Semigroup Theory [Lecture notes]

44 Pages·2008·0.272 MB·English

by  Vicky Gould, J Taylor

SEMIGROUP THEORY
VICKY G
1.
The Basic Concept
Definition: A semigroup is a pair (S,∗) where S is a non-empty set and ∗ is an associative
binaryoperationonS. [i.e.∗isafunctionS×S → S with(a,b) 7→ a∗bandforalla,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 induction
a a ...a is unambiguous. Thus we write an for
1 2 n
aa...a.
ntimes
Index Laws: We have that for all n,m ∈ N = {1,2,...} the following index laws hold
| {z }
anam = an+m
an m = anm.
Definition: A monoid M is a semig(cid:0)rou(cid:1)p with an identity, i.e. there exists 1 ∈ M such
that 1a = a = a1 for all a ∈ M.
Putting a0 = 1 then the index laws hold for all n,m ∈ N0.
Note. The identity of a monoid is unique.
Definition: A group G is a monoid such that for all a ∈ G there exists a b ∈ G with
ab = 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 table
e
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
2 VICKYG
Example 1.3. N is a monoid under ×. N forms a semigroup under + and N0 is a monoid
under + 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 onI×J. To verify thatT is a semigroup
we 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),
(cid:0) (cid:1)
for all (i,j),(k,ℓ),(m,n) ∈ T(cid:0)and hence m(cid:1)ultiplication is associative.
J
j ℓ
i (i,j) (i,ℓ)
I
k (k,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 be
looked at from ring theory because any ring where every element is an idempotent means
that the ring is commutative. However, the rectangular band does not have to be commu-
tative.
1.1. Adjoining an Element
Let S be a semigroup, which is not a monoid. Find a symbol not in S, call it “1”. We now
extend the definition of ∗ on S to S ∪{1} by
SEMIGROUP THEORY 3
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 S1 is defined by
S if S is a monoid,
S1 =
S ∪{1} if S is not a monoid.
(
So, S1 is “S with a 1 adjoined if necessary”.
Example 1.5. Let T be the rectangular band on {a}×{b,c}. Then T1 = {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| < ∞, then
maxA is the biggest element in A. We also use this notation to represent
a if a > b,
max{a,b} =
b if b > a.
(
We note some further things about max:
• max{a,0} = a if a ∈ N0,
• 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 (N0,max) is a monoid.
(cid:8) (cid:9) (cid:8) (cid:9)
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}.
(cid:26)
Put B = N0 ×N0. Now, on B we define 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.
4 VICKYG
Proof. With (a,b),(c,d) ∈ B and t = max{b,c} we have t−b > 0 and t−c > 0. Thus we
have a−b+t > a and d−c+t > d. Therefore, in particular (a−b+t,d−c+t) ∈ B so
multiplication 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}
(cid:0) (cid:1) (cid:0) (cid:1)
f −e+max{d−c+max{b,c},e} .
(cid:0)
(a,b) (c,d)(e,f) = (a,b) c−d+max{d,e},f −e+max{d,e} ,
(cid:1)
= (a−b+max{b,c−d+max{d,e}}
(cid:0) (cid:1) (cid:0) (cid:1)
f −e−c+d+max{b,c−d+max{d,e}} .
Now we have to show that (cid:1)
(cid:24)a−(cid:24)(cid:24)b−d+c+max d−c+max{b,c},e =(cid:24)a−(cid:24)(cid:24)b+max b,c−d+max{d,e} ,
(cid:24)f (cid:24)−(cid:24)e+max d−c+max{b,c},e =(cid:24)f (cid:24)−(cid:24)e−c+d+max b,c−d+max{d,e} .
(cid:8) (cid:9) (cid:8) (cid:9)
We can see that thes(cid:8)e equations are the s(cid:9)ame and so we only need(cid:8)to show (cid:9)
c−d+max d−c+max{b,c},e = max b,c−d+max{d,e} .
Now, we have from ((cid:8)⋆) that (cid:9) (cid:8) (cid:9)
max max{b,c},c−d+e = max b,c−d+max{d,e} .
The RHS of this equ(cid:8)ation is (cid:9) (cid:8) (cid:9)
max b,c−d+max{d,e} = max b,max{c−d+d,c−d+e} ,
= max b,max{c,c−d+e} ,
(cid:8) (cid:9) (cid:8) (cid:9)
= max{b,c,c−d+e},
(cid:8) (cid:9)
= max max{b,c},c−d+e .
Therefore multiplication is associative and hence B(cid:8)is a semigroup/mon(cid:9)oid. (cid:3)
SEMIGROUP THEORY 5
Example 1.7. T is a semigroup. See the examples class for proof.
X
1.2. Easy Facts / Definitions
A semigroup S is commutative if ab = ba for all a,b ∈ S. For example N with + is
commutative. 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).
Thuswehave(0,1)(1,0) 6= (1,0)(0,1). NoticethatinB; (a,b)(b,c) = (a,c). Itisimportant
to see that in general we have NO cancellation in Semigroups. Thus we have that
ac = bc 6⇒ a = b,
ca = cb 6⇒ 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) 6= (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 subsemigroup
of a group is cancellative). N0 is a cancellative monoid, which is not a group.
1.3. Zeros
A 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 S0 = S ∪ {0}; define a binary
operation · on S0 by
a·b = ab for all a ∈ S,
0·a = 0 = a·0 for all a ∈ S,
0·0 = 0.
6 VICKYG
Then · is associative, so S0 is a semigroup with zero. We say that “S is a semigroup with
a zero adjoined”.
2.
Reminder of Familiar Ideas
Example 2.1 (Function). Let A = {1,2,3} and B = {a,b,c} then we define a function
f : A → B by f(1) = a, f(2) = a, f(3) = c. We can also describe this using two row
notation
1 2 3
f = .
a b c
(cid:18) (cid:19)
We write xf rather than f(x). Here 1f = a = 2f, 3f = c.
Binary Operations
Let A be a set. A binary operation ∗ on A is a function from A×A → A. We write a∗b
for (a,b)∗.
Composition of Functions
Let f : A → B and g : B → C then we define the composition of these functions,
f ◦g : A → C, to be
a(f ◦g) = (af)g.
Now given another function h : C → D then the composition of these functions is associa-
tive. For any a ∈ A we have
a (f ◦g)◦h = a(f ◦g) h,
= (af)g h,
(cid:0) (cid:1) (cid:0) (cid:1)
= (af)(g ◦h),
(cid:0) (cid:1)
= a f ◦(g ◦h) .
(cid:0) (cid:1)
Identity Function
For any set X we define the identity function I : X → X such that for all x ∈ X
X
xI = x.
X
Also, if f : X → Y a function then
I ◦f = f = f ◦I .
X Y
Let X be a set such that X 6= ∅. Recall that S = {α : X → X | α is a bijection}
X
is a group under ◦, called the symmetric group on X. Usually write S for S where
n n
n = {1,2,...,n}.
SEMIGROUP THEORY 7
Theorem 2.1 (Cayley). If G is a group then there exists a 1:1 homomorphism θ : G → S .
G
Definition: T = {α | α is a function X → X}.
X
We write T for T . This has size nn because that’s how many ways there are of choosing
n n
a function on n elements. T is a monoid with identity I called the full transformation
X X
monoid / semigroup on X.
Claim. Clearly S ⊆ T and if |X| > 2 then S 6= T .
X X X X
Proof. Foranyn ∈ X letc ∈ T betheconstant map onn, i.e.yc = xforally ∈ X. Note
n X x
that Imc = {n} and if |X| > 2 then c is neither one-to-one nor onto, so c ∈ S \T .
n n x X X
Now notice αc = c for all α ∈ T .
x x X
We show this by letting y ∈ X, then
y(αc ) = (yα)c = x = yc .
x x x
Hence αc = c . Thus c is a right zero for T . Furthermore αc = βc for all α,β ∈ T
x x x X x x X
so T is not cancellative for |X| > 2. (cid:3)
X
If α ∈ S then α can be written in “two row” notation or as a product of disjoint cycles.
n
For example
1 2 3 4 5
α = ∈ S
2 3 1 5 4 5
(cid:18) (cid:19)
= (123)(45).
We can also express this pictorially as the cycle diagram of α. For T we can use “two
X
row” notation. For example if α ∈ T is given by 1α = 2,2α = 2,3α = 3,4α = 1,5α = 1.
5
Then,
1 2 3 4 5
α = .
2 2 3 1 1
(cid:18) (cid:19)
For example,
1 2 3 4 5
c = ∈ T .
2 2 2 2 2 2 5
(cid:18) (cid:19)
We have map diagrams for T.
3.
Subsemigroups / Submonoids
Definition: Let S be a semigroup and ∅ =6 T ⊂ S. Then T is a subsemigroup of S if
a,b ∈ T ⇒ ab ∈ T. If S is a monoid then T is a submonoid of S if T is a subsemigroup
and 1 ∈ T.
Example 3.1. (N,+) is a subsemigroup of (Z,+).
8 VICKYG
Example 3.2. R = {c | x ∈ X} is a subsemigroup of T called the right zero semigroup.
x X
Proof. c c = c for all x,y ∈ X. (cid:3)
x y y
Example 3.3 (Bicyclic Monoid). Put E(B) = {(a,a) | a ∈ N0}, the set of all idempotents
inB. WeclaimthatE(B)isacommutativesubmonoidofB. Clearlywehave(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. Morphisms
Definition: 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 and
1 θ = 1 .
S T
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),
(cid:0) (cid:1)
= (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 define α : T → T by
J
(i,j)α = c . Then we have
j
(i,j)(k,ℓ) α = (i,ℓ)α,
= c ,
(cid:0) (cid:1) ℓ
= c c ,
j ℓ
= (i,j)α(k,ℓ)α.
So, α is a morphism.
Definition: A bijective morphism is an isomorphism.
Isomorphisms preserve algebraic properties (e.g. commutativity). Suppose f : S → T is a
morphism then with A ⊆ S we have
SEMIGROUP THEORY 9
Af = {af | a ∈ A}.
If A = S then Sf is the image, Imf, of S. So, if A is a subsemigroup of S then Af will
be a subsemigroup of T. Now, conversely, if B ⊆ T then we define
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, Imα is a subsemigroup of T. If α is
1:1 then α : S → Imα 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 = S1. We need a 1:1 morphism S → T . For
X
s ∈ S, we define ρ ∈ T by xρ = xs. Now, define α : S → T by sα = ρ . First show
s X s X s
that α is 1:1. If sα = tα then ρ = ρ and so xρ = xρ for all x ∈ S1; in particular
s t s t
1ρ = 1ρ and so 1s = 1t hence s = t and α is 1:1.
s t
Let u,b ∈ S. For any x ∈ X we have
x(ρ ρ ) = (xρ )ρ ) = (xu)v = x(uv) = xρ .
u v u v uv
Hence ρ ρ = ρ and so uαvα = ρ ρ = ρ = (uv)α. Therefore α is a morphism. Hence
u v uv u v uv
α : S → T is an embedding. (cid:3)
X
Theorem 3.2 (The “Cayley Theorem” - for Monoids). Let S be a monoid then there exists
an embedding S ֒→ T for some X.
X
Proof. We know from the proof of the Cayley Theorem for Semigroups that α : S → T ,
X
given by sα = ρ where xρ = xs, is a semigroup embedding. In the proof above we had
s s
X = S1 ⇒ X = S because S is a monoid. We must check that α is a monoid morphism,
i.e. we need to check 1α = I . Now 1α = ρ and for all x ∈ X = S we have
X 1
xρ = x1 = x = xI
1 X
and so 1α = ρ = I . (cid:3)
1 X
Theorem 3.3 (The Cayley Theorem - for Groups). Let S be a group. Then there exists
an embedding S ֒→ S for some X.
X
Proof. Exercise. (cid:3)
10 VICKYG
3.2. Idempotents
Definition: e ∈ S is an idempotent if e2 = e. Also we define the set of idempotents in S
to be
E(S) = {e ∈ S | e2 = e}.
Now, E(S) may beempty, e.g.E(S) = ∅ (Nunder +) butE(S) mayalso beS. 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 so
E(S) = S.
Definition: If E(S) = S, then S is a band.
For the bicyclic semigroup B we have E(B) = (a,a) | a ∈ N0 – from exercises 1. If S is
a monoid then 1 ∈ E(S). If S is a cancellative monoid, then 1 is the only idempotent: for
(cid:8) (cid:9)
if e2 = e then ee = e1 and so e = 1 by cancellation. In particular for S a group we have
E(S) = {1}.
Lemma 3.1. Suppose ef = fe 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). (cid:3)
Definition: 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 ∈ N0 is a semilattice.
Example 3.7. A rectan(cid:8)gular band I ×(cid:9)J is not a semilattice (unless |I| = |J| = 1) since
(i,j)(k,ℓ) = (k,ℓ)(i,j) ⇔ i = k and j = ℓ.
Definition: Let a ∈ S. Then we define hai = {an | n ∈ N}, which is a commutative
subsemigroup of S. We call hai the monogenic subsemigroup of S generated by a.
Proposition. hai ∼= (N,+) or hai is finite.
Proof. If ai 6= aj for all i,j ∈ N with i 6= j then θ : hai → N defined by aiθ = i is an
isomorphism. Suppose that in the list of elements a,a2,a3,... there is a repetition, i.e.
ai = aj for some i < j. Let k be least such that ak = an for some n < k. Then k = n+r
for some n,r ∈ N – where n is the index of a, r is the period of a. Then the elements
a,a2,a3,...,an+r−1 are all distinct and an = an+r.
We can express this pictorially by a map diagram.