS 1 = aS { a}and similarly

S 1aS 1 = SaS aS Sa { a}.

For = I S then we have I is an ideal S 1IS 1 I .

Claim. aS 1 is the “smallest” right ideal containing a.

Proof. a = a1 aS 1 and (aS 1)S = a(S 1S ) aS 1. So, aS 1 is a right ideal containing a. If a I and I is a right ideal, then aS 1 IS 1 = I IS I . Then aS 1 is the principal right ideal generated by a. S 1a is the principal left ideal generated by a. S 1aS 1 is the smallestideal containing a – it’s the called the principal ideal generated by a.

If S is commutative then aS 1 = S 1a = S 1aS 1.



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Example 5.5. In a group G we have

aG 1 = G = G1a = G1aG 1

for all a G.Example 5.6. In N under addition we have

I n = {n, n + 1 , n + 2 , . . . } = “ n + N 1”

Example 5.7. B is simple, so

B(m, n )B = B 1(m, n )B 1 = B

Claim. (m, n )B = ( m, n )B 1 = (x, y) | x m, y N 0

Proof. We have

(m, n )B = (m, n )(u, v) | (u, v) B(x, y) | x m, y N 0 .

Let x m then

(m, n ) n + ( x − m), y = m − n + n + ( x − m), y ,= ( x, y).

Therefore (x, y) (m, n )B (x, y) | x m, y N 0 (m, n )B . hence we have proved

our claim. Dually we have B(m, n ) = (x, y) | x N 0, y n .

Lemma 5.1 (Principle Left Ideal Lemma) . The following statements are equivalent;i) S 1a S 1b,

ii) a S 1b,iii) a = tb for some t S 1,iv) a = b or a = tb for some t S .

Note. If S 1a = Sa and S 1b = Sb, then the Lemma can be adjusted accordingly.

Proof. It is clear that (i) (iii) (iv) (i) and so we prove (i) (ii).(i) (ii): If S 1a S 1b then a = 1a S 1a S 1b a S 1b.(ii) (i): If a S 1b, then as S 1a is the smallest left ideal containing a, and as S 1b is aleft ideal we have S 1a S 1b.

Lemma 5.2 (Principle Right Ideal Lemma) . The following statements are equivalent:i) aS 1 bS 1,

ii) a bS 1,



20 VICKY G

iii) a = bt for some t S 1,iv) a = b or a = bt for some t S .

Note. If aS = aS 1 and bS = bS 1 then aS bS a bS a = bt for some t S .

Denition : The relation L on a semigroup S is dened by the rule

aLb S 1a = S 1bfor any a, b S .

Note.(1) L is an equivalence.(2) If aLb and c S then S 1a = S 1b, so S 1ac = S 1bc and hence acLbc, i.e. L is right

compatible.(3) a right (left) compatible equivalence is a right (left ) congruence . Thus L is a right

congruence.Corollary 5.1. We have that

aLb s, t S 1 with a = sb and b = ta.

Proof. We start with aLb

aLb S 1a = S 1b S 1a S 1b and S 1b S 1a

s, t S 1

with a = sb, b = taby the Principle Left Ideal Lemma. We note that this statement about L can be used asa denition of L.

Remark.(1) aLb a = b or there exists s, t S with a = sb, b = ta .(2) If Sa = S 1a and Sb = S 1b, then aLb s, t S with a = sb, b = ta .

Dually, the relation R is dened on S by

aR b aS 1 = bS 1,

s, t S 1 with a = bs and b = at, a = b or s, t S with a = bs and b = at.

We can adjust this if aS 1 = aS as before. Now R is an equivalence ; it is left compatible and hence a left congruence .Denition : We dene the relation H = L ∩ R and note that H is an equivalence.The relations L, R , H are in fact three of Greens’ relations .



SEMIGROUP THEORY 21

Example 5.8. If S is commutative, L = R = H . In a group G,

G1a = G = G1b and aG1 = G = bG1 for all a, b G.

So aLb and aR b for all a, b G. Therefore L = R = ω = G × G and hence we must haveH = ω.

Example 5.9. In N under + we have

a + N 1 = {a, a + 1 , . . . }and so a + N 1 = b + N 1 a = b. Hence L = R = H = ι.

Example 5.10. In B we know

(m, n )B = (x, y) | x m, y N 0

and so we have

(m, n )B = ( p, q )B m = p.

Hence (m, n )R ( p, q ) m = p. Dually,

(m, n )L( p, q ) n = q.Thus ( m, n )H ( p, q ) (m, n ) = ( p, q ), which gives us H = ι.

5.2. L and R in T X

Claim. αT X β T X Ker β Ker α. [Recall Ker α = (x, y) X × X | xα = yα ].Proof. ( ) Suppose αT X β T X . Then α = βγ for some γ T X . Let (x, y) Ker β . Then

xα = x(βγ ) = ( xβ )γ = ( yβ )γ = y(βγ ) = yα.Hence (x, y) Ker α and so Ker β Ker α .

( ) Suppose Ker β Ker α. Dene γ : X → X by

(xβ )γ = xα

β

α

γ x

xβ

xα



22 VICKY G

Thus we have yγ = xo, y Im β where x0 is xed. If z = xβ = yβ , then ( x, y) Ker β Ker α so xα = yα. Hence γ is well-dened. So γ T X and βγ = α. Therefore α β T X sothat by the Principle Ideal Lemma, αT X β T X .

Corollary 5.2 (R − T X -Lemma) . αR β Ker α = Ker β .

Proof. We have

αR β αT X = β T X

αT X β T X and β T X αT X

Ker β Ker α and Ker α Ker β Ker α = Ker β.

Fact : T X α T X β Im α Im β (Exercise 4).

Corollary 5.3 (α − T X -Lemma) . αLβ Im α = Im β .

Consequently αHβ Ker α = Ker β and Im α = Im β .

Example 5.11. Let us dene

ε = 1 2 32 2 3 E (T 3)

Now we have Imε = {2, 3}. We can see that Ker has classes {1, 2}, {3}. So

αHε Im α = Im ε and Ker α = Ker Im α = {2, 3} and Ker α has classes {1, 2}, {3}.

So we have

α = 1 2 33 3 2 or α = ε = 1 2 3

2 2 3

ε αε ε αα α ε

which is the table of a 2-element group. Thus the H-class of ε is a group.



SEMIGROUP THEORY 23

5.3. Subgroups of SemigroupsS is a semigroup, H S . Then H is a subgroup of S if it is a group under (the restrictionof) the binary operation on S to H ; i.e.

• a, b H ab H • e H with ea = a = ae for all a H • a H b H with ab = e = ba

Remark.(1) S does not have to be a monoid. Even if S is a monoid, e does not have to be 1.

However, e must be an idempotent, i.e. e E (S ).(2) If H is a subgroup with identity e, then e is the only idempotent in H .

S

H e

Figure 2. e is the only idempotent in H .

(3) If e E (S ), then {e} is a trivial subgroup.(4) S X is a subgroup of T X ; {ε, α } (as above) is a subgroup of T 3. Notice

αH I X Im α = Im I X and Ker α = Ker I X , Im α = X and Ker α = ι, α is onto and α is 1:1, α S X .

Therefore S X is the H -class of I X .

Denition : La is the L-class of a; Ra is the R -class of a and H a is the H -class of a.

Now La = Lb aLb and H a = La ∩Ra . In B, L(2 ,3) = (x, 3) | x N 0 .

5.4. Maximal Subgroup TheoremLet e E (S ). Then H e is the maximal subgroup of S with identity e.

Lemma 5.3 (Principle Ideal for Idempotents) . Let a S , e E (S ). Then (i) S 1a S 1e ae = a

(ii) aS 1 eS 1 ea = a.



24 VICKY G

S H e

e

Figure 3. Existence of a Maximal Subgroup.

Proof. (We prove part (i) only because (ii) is dual). If ae = a, then a S 1e so S 1a S 1eby the Principle Ideal Lemma. Conversely, if S 1a S 1e then by the Principle Ideal Lemmawe have a = te for some t S 1. Then

ae = ( te)e = t(ee) = te = a.

Corollary 5.4. Let e ES . Then we have

aR e ea = a,aLe ae = a,aHe a = ae = ea.

Idempotents are left/right/two-sided identities for their R / L/ H -classes.

Proof. Let G be a subgroup with idempotent e. Then for any a G we have ea = a = ae

and there exists a− 1

G with aa− 1

= e = a− 1

a. Then

ea = a

aa − 1 = e aR e

ae = aa− 1a = e

aLe

aHe.

Therefore G He.

Theorem 5.1 (Maximal Subgroup Theorem) . Let e E (S ). Then He is the maximal subgroup of S with identity e.

Proof. We have shown G a subgroup with identity e G H e . We know H e is a subgroupwith identity, e. We know that e is an identity for He . Suppose a, b H e . Then bHe,so bR e hence abR ae (R is left compatible) so abR ae = aR e. Also, aLe abLeb = bLehence abHe so ab H e . It remains to show that for all a H e there exists b H e withab = e = ba.



SEMIGROUP THEORY 25

Let a H e by denition of H = R ∩ L there exists s, t S 1 with

at =

a R e

e = sa

a Le

.

We have e = ee = ate

e · ee = eee = a(ete) = ( ese)a.Let b = ete, c = ese so b, c S and eb = be = b, ec = ce = c. Also e = ab = ca. Now

b = eb = ( ca)b = c(ab) = ce = c.Finally,

eb = b ba = e

bR e

be = b ab = e

bLe

so bHe b H e . Hence He is a subgroup.

Let’s take 2 distinct idempotents e, f E (S ) with e = f . Since H e and H f are subgroupsH e = H f H e ∩ Hf = .

Corollary 5.5 (Maximal Subgroup Theorem) . Let a S . Then (i) (ii) (iii) (iv).

(i) a lies in a subgroup,(ii) aHe, some e E (S ),

(iii) Ha is a subgroup,

(iv) aHa2.

Proof. (i) (ii) a G G H e where e2 = e is the identity for G. Therefore a H e soaHe.

(ii) (iii) aHe H a = He and by MST H e is a subgroup.

(iii) (i) a H a .

(iii) (iv) If Ha is a subgroup, then certainly Ha is closed. Hence a, a 2 H a thereforeaHa2.

Theorem 5.2 (Green’s Theorem) . a S , then a lies in a subgroup iff aHa2

.Proof. see later.

Subgroups of T n

We use Green’s Theorem to show the following

Claim. α lies in a subgroup of T n the map diagram has no tails of length 2.



26 VICKY G

Proof. α lies in a subgroup αHα 2 αLα 2, αR α 2 Im α = Im α 2, Ker α = Ker α 2.We know Im α 2 Im α (as T n α 2 T n α). Let ρ be an equivalence on a set X . Put

X ρ = [x] | x X

for X = {1, 2, . . . , n } = n and α T n

|n/ Ker α | = | Im α | .

Claim (Mini Claimette) . For α T n , Im α = Im α 2 Ker α = Ker α 2.

Proof. Ker α Ker α 2 (αT n αT n ) so,

x Ker αy (x, y) Ker α (x, y) Ker α 2

Ker α classesKer α 2 classes

Figure 4. The classes of Ker α and Ker α 2.

Suppose Im α = Im α 2. If Ker α Ker α 2 then there exists ( u, v) Ker α 2 \ Ker α. Then

| Im α 2| =n

Ker α 2 <n

Ker α = | Im α |

a contradiction. Hence Ker α = Ker α 2.

Hence: α lies in a subgroup Im α = Im α 2. An arbitrary element of T n

α α

Im α \ Im α2

So, Im α = Im α 2 tails of length 2.



SEMIGROUP THEORY 27

Example 5.12.(1) We take an element of T 5 to be

α = 1 2 3 4 53 1 4 3 1 T 5.

This has map diagram

2

5

1 3 4

Now α has no tails with length 2 and therefore α doesn’t lie in any subgroup.(2) Let us take the constant element c1 T 5

α = 1 2 3 4 51 1 1 1 1 = c1.

This has the following map diagram1

2 3 4 5Now α has no tails of length 2, therefore α lies in a subgroup and hence α lies inHα . Note α 2 = α.

Now for any β

β H α β Hα,

β R α and β Lα, Ker β = Ker α and Im β = Im α, Ker β has classes {1, 2, 3, 4, 5} and Im β = {1}, β = α.

Therefore the maximal subgroup containing α is Hα = {α}.(3) Take the element



28 VICKY G

α = 1 2 3 4 52 3 3 3 5 .

This has map diagram

1

3 4

52

No tails of length 2. Therefore α lies in a subgroup. Hence α lies in a maximalsubgroup. Hence the maximal subgroup containing α is Hα . For any β

β H α β Hα, β R α and β Lα, Ker β = Ker α and Im β = Im α, Im β = {2, 3, 5} and Ker β has classes {1, 3}, {2, 4}, {5}.

We now gure out what the elements of Hα . We start with the idempotent. Weknow that the image of the idempotent is {2, 3, 5} and that idempotents are iden-tities on their images. Thus we must have

ε = 1 2 3 4 52 3 5 .

We also know that 1 and 3 go to the same place and 2 and 4 go to the same place.Thus we must have

ε = 1 2 3 4 53 2 3 2 5 .

We now have what the idempotent is and then the other elements of Hα are:

α = 1 2 3 4 52 3 2 3 5 = σ5

σ2 = 1 2 3 4 55 2 5 2 3 σ3 = 1 2 3 4 53 5 3 5 2

ρ = 1 2 3 4 55 3 5 3 2 ρ2 = 1 2 3 4 5

2 5 2 5 3

.These are all 6 elements because there are 3 options for the image



SEMIGROUP THEORY 29

32

22

= 6.

Check H α S3.

5.5. D and J Recall S 1aS 1 = {xay | x, y S 1}.

Denition : We say that a is J related to b if and only if

aJ b S 1aS 1 = S 1bS 1

s, t ,u,v S 1 with a = sbt b = uav

Note. If aLb, then S 1a = S 1b so

S 1aS 1 = S 1bS 1

and so aJ b, i.e. L J , dually R J .

Recall: S is simple if S is the only ideal of S . If S is simple and a, b S then

S 1aS 1 = S = S 1bS 1 so aJ band J = ω (the universal relation). Conversely if J = ω and InormS , then pick any a I and any s S . We have

s S 1sS 1 = S 1aS 1 I .Therefore I = S and S is simple. Thus we have that S is simple J = ω.

Similarly if S has a zero, then {0} and S \ { 0} are the only J -classes iff {0} and S are theonly ideals.

5.6. Composition of RelationsDenition : If ρ, λ are relations on A we dene

ρ ◦ λ = (x, y) A × A | z A with (x, z ) ρ and (z, y) λ .

Claim. If ρ, λ are equivalence relations and if ρ ◦ λ = λ ◦ ρ then ρ ◦ λ is an equivalencerelation. Also, it’s the smallest equivalence relation containing ρ λ.

Proof. Put ν = ρ ◦ λ = λ ◦ ρ• for any a A, aρaλa so aνa and ν is reexive.• Symmetric - an exercise.



30 VICKY G

• Suppose that aνbνc then there exists x, y A with

aρxλbλyρc.

From xλbλy we have xλy, so

aρxλyρc.Therefore xνc hence there exists z A such that aρxρzλc , therefore aρzλc andhence aνc. Therefore ν is transitive.

We have shown that ν is an equivalence relation. If (a, b) ρ then aρbλb so (a, b) ν .Similarly if (a, b) λ then aρaλb so (a, b) ν . Hence ρ λ ν .

Now, suppose ρ λ τ where τ is an equivalence relation. Let ( a, b) ν . Then we haveaρcλb for some c. Hence aτcτb so aτb as τ is transitive. Therefore ν τ .

Denition : D = R ◦ L , i.e. aDb c S with aR cLb.

Lemma 5.4 (The D Lemma). R ◦ L = L ◦ R

Proof. We prove , the proof of being dual. Suppose that aR ◦ L b. Then there existsc S with

aR cLbThere exists u, v,s, t S 1 with

a = cu(1)

c = av(2)

c = sb(3)

b = tc(4)

.

Put d = bu then we have

a =(1)

cu =(3)

sbu = sd,

d = bu =(4)

tcu =(1)

ta.

Therefore aLd. Also

b =(4)

tc =(2)

tav =(1)

tcuv =(4)

buv = dv.

Therefore bR d and hence a(L ◦ R )b.

Hence D is an equivalence relation – D = L R . Now by denition

H = L ∩ R L D ,H = L ∩ R R D .

As J is an equivalence relation and L R J we must have D J . This has HasseDiagram



32 VICKY G

cs = c · s,c = cs · s ′ .

Therefore cR cs = cρs .

Lemma 5.6 (Continuing Green’s Lemma) . For any c La we have ρs : H c → H cs is a bijection with inverse ρs ′ : H cs → H c. In particular – put c = a then

ρs : H a → H b and ρs : H b → H aare mutually inverse bijections. For any s S 1, λs : S → S is given by aλ s = sa .

Lemma 5.7 (Dual of Green’s Lemma) . Let a, b S be such that aLb and let t, t ′ S be such that ta = b and t′b = a. Then λt : Ra → Rb, λt ′ : Rb → Ra are mutually inverse L-class preserving bijections. In particular, for any c Ra we have λt : H c → H tc ,λ t ′ : H tc → H c are mutually inverse bijections. So, if c = a we have λt : H a → H b,λ t ′ : H b → H a are mutually inverse bijections.

Corollary 5.6. If aDh then there exists a bijection H a → H b

Proof. If aDb then there exists h S with aR hLb. There exists a bijection H a → H h byGreen’s Lemma and we also have that there exists a bijection H h → H b by the Dual of Green’s Lemma. Therefore there exists a bijection H a → H b.

Thus any two H-classes in the same D-class have the same cardinality.

Theorem 5.3 (Green’s Theorem – Strong Version) . Let H be an H-class of a semigroupS . Then either H 2 ∩ H = or H is a subgroup of S .

Proof. Suppose H 2 ∩ H = then there exists a,b,c H with ab = c. Since aR c, ρb :H a → H c is a bijection. But H a = H c = H so ρb : H → H is a bijection. Hence Hb = H .Dually, aH = H .Since b H , b = db for some d H . As bR d, d = bs for some s S 1 and thenbs = dbs d = d2. Hence H contains an idempotent (so by the Maximal SubgroupTheorem, it’s a subgroup).We have d is an identity for H . Let h, k H . Then hLd so

hd = h λh : H → H

is a bijection. Hence kλh = hk H . So H is closed. Let a H , then a2 H , henceaH = H = Ha. So, there exists a′ , a ′′ H with

d = aa ′ = a′′ aand we have

a ′′ = a ′′ d = a ′′ (aa ′) = ( a ′′ a)a ′ = da ′ = a ′ .



SEMIGROUP THEORY 33

Corollary 5.7.(i) aHa2 H a is a subgroup,

(ii) e E (S ) H e is a subgroup.

Proof.(i) We know H a is a subgroup a, a 2 H a so aHa2. If aHa2, then a2 H a ∩ (H a )2.

Hence H a ∩ (H a )2 = . So, by Green H a is a subgroup.

6. Rees Matrix SemigroupsConstruction: Let G be a group, I, Λ be non-empty sets and P = ( pλi ) a matrix. P is aΛ× I matrix over G { 0} such that every row / column of P contains at least one non-zeroentry.Denition : m0 = m0(G; I, Λ;P ) is the set

I × G × Λ { 0}with binary operation given by 0 n = 0 = n0 for all n m0 and

(i,a,λ )(k,b,µ) =0 if pλk = 0,(i,ap λk b, µ) if pλk = 0 .

Then m0 is a semigroup with zero 0 – a Rees Matrix Semigroup over G.Denition : a S is regular if there exists x S with

a = axa.S is regular if every a S is regular.If S is regular then aR b aS = bS there exists s, t S with a = bs and b = at , etc.

6.1. Rees Matrix FactsLet m0 = m0(G; I, Λ;P ) be a Rees Matrix Semigroup over a group G. Then

(1) (i ,a,λ ) is idempotent pλi = 0 and pλi = a− 1.(2) m0 is regular.(3) (i ,a,λ )R ( j, b, µ) i = j .(4) (i ,a,λ )L( j, b, µ) λ = µ.(5) (i ,a,λ )H ( j, b, µ) i = j and λ = µ.(6) The D = J -classes are {0} and m0 \ { 0} (so 0 and m0 are the only ideals).(7) m0 is 0-simple.(8) The rectangular property;

xyDx xyR xxyDy xyLy x, y m0



34 VICKY G

(9) Put H iλ = (i,a,λ ) | a G by (5) we have H iλ is an H-class (H iλ = H (i,e,λ )). If ρλi = 0 we know (i, ρ− 1

λi , λ) is an idempotent H iλ is a group, by the MaximalSubgroup Theorem. The identity is ( iρ− 1

λi , λ) and ( i ,a,λ )− 1 = ( i, ρ− 1λi a− 1, ρ− 1

λi , λ).

(10) If ρλi = 0 and ρµj = 0 then (exercise) H iλ H jµ (the bijection is ( i,a,λ ) →( j, a, µ )). The homomorphism is a bit more sophisticated.

Proof.(1) We have that

(i ,a,λ ) E (m0) (i,a,λ ) = ( i ,a,λ )( i,a,λ ), (i,a,λ ) = ( i,ap λi a, λ ), a = apλi a,

pλi = 0 and pλi = a− 1.

(2) 0 = 000 so 0 is regular. Let (i,a,λ ) m0 \ { 0} then there exists j I with pλj = 0and there exists µ Λ with pµi = 0. Now,

(i,a,λ )( j, p− 1λj a− 1 p− 1

µi , µ)( i,a,λ ) = ( i,a,λ )and hence m0 is regular.

(3) {0} is an R -class. If (i ,a,λ )R ( j, b, µ) then there exists ( k,c,ν ) m0 with

(i,a,λ ) = ( j, b, µ)(k,c,ν ) = ( j, bpµk c, ν )and so i = j . Conversely, if i = j , pick k with pµk = 0. Then

(i,a,λ ) = ( j, b, µ)(k, p− 1µk b− 1a, λ )

and consequently ( i,a,λ )R ( j, b, µ)(4) Dual.(5) This comes from (3) and (4) above.(6) {0} is a D-class and a J -class. If (i,a,λ ), ( j, b, µ) m0 then

(i,a,λ )R (i,a,µ )L( j, b, µ)so (i ,a,λ )D( j, b, µ) and so (i ,a,λ )J ( j, b, µ). Therefore D = J and {0} and m0 \{ 0}are the only classes.

(7) Let i I , then there exists λ Λ with pλi = 0 so (i, 1, λ)2 = 0. Therefore ( m0)2 = 0and so m0 is 0-simple.

(8) The 2 D-classes are zero and everything else. If xy = x = 0 then xyDx and xyR xalways. If xyR x then xyDx as R D . Suppose xyD and xy,x = 0. Then y = 0,so

x = ( i,a,λ ) y = ( j, b, µ)say. Then as xy = ( i,aρ λj b, µ) so xyR x. The result for L is dual.



SEMIGROUP THEORY 35

A nitary property is a property that captures nite nature.

Chain conditions

Denition : S a semigroup has M L if there are no innite chains

S 1a1 S 1a2 S 1a3 . . .if principal left ideals. M L is the descending chain condition (d.c.c.) on principal leftideals. Note that M R is the dual of this.

Claim (The Chain Claim) . S has M L if and only if any chain

S 1a1 S 1a2 . . .

terminates stabilizes with

S 1an = S 1an +1 = . . .

Proof. If every chain with terminates, then clearly we cannot have an ∞ strict chain

S 1a1 S 1a2 . . .So S has M L . Conversely; suppose S has M L and we have a chain

S 1a1 S 1a2 . . .

The strict inclusions are at the j i th steps

S 1a1 = S 1a2 = · · · = S 1a j 1 S 1a j 1 +1 = S 1a j 1 +2 = · · · = S 1a j 2 S 1a j 2 +1 = . . .Then S 1a j 1 S 1a j 2 . . . . As S has M L , this chain is nite with length n say. Then

S 1a j n +1 = S 1a j n +2 = . . .and our sequence has stabilised.

Denition : The ascending chain condition (a.c.c.) on principal ideals on left / right idealsM L (M R ) is dened as above but with the inclusions reversed. The analogue of the chainclaim holds.

Example 6.1. Every nite semigroup has M L , M R , M L , M R . A chain

S 1a1 S 1a2

is in particular, a chain of distinct subsets of S – but a nite semigroup S has at most 2 |s |

subsets.



SEMIGROUP THEORY 37

b = xay = ( xu)b(vy) = ( xu)(xubvy)(vy) = ( xu)2b(vy)2 = · · · = ( xu)n b(vy)n

for all n N . By ( ) there exists n with (xu)n L(xu)n +1 . Therefore

b = ( xu)n b(vy)n L(xu)n +1 b(vy)n = xu (xu)n b(vy)n = xub.Therefore bLxub, so

S 1b = S 1xub S 1ub S 1b.So S 1b = S 1ub and bLub. Dually, bR bv. Therefore a = ubvR ubLb. So aDb and J D .Consequently, D = J .

The Rectangular Property: Let S satisfy ( ). Then for all a, b S we have(i) aJ ab aDab aR ab,

(ii) bJ ab bDab bLab.

Proof. We prove (i), (ii) being dual. Now,

aJ ab aDabas D = J . Then aR ab aDab; as R D . If aJ ab then there exists x, y S 1 with

a = xaby = xa(by) = xn a(by)n

for all n. Pick n with (by)n R (by)n +1 . Then

a = xn a(by)n R xn a(by)n +1 = xn a(by)n by = aby

aR aby. Now aS 1 = abyS 1 abS 1 aS 1. Hence aS 1 = abS 1 and aR ab.

Lemma 6.1 (0-Simple Lemma). Let S have a 0 and S 2 = 0 . Then the following are equal (i) S is 0-simple,

(ii) SaS = S for all a S \ { 0},(iii) S 1aS 1 = S for all a S \ { 0},(iv) the J -classes are {0} and S \ { 0}.

Proof. (i) (iv) is a standard exercise.

(ii) (iii): Let a S \ { 0}. Then

S = SaS S 1aS 1 S and therefore S = S 1aS 1.

(iii) (iv): We know J 0 = {0}. Let a, b S \ { 0}. Then

S 1aS 1 = S = S 1bS 1

and hence aJ b. Therefore {0} and S \ { 0} are the only J -classes.



38 VICKY G

(i) (ii): Since S 2 = 0 and S 2 is an ideal, then S 2 = S . Therefore

S 3 = SS 2 = S 2 = S = 0 .

Let I = {x S | SxS = 0}. Clearly 0 I and hence I = . If x I and s S , then

0 SxsS SxS = 0.Therefore SxsS = 0 and so xs I . Dually sx I ; therefore I is an ideal. If I = S , then

S 3 = SIS,

=x I

SxS,

= 0 .

This is a contradiction, therefore I = S . Hence I = 0. Let a S \ { 0}. Then SaS is anideal and as a I we have SaS = 0. Hence SaS = S .

Corollary 6.1. Let S be completely 0-simple. Then S contains a non-zero idempotent.

Proof. Let a S \ { 0}. Then SaS = S , therefore there exists a u, v S with a = uav . So,

a = uav = u2av2 = · · · = un avn

for all n. Hence un = 0 for all n N . Pick n, m with un R un +1 , um Lum +1 . Notice

un +1 R un +2

as R is a left congruence. Similarly,

un +2 R un +3

we deduce that un R un + t for all t 0. Similarly um Lum + t for all t 0. Let s = max {m, n }.Then us R u2s , us Lu2s so us Hu2s = ( us )2. Hence us lies in a subgroup. Therefore us He forsome idempotent e. As us = 0 and H 0 = {0}, we have e = 0.

Theorem 6.2 (Rees’ Theorem - 1941). Let S be a semigroup with zero. Then S is com-pletely 0-simple S is isomorphic to a Rees Matrix Semigroup over a group.

Proof. If S = m

0

(G; I ; Λ;P ) where G is a group, we know m0

is completely 0-simple (byRees Matrix facts), hence S is completely 0-simple.

Conversely, suppose S is completely 0-simple. By the D = J Theorem, D = J (as S hasM R and M L , it must have ( )). As S is 0-simple, the D = J -classes are {0} and S \ { 0}.Let D = S \ { 0}. By the Corollary to the 0-simple Lemma, D contains e = e2.

Let {R i | i I } be the set of R-classes in D (so I indexes the non-zero R-classes). Let{Lλ | λ Λ} be the set of L-classes in D (so Λ index the non-zero L-classes).



40 VICKY G

(i,a,λ )(k,b,µ) θ =0θ if pλk = 0,(i,ap λk b, µ)θ if pλk = 0 ,

=0 if pλk = 0,r iapλk bq µ if pλk = 0 ,

= r iapλk bq µ ,= r iaq λ r k bq µ ,= ( i ,a,λ )θ(k,b,µ)θ.

Therefore θ is an isomorphism.

7. Regular SemigroupsDenition : a S is regular if a = axa for some x S . S is regular if every a S isregular.

Denition : a′ S is an inverse of a if a = aa ′a and a′ = a′aa ′ . We denote V (a) to bethe set of inverses of a.

Caution: Inverses need not be unique. In a rectangular band T = I × Λ then

(i, j )(k, ℓ)( i, j ) = ( i, j )

(k, ℓ)( i, j )(k, ℓ) = ( k, ℓ)for any (i, j ) and (k, ℓ). So every element is an inverse of every other element. If G is agroup then V (a) = {a− 1} for all a G.

Lemma 7.1 (Lemma R) . a is regular V (a) = .

Proof. If V (a) = , clearly a is regular. Conversely suppose that a is regular. Then thereexists x S with a = axa . Put a′ = xax . Then

aa ′a = a(xax )a = ( axa )xa = axa = a,

a′

aa′

= ( xax )a(xax ) = x(axa )(xax ) = xa(xax ) = x(axa )x = xax = a.So a′ V (a).

Note. If a = axa then

(ax )2 = ( ax )(ax) = ( axa )x = ax

so ax E (S ) and dually, xa E (S ). Moreover



SEMIGROUP THEORY 41

a = axa ax = ax aR ax,a = axa xa = xa aLxa.

a

xa

ax

Figure 5. The egg box diagram of E (S ).

Denition : S is inverse if |V (a)| = 1 for all a S , i.e. every element has a uniqueinverse.

Example 7.1.(1) Groups are inverse; V (a) = {a− 1},(2) A rectangular band T is regular; but (as every element of T is an inverse of every

other element) T is not inverse (unless T is trivial),(3) If S is a band then S is regular as e = e3 for all e S ; S need not be inverse,(4) B is regular because (a, b) = ( a, b)(b, a)(a, b) for all (a, b) B then B is inverse –

see later,(5) m0 is regular (see “Rees Matrix Facts”),(6) T X is regular (see Exercises),(7) (N , +) is not regular as, for example 1 = 1 + a + 1 for any a N .

Theorem 7.1 (Inverse Semigroup Theorem) . A semigroup S is inverse iff S is regular and E (S ) is a semilattice (i.e. ef = f e for all e, g E (S )).

Proof. ( ) Let a S . As S is regular, a has an inverse by Lemma R. Suppose x, y V (a).Then

a =(1) axa x =(2) xax a =(3) aya y =(4) yay,so ax,xa,ay,ya E (S ). This gives us that

x =(2)

xax =(3)

x(aya )x = ( xa )(ya)x = ( ya)(xa )x = y(axa )x

=(1)

yax =(3)

y(aya )x = y(ay)(ax) = y(ax)(ay) = y(axa )y =(1)

yay =(4)

y.



42 VICKY G

So |V (a)| = 1 and S is inverse. Conversely suppose S is inverse. Certainly S is regular.Let e E (S ). Then e = e′ as e = eee, e = eee. Let x = ( ef ) ′ . Consider fxe . Then

(fxe )2

= ( fxe )(fxe ) = f (xefx )e = f xeas x = ( ef ) ′ . So (fxe ) E (S ) and therefore fxe = ( fxe ) ′ . We want to show that fxeand ef are mutually inverse, i.e.

ef (fxe )ef = ef 2xe2f = efxef = ef,(fxe )ef (fxe ) = f xe 2f 2xe = f x (efx )e = fxe.

Therefore we have ef = ( fxe )′ = f xe E (S ). Therefore E (S ) is a band. Let e, f E (S ).Then

ef (fe )ef = ef 2

e2

f = efef = ef,fe (ef )fe = f e

similarly. Therefore we have ef = ( fe ) ′ = f e .

Example 7.2.(1) E (B) = (a, a ) | a N 0 because

(a, a )(b, b) = ( t, t ) = ( b, b)(a, a )where t = max {a, b}. Therefore B is inverse and note (a, b)′ = ( b, a.

(2) T X – we know T X is regular. For |X | 2 let x, y X with x = y we havec

x, c

y E (

T X ). Then c

xc

y = c

yc

x so

T X is not inverse.

(3) S a band – then S is regular. We have

S is inverse ef = f e for all e, f E (S ), ef = f e for all e, f S, S is a semilattice .

7.1. Green’s Theory for Regular D-classesIf e E (S ) then H e is a subgroup of S (by the Maximal Subgroup Theorem or Green’sTheorem). If eDf then |H e | = |H f | (by the Corollary to Green’s Lemmas). We will showthat H e = H f .

Lemma 7.2. We have that (i) If a = axa then ax, xa E (S ) and aR xa ,

(ii) If bR f E (S ), then b is regular,(iii) If bLf E (S ), then b is regular.

Proof.



SEMIGROUP THEORY 43

(i) we know this.(ii) If bR f then fb = b. Also, f = bs for some s S 1. Therefore b = f b = bsb and so

b is regular.

(iii) this is dual to (ii).

Lemma 7.3 (Regular D-class Lemma) . If aDb then if a is regular, so is b.

Proof. Let a be regular with aDb . Then aR cLb for some c S .

a

b

ce

f

Figure 6. The egg box diagram of D.

There exists e = e2 with eR aR c by (ii) above. By (ii), c is regular. By (i), cLf = f 2. By(iii), b is regular.

Corollary 7.1 (Corollary to Green’s Lemmas) . Let e, f E (S ) with eDf . Then H e = H f .

Proof. Suppose e, f E (S ) and eDf . There exists a S with eR aLf .

a

f

e

As eR a there exists s S 1 with e = as and ea = a. So a = asa . Put x = f se . Then

ax = afse = ase = e2 = eand so a = ea = axa . Since aLf there exists t S 1 with ta = f . Then

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