lecture notes on semigroups - semantic scholar · lecture notes on semigroups tero harju department...

Lecture Notes on

SEMIGROUPS

Tero HarjuDepartment of Mathematics

University of TurkuFIN-20014 Turku, Finland

19961

1 Small corrections in 2010

Contents

1 Basic Concepts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 51.2 Subsemigroups and Direct Products . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 91.3 Homomorphisms and transformations . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 111.4 Partial orders and semilattices . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 16

2 General Structure Results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.1 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 182.2 Homomorphism theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 212.3 Ideals and Rees congruences . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 24

3 Free Semigroups and Presentations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.1 Free semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 263.2 Presentations of semigroups . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 313.3 Embeddings into 2-generator semigroups . . . . . . . . . . . . .. . . . . . . . . . . . . 35

4 Combinatorial Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.1 The defect theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 374.2 Ehrenfeucht’s conjecture . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 404.3 Isbell’s Zig-Zag Theorem for Dominions . . . . . . . . . . . . . .. . . . . . . . . . . . . 49

5 Green’s Relations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 525.2 Green’s Lemma and Its Corollaries . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 56

6 Inverse Semigroups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.1 Regular Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 616.2 Inverse Semigroups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 636.3 Representations by Injective Partial Mappings . . . . . . .. . . . . . . . . . . . . . . 676.4 Congruences of Inverse semigroups . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 70

Contents 3

6.5 Free Inverse Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 766.6 Concrete Free Inverse Semigroups . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 80

Books on semigroups:

A.H. Clifford and G.B. Preston,The algebraic theory of semigroups, Vol. I & II, Math-ematical Surveys of the Amer. Math. Soc.7 (1961 & 1967). [A two-volume ‘handbook’on semigroups from their algebraic point of view.]

P.M. Higgins, Techniques of semigroup theory, Oxford University Press, 1992 [Goes tothe advanced topics rather fast. Contains up-to-date proofs for free inverse semigroups,topics on biordered sets, Isbell’s zig-zags, and some combinatorics of transformationsemigroups.]

J.M. Howie, An introduction to semigroup theory, Academic Press, 1976. [An easyintroduction. See the next entry.]

J.M. Howie, Fundamentals of semigroup theory, 1995. [Probably the best introduction.Contains more advanced material such as Isbell’s zig-zags,and the free inverse semi-groups].

G. Lallement, Semigroups and combinatorial applications, Wiley, 1979. [Concentrateson combinatorial topics such as codes and free word semigroups, formal languages andrelated topics.]

E.S. Lyapin,Semigroups, Translations of Math. Monographs, Vol 3, Amer. Math. Soc.,1974. [Easy to read.]

M. Petrich,Introduction to semigroups, Merrill, 1973. [More advanced topics from thestructural point of view.]

M. Petrich,Lectures in semigroups, Wiley, 1977. [Bands, varieties, and lattices of vari-eties.]

M. Petrich,Inverse semigroups, Wiley, 1984. [The bible on inverse semigroup.]

M. Petrich and N.R. Reilly,Completely regular semigroups, in preparation. [A largebook on one of the popular themes in semigroup theory.]

Related books:

J. Berstel and D. Perrin,Theory of Codes, Academic Press, 1985. [Formal languages,codes, word semigroups.]

S. Burris and H.P. Sankappanavar,A course in universal algebra, Springer, 1981. [Agood introduction to general properties of algebras.]

P.M. Cohn,Universal algebra, Reidel, 1981. [A more comprehensive treatment of uni-versal algebras.]

S. Eilenberg,Automata, Languages, and Machines, Vol. A, Academic Press, 1974. [Oneof the basic books on automata.]

M. Lothaire,Combinatorics on Words, Addison-Wesley, 1983. [Combinatorial topic onwords and related areas.]

Contents 4

R.C. Lyndon and P. Schupp,Combinatorial group theory, Springer, 1977. [Containssome theory of free groups among other topics.]

W. Magnus, A. Karrass and D. Solitar,Combinatorial group theory, Dover, 1976 [Con-tains some theory of free groups among other topics.]

A. Salomaa and G. Rozenberg (eds),Handbook of formal languages, Vol. I, Springer,1996. [A collection of survey articles by various authors oncombinatorial topics onwords, and formal languages in general. Includes two proofsof Ehrenfeucht’s conjec-ture; one of which is in these Lecture Notes. Includes also many other interesting resultsof combinatorial nature.]

1

Basic Concepts

1.1 Definitions and examples

Semigroups

LetS be a set andσ : S×S → S a binary operation that maps each ordered pair(x, y) ofS to an elementσ(x, y) of S. The pair(S, σ) (or justS, if there is no fear of confusion)is called agroupoid. The mappingσ is called aproduct of (S, σ). We shall mostly writesimply xy instead ofσ(x, y). If we want to emphasize the place of the operation, thenwe often writex · y. The elementxy (= σ(x, y)) is theproduct of x andy in S.

We also use other notations for the product, when this is natural or otherwise conve-nient. In particular, the following symbols may be used:·, +, ⋆, ◦, ⊕, ⊗. If we do notexplicitly (or implicitly) mention the (notation for the) product, we choose the dot:x · y.

A groupoidS is asemigroup, if the operationσ is associative: for all x, y, z ∈ S,

x · (y · z) = (x · y) · z .

That is,σ(x, σ(y, z)) = σ(σ(x, y), z) using the unfriendly notation. This means that theorder in which the operationσ is carried out is irrelevant, and therefore we may write

x1x2 . . . xn = x1 · (x2 · (· · · · xn)) . . . ) .

Of course, the order of the operandsx1, . . . , xn (where we can have repeated elements,xi = xj), is important.

For an elementx ∈ S we letxn be the product ofx with itself n times. So,x1 = x,x2 = x · x, andxn+1 = x · xn for n ≥ 1.

• S is afinite semigroup if it has only a finitely many elements.• S is acommutative semigroup, if it satisfies

∀x, y ∈ S : x · y = y · x .

We letN = {0, 1, . . . } be the set of all non-negative integers, andN+ = {1, 2, . . . }the set of all positive integers.

Example 1.1.(1) (N, ·) is a semigroup for the usual multiplication of integers,n · m.Also (N,+) is a semigroup, when+ is the ordinary addition of integers. Define(N, ⋆)by

n ⋆ m = max{n,m} .

1.1 Definitions and examples 6

(N, ⋆) is a semigroup, since

n ⋆ (m ⋆ k) = max{n,max{m,k}} = max{n,m, k}

= max{max{n,m}, k} = (n ⋆ m) ⋆ k .

(2) The previous case generalizes to other familiar sets of numbers:(Z, ·), (Z,+),(Q, ·), (Q,+), (R, ·), (R,+), (C, ·), (C,+) of integers, rationals, reals and complexnumbers. In the multiplicative cases one may remove the zeroelement0 from the do-main, and still obtain a semigroup.

(3) Consider the upper triangular integer matrices

S =

{(1 n0 1

) ∣∣ n ≥ 1

}.

ThenS is a semigroup with the usual matrix product. All the semigroups in the aboveexamples have been commutative.

(4) Let TX be the set of all functionsα : X → X. Then(TX , ◦) is a semigroup,where◦ is the composition of functions:(β ◦ α)(x) = β(α(x)) for all x ∈ X.

(5) LetS = {a, b, c} be a set of three elements, and define the product in the table.

ThenS is a finite semigroup. In order to check this we should runthrough all the triples:x · (y · z) = (x · y) · z. In this case, thereare some helpful restrictions inS. For instance, ifz = c, then theproduct is alwaysc no matter whatx andy are. ⊓⊔

· a b c

a a b cb b a cc c b c

Monoids and groups

Let S (that is,(S, ·)) be a semigroup. An elementx ∈ S is a left identity of S, if

∀y ∈ S : x · y = y .

Similarly, x is aright identity of S, if

∀y ∈ S : y · x = y .

If x is both a left and a right identity ofS, thenx is called anidentity of S. A semigroupis amonoid, if it has an identity.

Lemma 1.1.A semigroupS can have at most one identity. In fact, ifS has a left identityx and a right identityy, thenx = y. In particular, the identity of a monoid is unique.

Proof. By the definitions,y = xy = x. ⊓⊔


The identity of a monoidS is usually denoted by1S , or just by1, for short. Of course,if the monoidS has an established identity, we do not change its notation. So, the identityof the monoid(N,+) is 0.

For a semigroupS we define a monoidS1 by adjoining an identity toS, if S doesnot have one:

S1 =

{S if S is a monoid,

S ∪ {1} if S is not a monoid,

where1 is a (new) identity element.A monoidG is agroup, if everyx ∈ G has a (group) inversex−1 ∈ G:

x · x−1 = 1 = x−1 · x .

Although semigroups and groups seem to be rather close to each other, semigroups lackthe basic symmetry properties of groups, and therefore the general theory of semigroupsdiffers quite a lot from the theory of groups. The theory of semigroups is more applicablein theoretical computer science and also in some parts of combinatorial mathematics.

Zeros and idempotents

An elementx ∈ S is a left zero, if

∀y ∈ S : x · y = x .

Similarly, x is aright zero, if

∀y ∈ S : y · x = x .

A zero element is both a left and a right zero ofS.

Lemma 1.2.A semigroup can have at most one zero.

Proof. Exercise. ⊓⊔

An elemente ∈ S is an idempotent, if e2 = e. The set of all idempotents ofS isdenoted byE = ES .

Example 1.2.(1) Consider the finite semigroupS as defined above in Example 1.1(5).The elementa is an identity ofS, and henceS is a monoid. The elementc is a right zero.There are no left zeros, and henceS does not have a zero. Botha andc are idempotents.Indeed, all left and right identities and zeros are idempotents. The elementb is not anidempotent, sinceb2 = a.

(2) The matrix semigroup from Example 1.1(3) has no idempotents let alone identi-ties or zeros.

(3) Consider the set of alln-by-n-matrices with integer entries. This is a monoidZn×n. For instance,


Z2×2 =

{(a bc d

) ∣∣ a, b, c, d ∈ Z

}.

The identity matrixI is the identity ofZ2×2 and the zero matrix0 is its zero:

I =

(1 00 1

)and 0 =

(0 00 0

).

This monoid has quite many idempotents. As an example,

(−1 −12 2

)is an idempotent.

(4) LetGL(n,R) denote the monoid of all nonsingularn-by-n-matricesA with realentries – so thatdet(A) 6= 0. Apart from the identity matrixGL(n,R) has no otheridempotents, because it is a group, called thegeneralized linear groupoverR. Alsothe monoidSL(n,R) of all n-by-n-matricesA in Rn×n such thatdet(A) = 1 forms agroup. This is thespecial linear groupoverR.

(5) Let us define a product in a setS by

∀x, y ∈ S : x · y = x .

We obtain thus a semigroup, where each elementx ∈ S is a right identity and at thesame time a left zero. Moreover, all the elements in this semigroup are idempotents,S = ES . ⊓⊔

Lemma 1.3.If G is a group, thenEG = {1}.


Example 1.3.For any two subsetsA,B ⊆ S of a semigroupS, define their product by

A ·B = {a · b | a ∈ A, b ∈ B} .

The so obtained binary operation is associative, and hence the subsets ofS form a semi-group, theglobal semigroupof S. We denote this new semigroup by2S . For a subsetA ⊆ S, we let, as usual,A2 = A · A. ⊓⊔

Cancellation semigroups

A semigroupS is left cancellative, if

zx = zy =⇒ x = y ,

andS is right cancellative, if

xz = yz =⇒ x = y .

If S is both left and right cancellative, then it iscancellative.

1.2 Subsemigroups and Direct Products 9

Example 1.4.(1) All groups are cancellative semigroups.

(2) Consider the matrix semigroupZ2×2. Here(1 00 0

)(1 11 1

)=

(1 00 0

)(1 10 0

),

and henceZ2×2 is not cancellative. However, if we take all matricesA with det(A) = 1,then this semigroup is cancellative. ⊓⊔

1.2 Subsemigroups and Direct Products

Subsemigroups

Let A 6= ∅ be a (nonempty) subset of a semigroup(S, ·). We say that(A, ·) is a sub-semigroupof S, denoted byA ≤ S, if A is closed under the product ofS:

∀x, y ∈ A : x · y ∈ A ,

that is,A ≤ S ⇐⇒ A2 ⊆ A .

A subsemigroup uses the operation of its mother semigroup, and hence ifA is asubsemigroup ofS, then certainly the operation ofA is associative, and thusA is asemigroup by its own right.

Example 1.5.Consider the additive semigroupS = (Q,+) of rationals. Now,(N,+) isa subsemigroup ofS, but (N, ·) is not, because its operation (multiplication) is not theoperation ofS. ⊓⊔

Lemma 1.4.Let Ai ≤ S be subsemigroups ofS for all i ∈ I. If their intersection isnonempty, then ⋂

i∈I

Ai ≤ S .

Proof. Suppose the intersection is nonempty. Ifx, y ∈ A = ∩i∈IAi, thenx, y ∈ Ai foreachi ∈ I, and hencexy ∈ A as required. ⊓⊔

For a subsetX ⊆ S with X 6= ∅ denote

[X]S =⋂

{A | X ⊆ A, A ≤ S} .

By Lemma 1.4,[X]S is a subsemigroup ofS, called thesubsemigroup generated byX.It is the smallest subsemigroup ofS that containsX. Again, [X]S is sometimes writtensimply [X], if the semigroupS is clear from the context.

WhenX is a singleton set,X = {x}, then we write[x]S rather than[{x}]S . Moregenerally, ifX = {x1, x2, . . . } is finite or infinite, then we write[x1, x2, . . . ]S insteadof [{x1, x2, . . . }]S .

1.2 Subsemigroups and Direct Products 10

Theorem 1.1.LetX 6= ∅,X ⊆ S for a semigroupS. Then

[X]S =

∞⋃

n=1

Xn = {x1x2 . . . xn | n ≥ 1, xi ∈ X} .

Proof. Write A = ∪∞n=1X

n. It is easy to show thatA ≤ S. Also,Xn ⊆ [X]S for alln ≥ 1, since[X]S ≤ S, and hence the claim follows. ⊓⊔

Example 1.6.(1) Let S be the finite semigroup from Example 1.1(5). Now,[a]S ={a, a2, a3, . . . } = {a}, and hence{a} is a subsemigroup ofS. Also, [c]S = {c}. On theother hand,[b]S = {a, b} = [a, b]S , and[b, c]S = {a, b, c} = S. Here{b, c} generatesthe full semigroupS.

(2) Consider the matrix semigroupS = Z2×2, and let

M =

(0 1

−1 0

).

Now,

M2 =

(−1 00 −1

), M3 =

(0 −11 0

), M4 =

(1 00 1

), M5 =M,

and hence[M ]S = {I,M,M2,M3} is a finite subsemigroup ofS. In fact, [M ]S is asubgroup ofS, that is, a subsemigroup which is a group. Note thatZ2×2 is not a groupitself. ⊓⊔

We say that a subsetX of a monoidM generatesM (as a monoid), if [X]M = Mor [X]M = M \ {1M}. Hence in a monoid the identity element is always taken intogranted – a generator set need not produce it.

Index and period

If X = {x} is a singleton subset of a semigroupS, then[x]S = {x, x2, x3, . . . }, is calleda monogenic semigroup(also called acyclic semigroup). Hence[x]S = {xn | n ≥ 1}consists of the positive powers ofx. There are two possibilities for[x]S : Either it isan infinite monogenic semigroup, wherexn 6= xm for all n 6= m, or there existsan integerk such that[x]S = {x, x2, . . . , xk−1}. In the latter casexk = xr for some1 ≤ r ≤ k − 1. Herer is called theindex andp = k − r theperiod of x. You shouldnotice thatxr+p = xr, xr+p+1 = xr+1, . . . ,xr+p+p = xr+p = xr, and so forth, that is,

∀n ≥ r : xn = xn+p .

Lemma 1.5.Let r be the index andp the period of an elementx ∈ S. Then

Kx = {xr, xr+1, . . . , xr+p−1}

is a subgroup (i.e., a subsemigroup that is a group) ofS.

Proof. A nontrivial exercise. ⊓⊔

1.3 Homomorphisms and transformations 11

Direct products

Thedirect product S × T of two semigroupsS andT is defined by

(x1, y1) · (x2, y2) = (x1x2, y1y2) (xi ∈ S, yi ∈ T ) .

It is easy to show that the so defined product is associative onthe pairs(x, y) ∈ S × T ,and hence the direct product is, indeed, a semigroup.

Example 1.7.LetS = (N,+) andT = (N, ·). Then in the direct productS×T we have(n, r) · (m, s) = (n+m, rs). ⊓⊔

The direct product is a convenient way of combining two semigroup operations. Thenew semigroupS × T inherits properties of bothS andT .

The mappingsπ1 : S × T → S andπ2 : S × T → T such thatπ1(x, y) = x andπ2(x, y) = y are called theprojections of the direct productS × T .

In generalS×T 6= T ×S, as already verified in the previous example. However, thedirect product operation is associative on semigroups:S× (T ×U) = (S×T )×U , andhence we can defineS1×S2×· · ·×Sn as the (finite) direct product of the semigroupsSi.

1.3 Homomorphisms and transformations

Definition

Let (S, ·) and(P, ⋆) be two semigroups. A mappingα : S → P is ahomomorphism, if

∀x, y ∈ S : α(x · y) = α(x) ⋆ α(y) .

Thus a homomorphism respects the product ofS while ‘moving’ elements toP (whichmay have a completely different operation as its product). However, a homomorphismmay alsoidentifyelements:α(x) = α(y).

Example 1.8.(1) LetS = (N,+) andP = (N, ·), and defineα(n) = 2n for all n ∈ N.Now,

α(n +m) = 2n+m = 2n · 2m = α(n) · α(m) ,

and henceα : S → P is a homomorphism.

(2) DefineS andP by the following tables:

· a b c d

a a b c db b a d cc c d c dd d c d c

⋆ e f g

e e f gf f e gg g g g


These are semigroups – which is never too obvious! Define a mappingα : S → P ,that identifies the elementsc andd of S, as follows

α(a) = e, α(b) = f, α(c) = g = α(d) .

Now,a is the identity ofS, and its imagee = α(a) is the identity ofP . Therefore for allx ∈ S,α(a ·x) = α(x) = e⋆α(x) = α(a)⋆α(x), and similarlyα(x ·a) = α(x)⋆α(a).Also, the other cases can be checked easily to ensure thatα is a homomorphism.

(2) LetS be the semigroup of integersS = (Z, ·) under multiplication, and letP bethe semigroup of integersP = (Z,+) under addition. Define a mappingα : S → P byα(n) = n for all n ∈ Z. Thenα is nota homomorphism, because6 = α(6) = α(2·3) 6=α(2) + α(3) = 5.

(3) If α : S → P is a homomorphism, thenα(xn) = (α(x))n for all x ∈ S andn ≥ 1. ⊓⊔

Below we write for a subsetX ⊆ S and for a mappingα : S → P ,

α(X) = {α(x) | x ∈ X} .

Lemma 1.6.Letα : S → P be a homomorphism. IfX ⊆ S, thenα([X]S) = [α(X)]P .

Proof. If x ∈ [X]S , then, by Lemma 1.1,x = x1x2 . . . xn for somexi ∈ X. Sinceα isa homomorphism,

α(x) = α(x1)α(x2) . . . α(xn) ∈ [α(X)]P ,

and soα([X]S) ⊆ [α(X)]P . On the other hand, ify ∈ [α(X)]P , then, again byLemma 1.1,y = α(x1)α(x2) . . . α(xn) for someα(xi) ∈ α(X) (xi ∈ X). The claimfollows now, sinceα is a homomorphism:y = α(x1x2 . . . xn), wherex1x2 . . . xn ∈[X]S . ⊓⊔

Corollary 1.1. If A ≤ S andα : S → P is a homomorphism, thenα(A) ≤ P .

Proof. Immediate from the previous lemma. ⊓⊔

Lemma 1.7.If α : S → P andβ : P → T are homomorphisms, then so isβα : S → T .

Proof. Indeed, for allx andy,

βα(x · y) = β(α(x · y)) = β(α(x) · α(y)) = β(α(x)) · β(α(y)) = βα(x) · βα(y).

⊓⊔

For a mappingα : S → P we denote byα↾X the restriction of α to the subsetX ⊆ S, that is,α↾X : X → P is defined by

(α↾X)(x) = α(x) (x ∈ X) .

The following result states that two homomorphisms are the same if they map the gen-erators in the same way.


Theorem 1.2.Letα, β : S → P be two homomorphisms and letX ⊆ S. Then

α↾X = β↾X ⇐⇒ α↾[X]S = β↾[X]S .


Isomorphism and embeddings

A homomorphismα : S → P is

• anembeddingor amonomorphism, denotedα : S → P , if it is injective, that is, ifα(x) = α(y) impliesx = y;

• anepimorphism, denotedα : S ։ P , if it is surjective, that is, if for ally ∈ P thereexists anx ∈ S with α(x) = y;

• an isomorphism, denotedα : S P , if it is both an embedding and an epimor-phism;

• anendomorphism, if P = S;• anautomorphism, if it is both an isomorphism and an endomorphism.

Lemma 1.8.If α : S → P is an isomorphism, then also the inverse mapα−1 : P → Sis an isomorphism.

Proof. First of all the inverse mappingα−1 exists, becauseα is a bijection. Furthermore,αα−1 = ι, and thus becauseα is a homomorphism,

α(α−1(x) · α−1(y)) = α(α−1(x)) · α(α−1(y)) = xy ,

and soα−1(x) · α−1(y) = α−1(xy), which proves the claim. ⊓⊔

A semigroupS is embeddablein another semigroupP , if there exists an embeddingα : S → P , andS is isomorphic to P , denotedS ∼= P , if there exists an isomorphismα : S P . Two isomorphic semigroups share theiralgebraicproperties (but they maydiffer in their combinatorialproperties).

Example 1.9.(1) Let S be an infinite monogenic semigroup, that is,S is generated byone elementS = [x] = {x, x2, x3, . . . }. Defineα : S → (N+,+) by α(xn) = n.Now, α is a homomorphism, sinceα(xn · xm) = α(xn+m) = n + m = α(xn) +α(xm). Moreover,α is a bijection, and thus an isomorphism. Therefore,S ∼= (N+,+). Inconclusion,each infinite monogenic semigroup is isomorphic to the additive semigroupof positive integers.

(2) If P ≤ S, then theinclusion mappingι : P → S, ι(x) = x, is a homomorphism,and it is injective (but need not be surjective). Thereforeι is an embedding (ofP intoS).In particular, theidentity function ι : S → S, ι(x) = x, is always an automorphism,the trivial automorphism , of S. ⊓⊔


The endomorphisms of a semigroupS are closed under composition, that is, ifα, β : S → S are endomorphisms, then so isβα. The same holds true for automor-phisms.

Theorem 1.3. 1. The endomorphisms of a semigroupS form a monoid.2. The automorphisms of a semigroupS form a group.


The full transformation semigroup

Let X be again a set, and denote byTX the set of all functionsα : X → X. ThenTX is the full transformation semigroup onX with the operation of composition offunctions.

Example 1.10.Let X = {1, 2, 3}. There are altogether33 = 27 functions inTX . Amappingα : X → X, which is defined byα(1) = 2, α(2) = 3 andα(3) = 3, can berepresented conveniently in two different ways:

α =

(1 2 32 3 3

)

or1

α→ 2

α→ 3 .

Let then

β =

(1 2 31 1 2

),

andS = [α, β]TXthe subsemigroup ofTX generated byα andβ. We have then

α2 =

(1 2 33 3 3

)β2 =

(1 2 31 1 1

)βα =

(1 2 31 2 2

)

αβ =

(1 2 32 2 3

)βα2 =

(1 2 32 2 2

)= α2β .

One may check that there are no other elements in this semigroup, and henceS has sevenelements,S = {α, β, α2, β2, αβ, βα, βα2}. ⊓⊔

Representations

A homomorphismϕ : S → TX is a representation of the semigroupS. We say thatϕis afaithful representation, if it is an embedding,ϕ : S → TX .

The following theorem states that semigroups can be thoughtof as subsemigroupsof the full transformation semigroups, that is, for each semigroupS there exists a setXsuch thatS ∼= P ≤ TX for a subsemigroupP of transformations.


Theorem 1.4.Every semigroupS has a faithful representation.

Proof. Let X = S1, that is, add the identity1 to S if S is not a monoid. Consider thefull transformation semigroupT = TS1. For eachx ∈ S define a mapping

ρx : S1 → S1, ρx(y) = xy (y ∈ S1) .

Thusρx ∈ T , and for allx, y ∈ S and for allz ∈ S1,

ρxy(z) = (xy)z = ρx(yz) = ρx(ρy(z)) = ρxρy(z) ,

and henceρxy = ρxρy. Consequently, the mapping

ϕ : S → T, ϕ(x) = ρx

is a homomorphism. For injectivity we observe that

ϕ(x) = ϕ(y) =⇒ ρx = ρy =⇒ ρx(1) = ρy(1) =⇒ x = y .

⊓⊔

In the previous proof the special element1 is needed (only) to ensure injectivity ofϕ. It is needed, because there exists a semigroupS (without identity) that has two leftidentitiesx 6= y. In such a case,xz = yz for all z ∈ S.

Hence, loosely speaking (by Theorem 1.4) the theory of semigroups can be taught ofas the theory of transformations. This situation is reflected in a more restricted mannerin the theory of groups, where we have the following fundamental theorem of represen-tations.

Theorem 1.5 (Cayley).Each group can be represented faithfully as a (semi)group ofpermutations (bijectionsX → X).


Example 1.11.Let S = {e, a, b, c} (see the table).By the above definitions we obtain the following faithful rep-resentationϕ : S → T{e,a,b,c} for S. Notice that the mappingsρx are just the rows of the multiplication table on the right.

· e a b c

e e a b ca a e c bb b b b bc c c c c

ϕ(e) =

(eabceabc

), ϕ(a) =

(eabca e cb

),

ϕ(b) =

(eabcb b bb

), ϕ(c) =

(eabcc c cc

).

The representationϕ is by no means unique. Considerϕ1 : S → T{1,2} defined by

ϕ1(e) =

(1 21 2

), ϕ1(a) =

(1 22 1

), ϕ1(b) =

(1 21 1

), ϕ1(c) =

(1 22 2

).

One may check that this simplerϕ1 also representsS by checking that the transforma-tionsϕ1(x) satisfy the multiplication table ofS. ⊓⊔

1.4 Partial orders and semilattices 16

1.4 Partial orders and semilattices

Equivalences

LetX be a nonempty set. A subsetδ ⊆ X ×X is arelation onX. We shall write both(x, y) ∈ δ andxδy to denote that the (ordered) pair(x, y) is in relationδ. LetB(X) bethe set of all relations onX. This set forms a monoid under the operation of composition:

ρ ◦ δ = {(x, y) ∈ X ×X | ∃z : (x, z) ∈ ρ, (z, y) ∈ δ} .

The identity element ofB(X) is theidentity relation ι = ιX = {(x, x) | x ∈ X}.(It is the identity function onX). Theuniversal relation of B(X) is the relationω =ωX = X ×X = {(x, y) | x, y ∈ X}.

Let δ ∈ B(X) and assumeY ⊆ X. We adopt the following notations:

xδ = {y | (x, y) ∈ δ}, Y δ =⋃

y∈Y

yδ, ran(δ) = Xδ, dom(δ) = Xδ−1 ,

whereδ−1 = {(y, x) | (x, y) ∈ δ}. In particular,δ ⊆ dom(δ) × ran(δ).A relation δ ∈ B(X) is anequivalence relation, if its is reflexive (ιX ⊆ δ), sym-

metric (δ−1 = δ) andtransitive (δ2 ⊆ δ). The setsxδ are calledequivalence classes(with respect toδ), and they form apartition of the setX:

X =⋃

x∈X

xδ and xδ ∩ yδ 6= ∅ ⇐⇒ xδ = yδ .

A relationδ ∈ B(X) is apartial order , if it is reflexive,antisymmetric (δ∩δ−1 ⊆ ι)and transitive. A partial order is often denoted by the symbols ≤ or ⊆ (with or withoutindices).

Semilattice of idempotents

The idempotentsES of a semigroup can be given a partial order (ifES 6= ∅) as follows.Define fore, f ∈ ES :

e ≤ f ⇐⇒ ef = e = fe .

Lemma 1.9.The relation≤ is a partial order onES .

Proof. First of all, for alle ∈ ES , e2 = e, and soe ≤ e (reflexive). Ife ≤ f andf ≤ e,thene = ef = f (antisymmetric). Ife ≤ f andf ≤ h, then

e = ef = efh = eh and e = fe = hfe = he ,

and so alsoe ≤ h (transitive). ⊓⊔

1.4 Partial orders and semilattices 17

If S is a commutative semigroup and all its elements are idempotents (S = ES), thenS is called asemilattice. Hence, in this case, for allx, y ∈ S: x2 = x andxy = yx.

The relation≤ on idempotents is defined on the whole of a semilatticeS. An elementg is a lower bound of elementse andf , if g ≤ e andg ≤ f .

Lemma 1.10.Let S be a semilattice. Thenef ∈ S is the greatest lower bound of theelementse andf of S.

Proof. Let e, f ∈ S. Thenef = fe = ffe = eff , and so(ef)f = ef = f(ef), whichmeans thatef ≤ f . Similarly, ef ≤ e holds, and thusef is a lower bound ofe andf . Ifg is also a lower bound ofe andf , theng = gf = ge, and therefore

g(ef) = (ge)f = gf = g ,

and henceg ≤ ef . Henceef ∈ S is the greatest lower bound of the elementse andf inthe semilatticeS. ⊓⊔

2

General Structure Results

2.1 Quotients

Congruences

An equivalence relationρ on a semigroupS is a left congruence, if

∀x, y, z ∈ S : xρy =⇒ (zx)ρ(zy) .

Similarly, ρ is aright congruence, if

∀x, y, z ∈ S : xρy =⇒ (xz)ρ(yz)

andρ is acongruence, if it both a left and a right congruence.Recall that an equivalence relationρ partitions the domainS into the equivalence

classesxρ (x ∈ S). An equivalence class of a congruence is called acongruence class.If ρ is a congruence, then it respects the product ofS, that is, if the elementsx1, y1 andx2, y2 are in the same equivalence class (i.e.,x1ρ = y1ρ andx2ρ = y2ρ), thenx1x2andy1y2 are in the same equivalence class. A formal statement of thisis in the followinglemma.

x1

y1x2

y2

x1 · x2 y1 · y2

Lemma 2.1.An equivalence relationρ on S is a congruence if and only if for allx1, x2, y1, y2 ∈ S:

x1ρy1x2ρy2

}=⇒ (x1x2)ρ(y1y2) .

Proof. Suppose first thatρ is a congruence. Ifx1ρy1 andx2ρy2, then, by the definition,(x1x2)ρ(x1y2) and(x1y2)ρ(y1y2) and hence also(x1x2)ρ(y1y2) by transitivity. In theconverse direction the claim is trivial. ⊓⊔

2.1 Quotients 19

Syntactic congruences

We say that a congruenceρ saturatesa subsetX ⊆ S, if X is a union of congruenceclasses ofρ.

Lemma 2.2.A congruenceρ saturatesX ⊆ S if and only if

X =⋃

x∈X

xρ . (2.1)

Proof. Sincex ∈ xρ, X is always included in the union of (2.1). Consequently, ifρsaturatesX, thenX equals the union in (2.1). In converse the claim is also trivial. ⊓⊔

LetX be a subset of a semigroupS. Define a relationΓX as follows:(x, y) ∈ ΓX ifand only if

∀u, v ∈ S1 : uxv ∈ X ⇐⇒ uyv ∈ X .

Lemma 2.3.For all X ⊆ S the relationΓX is a congruence, called thesyntactic con-gruence of X in S. It is the largest congruence that saturatesX.

Proof. We leave it as an exercise to show thatΓX is an equivalence relation.Write Γ = ΓX . Assume thatxΓy holds, and letz ∈ S and u, v ∈ S1. Now,

u(zx)v = (uz)xv andu(zy)v = (uz)yv, and hence (when the definition ofΓX isapplied touz ∈ S1 andv ∈ S1), u(zx)v ∈ X implies thatu(zy)v ∈ X. The sameargument gives that ifu(zy)v ∈ X, thenu(zx)v ∈ X; and also thatu(xz)v ∈ X if andonly if u(yz)v ∈ X. HencezxΓzy andxzΓyz. This shows thatΓ is a congruence.

Certainly,X is included in the union of allxΓ (x ∈ X). Moreover, ify ∈ xΓ , thenby choosingu = 1 = v in the definition ofΓ we obtain:x ∈ X impliesy ∈ X. HencexΓ ⊆ X for all x ∈ X. We have shown thatX is the union as required in the claim.

We still have to show thatΓ is the largest such congruence. Assumeρ is any congru-ence ofS that saturatesX. By Lemma 2.2,

X =⋃

x∈X

xρ .

Assumexρy holds, and letu, v ∈ S1 be any elements. By Lemma 2.3, also(ux)ρ(uy),and, by the same lemma,(uxv)ρ(uyv). Hence(uxv)ρ = (uyv)ρ, from which it followsthatuxv ∈ X if and only if uyv ∈ X, sinceρ saturatesX. ThusxΓy, and soρ ⊆ Γ asrequired. ⊓⊔

2.1 Quotients 20

Quotients

Let ρ be a congruence ofS, and let

S/ρ = {xρ | x ∈ S}

be the set of all congruence classes ofρ. We define a new semigroup, thequotientsemigroup (of S modulo ρ), on the domainS/ρ by

xρ · yρ = (xy)ρ .

This is a well defined binary operation by Lemma 2.1, and its associativity is easilychecked:

xρ · (yρ · zρ) = xρ · (yz)ρ = (x(yz))ρ = ((xy)z)ρ = (xy)ρ · zρ = (xρ · yρ) · zρ .

The quotientS/ρ is thus obtained fromS by contracting each congruence classxρinto a single element.

x1

x2

y1z1z2z3

u1 u2

S

a = xiρ•

b = yiρ•

c = ziρ•

d = uiρ•

S/ρ

Example 2.1.(1) Consider the finite semigroupS defined in the leftmost table. Hereeandf are idempotents. Indeed,e is an identity ofS.

· e a f b

e e a f ba a e b ff f b f bb b f b f

· c d g

c c d gd d c gg g g g

Let ρ be the relation, where apart from the relationsxρx (x ∈ S), we have onlyfρb and bρf . Thenρ is a congruence ofS, and its congruence classes (with a shorthand notations) are:c = {e}, d = {a} andg = {f, b}. The multiplication table for thequotient semigroupS/ρ is given in the second table.

Also, the symmetric relationρ′ (with ιS ⊆ ρ′) so thateρ′a andfρ′b, is a congruence.It has only two congruence classes,{e, a} and{f, b}, and hence the quotientS/ρ′ is asemigroup having two elements.

2.2 Homomorphism theorem 21

The symmetric relationρno (with ιS ⊆ ρno) so thataρnob is not a congruence, be-causea · a = e anda · b = f in S, but eρnof does not hold. In this case,ρno does notrespect the product ofS: aρnob holds, but(aa)ρno(ab) does not hold.

(2) Consider the semigroupS = (Z,+). If ρ is a congruence onS, thennρm impliesthat(n+k)ρ(m+k) holds for all integersk, by the definition of a congruence. Supposethat k is the smallest non-negative integer such thatnρ(n + k) for somen ∈ Z. Inparticular,(n−n)ρ(n+k−n), i.e.,0ρk. Denote bym the remainder ofmwhen dividedby k: 0 ≤ m ≤ m andm ≡ m (mod k). By abovemρm. Also the converse holds, andthus the congruences of(Z,+) are exactly the ordinary number theoretic congruences,ρ equalsmod k (k ≥ 0). ⊓⊔

We prove now that the congruences of a semigroupS are closed under intersections.

Lemma 2.4. 1. If ρi is a congruence ofS for eachi ∈ I, then so is the intersection⋂i∈I ρi.

2. Letδ ⊆ S × S be a relation, then

δc =⋂

{ρ | δ ⊆ ρ, ρ a congruence ofS}

is the smallest congruence ofS that containsδ.

Proof. Denoteρ = ∩i∈Iρi for the congruencesρi (i ∈ I). Assumexρy andz ∈ S,then alsoxρiy for all i ∈ I, and hence(zx)ρi(zy) and(xz)ρi(yz), which implies that(zx)ρ(zy) and(xz)ρ(yz). Thereforeρ is a congruence ofS.

For the other claim we observe first thatδc is a congruence, since it is an intersectionof congruences. Ifρ is any congruence such thatδ ⊆ ρ, thenρ takes part in the intersec-tion, and thusδc ⊆ ρ. The claim follows from this. ⊓⊔

2.2 Homomorphism theorem

Natural homomorphisms

For a congruenceρ, define a mappingρ♯ : S → S/ρ as follows

ρ♯(x) = xρ .

Theorem 2.1.Let ρ be a congruence ofS. The mappingρ♯ is an epimorphism, calledthenatural epimorphism.

Proof. Let x, y ∈ S. Now, ρ♯(xy) = (xy)ρ = xρ · yρ = ρ♯(x)ρ♯(y), and henceρ♯ isa homomorphism. On the other hand, ifu = xρ ∈ S/ρ, then clearlyρ♯(x) = u, andthereforeρ♯ is surjective. ⊓⊔


For a homomorphismα : S → P we define itskernel as the relation

ker(α) = {(x, y) | α(x) = α(y)} .

One could also setker(α) = αα−1, whereα−1(y) = {x ∈ S | α(x) = y}, andαα−1

is thought as a product of relations. (Functions are read from right to left, and relationsfrom left to right; this causes some complications while writing compositions).

Lemma 2.5.For each homomorphismα : S → P , ker(α) is a congruence ofS.

Proof. We leave it as an exercise to show thatker(α) is an equivalence relation. Let(x, y) ∈ ker(α), that is,α(x) = α(y). Now, for all z ∈ S,

α(zx) = α(z)α(x) = α(z)α(y) = α(xy) ,

and similarly,α(xz) = α(yz). This proves the claim. ⊓⊔

Lemma 2.6.If ρ is a congruence ofS, thenρ = ker(ρ♯).

Proof. Indeed,

xρy ⇐⇒ xρ = yρ ⇐⇒ ρ♯(x) = ρ♯(y) ⇐⇒ (x, y) ∈ ker(ρ♯) .

⊓⊔

Corollary 2.1. Every congruence is a kernel of some homomorphism.

Example 2.2.Often homomorphisms are much easier to handle that congruences. Con-siderS = (N,+). If α : S → P is a homomorphism, then

α(n) = α(1 + 1 + · · ·+ 1) = α(1) + α(1) + · · · + α(1) = nα(1) .

In particular,α(0) = 0. This implies thatα depends only on1. (Of course, this followsalready from the fact that1 generates(N,+)). It is now easy to show thatker(α) equalsmod k for somek. ⊓⊔

Homomorphism theorem

Theorem 2.2.Letα : S → P be a any homomorphism. There exists a unique embeddingβ : S/ ker(α) → P such that the following diagram commutes.

S P

RS/ ker(α)

�

-α

ker(α)♯ β

A commuting diagram:α = β ◦ ker(α)♯.


Proof. Let ρ = ker(α), andρ♯ : S → S/ρ the corresponding natural homomorphism.Defineβ : S/ρ→ P by

β(xρ) = α(x) (x ∈ S) .

First of all,β is well defined, that is, it is a function, because

xρ = yρ ⇐⇒ (x, y) ∈ ker(α) ⇐⇒ α(x) = α(y) ⇐⇒ β(xρ) = β(yρ) , (2.2)

and soβ(xρ) has a specific value (inP ), which is independent of the choice of therepresentative of the congruence classxρ.

Secondly,β is a homomorphism, since

β(xρ · yρ) = β((xy)ρ) = α(xy) = α(x)α(y) = β(xρ)β(yρ) .

Thirdly, β is injective by (2.2).Finally, β is unique, since ifγ : S/ρ→ P is another such embedding, thenα = γρ♯,

and henceα(x) = γ(xρ) for all x ∈ S. But this means thatγ = β. ⊓⊔

The previous theorem has the following improvement.

Theorem 2.3 (Homomorphism theorem).Let α : S → P homomorphism andρ ⊆ker(α) a congruence ofS. Then there exists a unique homomorphismβ : S/ρ→ P suchthat the following diagram commutes.

S P

RS/ρ

�

-α

ρ♯ β

Proof. The proof is quite similar to the previous one. Here we observe that the mappingβ defined byβ(xρ) = α(x) is well defined, since

xρ = yρ =⇒ xρy =⇒ (x, y) ∈ ker(α) =⇒ α(x) = α(y) ,

sinceρ ⊆ ker(α). ⊓⊔

The homomorphism theorem, as well as the next isomorphism theorem, are standard(universal) algebraic results, that is, they hold in all algebras (groups, rings, Booleanalgebras, and so on).

Below α(S) is the image semigroup ofS in P . In particular,α(S) ≤ P , byLemma 1.1.

Theorem 2.4 (Isomorphism theorem).Letα : S → P be a homomorphism. Then

α(S) ∼= S/ ker(α) .

2.3 Ideals and Rees congruences 24

Proof. The homomorphismα is surjective as a homomorphismS → α(S) ≤ P . WhenTheorem 2.2 is applied toα : S → α(S) we obtain a unique embeddingβ : S/ ker(α) →α(S). Moreover,β is surjective, sinceα is surjective ontoα(S) andα = βγ for thehomomorphismγ = ker(α)♯. Consequently,β is a bijective homomorphism, i.e., anisomorphism. ⊓⊔

2.3 Ideals and Rees congruences

Ideals

A nonempty subsetI of a semigroupS is

• a left ideal, if SI ⊆ I;• a right ideal , if IS ⊆ I;• an ideal, if it is both a left and a right ideal.

Lemma 2.7.A nonempty subsetI ⊆ S is

1. a left ideal ofS, if for all a ∈ I andx ∈ S, sa ∈ I;2. a right ideal ofS, if for all a ∈ I andx ∈ S, as ∈ I;3. an ideal ofS, if for all a ∈ I andx ∈ S, as, sa ∈ I.

Proof. Exercises. ⊓⊔

Lemma 2.8.If I is a (left or a right) ideal ofS, thenI is a subsemigroup ofS.

Proof. If SI ⊆ I or IS ⊆ I, then certainlyII ⊆ I, sinceI ⊆ S. ⊓⊔

Lemma 2.9.If I and J are left (right) ideals of a semigroupS with I ∩ J 6= ∅, thenI ∩ J is a left (right) ideal.

Proof. Let a ∈ I ∩ J andx ∈ S. Thensa ∈ I andsa ∈ J , sinceI andJ are ideals, andsosa ∈ I ∩ J . Hence(I ∩ J)S ⊆ (I ∩ J). ⊓⊔

Rees congruences

Let I be an ideal ofS. Define theRees congruence(with respect toI) as follows:

ρI = I × I ∪ ι .

HencexρIy holds if and only if eitherx, y ∈ I or x = y. This means thatρI contractsthe idealI and leaves the rest ofS as it was.

Lemma 2.10.The relationρI is a congruence for every idealI of S.


2.3 Ideals and Rees congruences 25

The quotient semigroupS/ρI will be denoted simplyS/I, and it is called theReesquotient (of the idealI). S/I has the elementsI and{x} for x ∈ S \ I. In order tosimplify notation we identify the element{x} (= xρI) with the elementx ∈ S \ I. Theproduct of elements inS/I is as follows:x · y = xy for x, y /∈ I, andIx = I = xI forall x. ThereforeI is a zero element of the semigroupS/I.

Minimal ideals

An idealI of S is said to beminimal , if for all idealsJ of S, J ⊆ I implies thatJ = I.

Lemma 2.11.If I is a minimal ideal, andJ is any ideal ofS, thenI ⊆ J .

Proof. First of all, I ∩ J 6= ∅. Indeed, by the definition of an idealIS ⊆ I, and hencealsoIJ ⊆ I. Similarly, IJ ⊆ J , and soIJ ⊆ I ∩ J . Here, of course,IJ 6= ∅. Now,I ∩ J ⊆ I implies thatI ∩ J = I, sinceI is a minimal ideal, andI ∩ J is an ideal. Itfollows thatI ⊆ J as required. ⊓⊔

By the above

Theorem 2.5.If a semigroupS has a minimal ideal, then it is unique.

Every finite semigroupS does have a minimal ideal. Indeed, consider an idealI,which has the least number of elements. Such an ideal exists becauseS is finite andS isits own ideal.

Example 2.3.The ideals of the semigroup(N,+) are exactly the setsn+ N = {n+ k |k ≥ 0} (Exercise). Further,m+N ⊆ n+N if and only ifm ≥ n. Therefore(N,+) hasno minimal ideals. ⊓⊔

Simple semigroups

A semigroupS is said to besimple, if it has no idealsI 6= S.

Lemma 2.12.S is simple if and only ifS = SxS for all x ∈ S.

Proof. Clearly, for anyx ∈ S, SxS is an ideal ofS, and hence ifS is simple, thenS = SxS for all x ∈ S.

In converse, assumeS = SxS for all x ∈ S. Let I is an ideal ofS and letx ∈ I.HenceS = SxS ⊆ I, and soI = S, from which it follows thatS is simple. ⊓⊔

Example 2.4.By the above, a semigroupS is simple if and only if for alls, r ∈ S theequationxsy = r has a solution inS. Using this one can show that the semigroup of all2-by-2-matrices (

x 0y 1

)(x, y ∈ Q with x, y > 0)

is a simple semigroup. This is an exercise. ⊓⊔

3

Free Semigroups and Presentations

3.1 Free semigroups

Word semigroups

Let A be a set of symbols, called analphabet. Its elements areletters. Any finite se-quence of letters is aword (or a string)overA. The set of all words overA, with at leastone letter, is denoted byA+. For clarity, we shall often writeu ≡ v, if the wordsu andv are the same (letter by letter).

The setA+ is a semigroup, theword semigroupoverA, when the product is definedas thecatenation of words, that is, the product of the wordsw1 ≡ a1a2 . . . an, w2 ≡b1b2 . . . bm (ai, bi ∈ A) is the wordw1 · w2 = w1w2 ≡ a1a2 . . . anb1b2 . . . bm.

When we join theempty word 1 (which has no letters) intoA+, we have thewordmonoidA∗,A∗ = A+ ∪ {1}. Clearly,1 · w = w = w · 1 for all wordsw ∈ A∗.

Example 3.1.LetA = {a, b} be abinary alphabet. Thena, b, aa, ab, ba, bb, aaa, aab,. . . are words inA+. Now, ab · bab ≡ abbab. As usual,wk means the catenation ofwwith itself k times, and so, for example,v ≡ ab3(ba)2 ≡ abbbbaba ≡ ab4aba. ⊓⊔

Free semigroups

Let S be a semigroup. A subsetX ⊆ S generatesS freely, if S = [X]S and everymappingα0 : X → P (whereP is any semigroup) can be extended to a homomorphismα : S → P such thatα↾X = α0. Here we say thatα is ahomomorphic extensionofthe mappingα0. If S is freely generated by some subset, thenS is afree semigroup.

Example 3.2.(1) (N+,+) is free, forX = {1} generates it freely: Letα0 : X → P be ahomomorphism, and defineα : N+ → P by α(n) = α0(1)

n. Now,α↾X = α0, andα isa homomorphism:α(n +m) = α0(1)

n+m = α0(1)n · α0(1)

m = α(n) · α(m).(2) On the other hand,(N+, ·) is not free. For, supposeX ⊆ N+, chooseP =

(N+,+), and letα0(n) = n for all n ∈ X. If α : (N+, ·) → P is any homomorphism,thenα(n) = α(1 · n) = α(1) + α(n), and thusα(1) = 0 /∈ P . So certainly noα can bean extension ofα0.

(3) Let S = {a, a2} be a monogenic semigroup with index 2 and period 1, that is,a3 = a2. If X generatesS, then necessarilya ∈ X (sincea is not a product of twoelements ofS). Let α0 : S → P = (N+,+) be such thatα0(a) = 1. If α : S → P isan extension ofα0, thenα(aa) = α(a) + α(a) = α0(a) + α0(a) = 2 and similarlyα(aaa) = 3. However,aa = aaa in S, and this is a contradiction. ThusS is not free.

⊓⊔

3.1 Free semigroups 27

The word semigroups are free:

Theorem 3.1.For any alphabetA, A+ is a free semigroup, and it is freely generatedbyA.

Proof. Clearly,A generatesA+, since all words are strings of letters fromA. Let thenS be any semigroup, andα0 : A→ S any mapping. We defineα : A+ → S by

α(a1a2 . . . an) = α0(a1)α0(a2) . . . α0(an) (n ≥ 1, ai ∈ A) .

Clearly,α↾A = α0, andα is a homomorphism by its definition. ⊓⊔

Theorem 3.2.If S is freely generated byX andα0 : S → P is a mapping, thenα0 hasa unique homomorphic extensionα : S → P .

Proof. By the definition, eachα0 has an extension. For the uniqueness suppose bothα : S → P andβ : S → P are homomorphic extensions ofα0. Now, for all x ∈ S,x = x1x2 . . . xn for somexi ∈ X, sinceX generatesS. Therefore,

α(x) =α(x1)α(x2) . . . α(xn) = α0(x1)α0(x2) . . . α0(xn)

= β(x1)β(x2) . . . β(xn) = β(x) ,

and soα = β. ⊓⊔

The next result states that every semigroupS is a homomorphic image of a wordsemigroup, and therefore the word semigroups are basic semigroups wherefrom all othersemigroups can be derived.

Two setsX andY have thesame cardinality, denoted|X| = |Y |, if there is abijectionα : X → Y . In this case the functionα−1 : Y → X is also a bijection, and sothere is a 1-to-1 correspondence between the elements ofX andY . If X is finite, then|X| = |Y | if and only ifX andY have exactly the same number of elements. However,for infinite sets we cannot ‘count’ the number of elements. Indeed,|N| = |Q|, althoughN ⊂ Q properly.

Theorem 3.3.For each semigroupS there exists an alphabetA and an epimorphismψ : A+ ։ S.

Proof. Let X be any generating set ofS; you may even chooseX = S. Let A be analphabet with|A| = |X|, and letψ0 : A → X be a bijection. By definition of free-ness,ψ0 has a homomorphic extensionψ : A+ → S. This extension is surjective, since[ψ(X)]S = ψ([X]S) = ψ(S), forX generatesS. ⊓⊔

Corollary 3.1. Every semigroup is a quotient of a free semigroup. Indeed,

S ∼= A+/ ker(ψ)

for a suitable epimorphismψ.


There are other free semigroups than the word semigroups, but all of them are iso-morphic to word semigroups.

Theorem 3.4.A semigroupS is free if and only if it is isomorphic to a word semigroupA+ for some alphabetA.

Proof. SupposeS is freely generated by a subsetX ⊆ S, and letA be an alphabetwith |A| = |X|. Let ψ0 : A → X be a bijection. SinceA generatesA+ freely there isan epimorphic extensionψ : A+ ։ S as in the previous theorem. Also, the mappingψ−10 : X → A is a bijection, which has an epimorphic extensionβ : S ։ A+, sinceS is

freely generated byX. The compositionβψ : A+ → A+ is an epimorphism, for which

βψ↾A = βψ0 = (β↾X)ψ0 = ψ−10 ψ0 = ιA .

Now, the mappingιA : A → A extends uniquely to the identity homomorphismιA+ : A+ → A+, and henceβψ↾A extends also uniquely toιA+ , that is,βψ = ιA+ .It follows from this thatβ = ψ−1 and thatψ is a bijection, and thus an isomorphism.

On the other hand, suppose that there exists an isomorphismψ : A+ → S. Inthis case,S = [ψ(A)]S , andψ has an inverse mappingψ−1 : S → A+, which isa homomorphism (an isomorphism, by Lemma 1.8). Denoteψ0 = ψ↾A andX =ψ(A). Let P be any semigroup, andα0 : X → P any mapping. Now, the mappingα0ψ0 : A → P extends uniquely to a homomorphismγ : A+ → P . Consider the map-ping β = γψ−1 : S → P . This is a homomorphism, since bothψ−1 andγ are. Further,for eachx ∈ X,

β(x) = γ(ψ−1(x)) = α0ψ0ψ−10 (x) = α0(x)

and henceβ↾X = α0, that is,β is a homomorphic extension ofα0. By the definition,Sis freely generated byX. ⊓⊔

The above proof reveals also

Corollary 3.2. If S is freely generated by a setX, thenS ∼= A+, where|A| = |X|.

Corollary 3.3. If S andR are free semigroups generated byX andY , respectively, suchthat |X| = |Y |, thenS ∼= R.

Corollary 3.4. Every free semigroup is cancellative.

Proof. This is immediate, sinceA+ is cancellative. ⊓⊔

A criterion for freeness

LetX ⊆ S. We say thatx = x1x2 . . . xn is afactorization of x overX, if eachxi ∈ X.Now, if X generatesS, then every elementx ∈ X has a factorization overX. Usually,this factorization is not unique, that is, we may havex1x2 . . . xn = x = y1y2 . . . ym forsomexi ∈ X (i = 1, 2, . . . , n) and for some differentyi ∈ X (i = 1, 2, . . . ,m).


Theorem 3.5.A semigroupS is freely generated byX if and only if everyx ∈ S has aunique factorization overX.

Proof. We observe first that the claim holds for the word semigroupsA+, for whichAis the only minimal generating set. LetA be an alphabet such that|A| = |X| and letα0 : X → A be a bijection.

SupposeX generatesS freely, and that there is anx ∈ S, for which

x = x1x2 . . . xn = y1y2 . . . ym (xi, yi ∈ X) .

For the homomorphic extensionα of α0,

α(x) = α0(x1)α0(x2) . . . α0(xn) = α0(y1)α0(y2) . . . α0(ym)

in A+. SinceA+ satisfies the condition of the theorem, andα0(xi), α(yi) are lettersfor eachi, we must have thatα0(xi) = α0(yi) for all i = 1, 2, . . . , n (andm = n).Moreover,α0 is injective, and soxi = yi for all i, and thus alsoS satisfies the claim.

Suppose then thatS satisfies the uniqueness condition. Denoteβ0 = α−10 and let

β : A+ → S be the homomorphic extension ofβ0. Now,β is surjective, sinceX gener-atesS. It is also injective, because ifβ(u) = β(v) for some wordsu 6= v ∈ A+, thenβ(u) would have two different factorizations overX. Henceβ is an isomorphism, andthe claim follows from this. ⊓⊔

Let thebaseof a semigroupS be defined by

Base(S) = S \ S2 = {x ∈ S | ∀y, z ∈ S : x 6= yz} ,

consisting of those elementsx ∈ S which are not products of any two or more elementsof S. The following result is an exercise.

Theorem 3.6.A semigroup is freeS if and only ifBase(S) generates it freely.

Example 3.3.(1) Let A = {a, b, c} be an alphabet. The wordsab, bab, ba generate asubsemigroup ofA+. This semigroupS = [ab, bab, ba] is not free, since the elementw = babab has two different factorizations inS: w = ba · bab = bab · ab.

(2) If A is as above, butS = [aab, ab, aa], thenS is a free subsemigroup ofA+.Indeed, if there were two different factorizationsw = u1u2 . . . un = v1v2 . . . vm of awordw ∈ S, then eitheru1 = v1 (and, by cancellation, there would be a shorter wordu2 . . . un = v2 . . . vm with two different factorizations) oru1 = aa andv1 = aab (orsymmetrically,u1 = aab andv1 = aa). But in this caseu2 cannot be found, because itshould start with the letterb.

(3) Let thenA = {a, b} be a binary alphabet, and consider

S = [abn | n ≥ 1] = [ab, ab2, ab3, . . . ] .

Clearly,Base(S) = S \ S2 = {abn | n ≥ 1}, andS is freely generated byBase(S).

(4) ForS = (R, ·), we haveBase(S) = ∅. In particular, this semigroup cannot befree. ⊓⊔


Free monoids

Observe that no monoidM can be a free semigroup, because1 = 1·1. Hence the identity1 cannot be among the free generator ofM . It follows that1 is a product of generatorsof M , 1 = x1x2 . . . xn, but nowx = 1 · x gives two factorizations of an elementx.

We say thatM is a free monoid (freely generated by a subsetX with 1 /∈ X),if X ∪ {1} is a generator set forM , and every mappingα0 : X → P (whereP is amonoid) extends to a monoid homomorphismα : M → P such thatα↾X = α0 andα(1M ) = 1P .

Again, if M is free andα0 : X → P a mapping to a monoidP , then its extensionα : M → P is unique.

We have immediately:

Theorem 3.7.If S is a free semigroup, thenS1 is a free monoid.

Corollary 3.5. The word monoidA∗ is a free monoid for all alphabetsA.

The converse of Theorem 3.7 holds, too.

Theorem 3.8.A monoidM is a free monoid if and only ifM \{1M} is a free semigroup.

Proof. Suppose first thatM is a free monoid, freely generated byX. ThenM \ {1M}is a subsemigroup ofM , for otherwise,1M = x1x2 . . . xn for somexi ∈ X, and amappingα0 : X → A∗ with α0(xi) ∈ A does not extend. The rest of the claim followsfrom the definition of a free semigroup.

The converse claim is an exercise. ⊓⊔

The other results for free semigroups can also be modified forfree monoids:

Theorem 3.9. 1. A monoidM is freely generated byX if and only if each nonidentityx ∈M \ {1M} has a unique factorization overX.

2. Every monoid is an epimorphic image of a word monoidA∗.3. A monoidM is free if and only if it is isomorphic toA∗ for some alphabetA.

Free matrix monoids

Let SL(2,N) be the set of all matricesM ∈ N2×2 with det(M) = 1, that is,

M =

(n11 n12n21 n22

), n11n22 − n12n21 = 1 (nij ∈ N) .

This is a monoid, called aspecial linear monoid, with the identity matrixI as its iden-tity. Note that ifdet(M1) = 1 = det(M2), then alsodet(M1M2) = 1.

3.2 Presentations of semigroups 31

Theorem 3.10.SL(2,N) is a free monoid, freely generated by the matrices

A =

(1 10 1

)and B =

(1 01 1

).

Proof. The matricesA andB have the following inverse matrices

A−1 =

(1 −10 1

)and B−1 =

(1 0

−1 1

)

with integer entries. Note that these matrices arenot inSL(2,N).LetM ∈ SL(2,N) be any matrix as above withM 6= I. Then

A−1M =

(n11 − n21 n12 − n22

n21 n22

)

B−1M =

(n11 n12

n21 − n11 n22 − n12

).

Either one of these is inSL(2,N): A−1M ∈ SL(2,N), if n11 ≥ n21; andB−1M ∈SL(2,N), if n11 < n21.

Moreover, the trace ofM , tr(M) = n11 + n22, is greater than the trace ofA−1M orB−1M whichever is inSL(2,N):

tr(A−1M) = tr(M)− n21 ,

tr(B−1M) = tr(M)− n12 .

Therefore, for eachM ∈ SL(2,N) with M 6= I, there exists a unique sequenceC1, C2, . . . , Ck (Ci = A or B) such thatC−1

k C−1k−1 . . . C

−11 M = I, that is,M =

C1C2 . . . Ck. By Theorem 3.5 and Theorem 3.8,SL(2,N) is a freely generated byAandB. ⊓⊔

3.2 Presentations of semigroups

Generators and relations

LetS be a semigroup. By Theorem 3.3 there exists an epimorphismψ : A+ ։ S, whereA+ is a suitable word semigroup. By Theorem 2.4,

S ∼= A+/ ker(ψ) ,

since nowψ(A+) = S. We say thatψ is ahomomorphic presentationof S. The lettersin A are calledgenerator symbolsor justgeneratorsof S, and if (u, v) ∈ ker(ψ) (thatis,ψ(u) = ψ(v)) thenu = v is called arelation or anequality in S. (In order to avoidconfusion we write usuallyu ≡ v, if the wordsu, v ∈ A+ are equal).


Define apresentationof S in generators (generator symbols)A = {a1, a2, . . . } andrelationsR = {ui = vi | i ∈ I},

S = 〈a1, a2, · · · | ui = vi (i ∈ I)〉 or S = 〈A | R〉 ,

if ker(ψ) is the smallest congruence ofA+ that contains the relation{(ui, vi) | i ∈ I}.In particular,

ψ(ui) = ψ(vi) for all ui = vi in R . (3.1)

The setR of relations is supposed to be symmetric, that is, ifu = v is inR, thenv = ualso holds.

One should remember that the wordsw ∈ A+ are not elements ofS but of the wordsemigroupA+, which is mapped ontoS. We say that a wordw ∈ A+ presents theelementψ(w) of S. The same element can be presented by many different words. Infact, if ψ(u) = ψ(v), then bothu andv present the same element ofS.

Let S = 〈A | R〉 be a presentation. We show thatS satisfies a relationu = v (thatis,ψ(u) = ψ(v)) if and only if there exists a finite sequenceu = u1, u2, . . . , uk+1 = vof words such thatui+1 is obtained fromui by substituting a factorui by vi for someui = vi in R.

To be more precise, we say that a wordv is directly derivable from the wordu, if

u ≡ w1u′w2 and v ≡ w1v

′w2 for someu′ = v′ in R . (3.2)

Clearly, if v is derivable fromu, thenu is derivable fromv, sinceR is supposed to besymmetric, and, in the notation of (3.2),

ψ(u) = ψ(w1u′w2) = ψ(w1)ψ(u

′)ψ(w2) = ψ(w1)ψ(v′)ψ(w2) = ψ(w1v

′w2) = ψ(v) ,

and henceu = v is a relation inS.The wordv isderivable fromu, if there exists a finite sequenceu ≡ u1, u2, . . . , uk ≡

v such that for allj = 1, 2, . . . , k − 1, uj+1 is directly derivable fromuj. Now, if v isderivable fromu, then alsoψ(u) = ψ(v), sinceψ(u) = ψ(u1) = · · · = ψ(uk) = ψ(v),and thereforeu = v is a relation inS. This can be written as

u ≡ u1 = u2 = · · · = uk ≡ v .

Theorem 3.11.LetS = 〈A | R〉 be a presentation (withR symmetric). Then

Rc = {(u, v) | u = v or v is derivable fromu} .

Henceu = v in S if and only ifv is derivable fromu.

Proof. Let the relationρ be defined by

uρv ⇐⇒ u = v or v is derivable fromu .

It is clear thatιS ⊆ ρ, and henceρ is reflexive. SinceR is symmetric, so isρ. Thetransitivity ofρ is easily verified, and henceρ is an equivalence relation.


If w ∈ A+ andv is derivable fromu, then clearly alsowv is derivable fromwu andvw is derivable fromuw. This proves thatρ is a congruence.

Let θ be any congruence such thatR ⊆ θ. Supposev is directly derivable fromu:u = w1u

′w2, v = w1v′w2 with u′ = v′ in R. SinceR ⊆ θ, also(u′, v′) ∈ θ, and since

θ is a congruence, also(w1u′w2, w1v

′w2) ∈ θ, that is,uθv. Therefore, by transitivityof ρ andθ, ρ ⊆ θ. In conclusion,ρ is the smallest congruence that containsR, that is,ρ = Rc. ⊓⊔

Theorem 3.12.Let A be an alphabet andR ⊆ A+ × A+ a symmetric relation. ThesemigroupS = A+/Rc, whereRc is the smallest congruence containingR, has thepresentation

S = 〈A | u = v for all (u, v) ∈ R〉 .

Moreover, all semigroups having a common presentation are isomorphic.

Proof. The claim is immediate from the above. ⊓⊔

Example 3.4.(1) Consider the following presentation

S = 〈a, b | aa = ab, ba = aab, bbb = aba〉 .

In this presentation there are two generators and three relations. For instance,S satisfiesthe relationbaabbaa = bbaaaba, sinceu1 = baabbaa ≡ b · aab · baa = b · ba · baa ≡ u2andu2 = bbabaa ≡ bba · ba ·a = bba ·aab ·a ≡ bbaaaba. Also,aaab = aabb = abbb =aaba = baa = bab in S and henceaaab = bab in S.

(2) A word semigroup does not need any relations:A+ = 〈A | ∅〉.

(3) The presentation

S = 〈a, b, c, d, e | ac = ca, ad = da, bc = cb, bd = db,

eca = ce, edb = de, cca = ccae〉

is calledTzeitin’s semigroup. This surprisingly simple semigroup has anundecidableword problem, that is, there exists no algorithm that determines whetheru = v is arelation in this semigroup, whereu, v ∈ {a, b, c, d, e}+ are given (input) words. Hencein Tzeitin’s semigroup one cannot effectively decide whether a word is derivable fromanother given word. ⊓⊔

All semigroups (and monoids and groups) have presentations. Indeed,S = 〈A |ker(ψ)〉 is one such presentation, whenψ : A+ → S is the presenting epimorphisms.Usually this presentation is all too complicated. We are mostly interested in semigroupsthat have afinite presentation, that is, a presentationS = 〈A | R〉, whereA is afinite alphabet andR is finite set of relations. However, not all semigroups have suchpresentations.


Monoid presentations

All monoids have a semigroup presentation, but it is more convenient to use monoidpresentations for these in order to take advantage of the identity element:

M = 〈a1, a2, · · · | ui = vi (i ∈ I)〉

is amonoid presentation, if ui, vi ∈ A∗ whereA = {a1, a2, . . . } is an alphabet. In amonoid presentation we may thus have relations of the formu = 1, which means thatthe wordu can be erased from another word or added somewhere in betweentwo letters.

Example 3.5.(1) Let M = 〈a, b | ab = ba〉 be a monoid presentation. HenceM ∼=A∗/Rc, whereA = {a, b} andR has a single relationab = ba. There is an epimorphismψ : A∗ ։M andM is generated by the elementsx = ψ(a) andy = ψ(b). The monoidM is commutative, because of the relationab = ba. If the generators ofM commutewith each other, thenM is commutative.Furthermore, each elementz ∈M has anormalform: Supposez = z1z2 . . . zn with zi = ψ(ai) (ai = a or b), and so

z = ψ(a1)ψ(a2) . . . ψ(an) = ψ(a1a2 . . . an) = ψ(akbm) = ψ(a)kψ(b)m

for somek,m ≥ 0. The monoidM is afree commutative monoid, and it can be shownthat every commutative monoid, which is generated by two elements, is an epimorphicimage ofM .

(2) The (monoid) presentation

〈a, b | aba = 1〉

defines a group. In fact, this group is isomorphic to(Z,+). Indeed, letM be a monoidwith the above presentation, i.e.,M ∼= A∗/Rc, whereA = {a, b} andR = {aba = 1},and letψ : A∗ ։ M be the corresponding epimorphism. ThenM is generated by theelementsx = ψ(a) andy = ψ(b). Furthermore,ab = ab · aba ≡ aba · ba = ba,and hencexy = yx. It follows thatM is a commutative monoid. Consequently, everyelementz ∈ M has the formz = ψ(anbm) for somen,m ≥ 0. Also, a · ba = 1 andba · a = ab · a = 1 means thatba is a group inverse ofa. Similarly,a2 is a group inverseof b, b = (aa)−1. This means that all elements ofM have the formz = ψ(ak) for k ∈ Z.It is now easy to show thatα : M → (Z,+) defined byα(a) = −1 andα(b) = 2 is anisomorphism.

(3) Define two mappingsα, β : N → N as follows:

α(n) =

{0 if n = 0

n− 1 if n ≥ 1and β(n) = n+ 1 (n ≥ 0) .

Consider thebicyclic monoid B = [α, β] generated by these two transformations. Weobserve thatαβ = ι, but

βα(n) =

{1 if n = 0

n if n ≥ 1,

3.3 Embeddings into 2-generator semigroups 35

and, more generally,

βkαk(n) =

{k if n < k

n if n ≥ k.

LetA = {a, b} be an alphabet, and define a homomorphismψ : A∗ → B by ψ(a) = αandψ(b) = β. By the extension propertyψ becomes uniquely defined by the images ofthe generator symbolsa andb. Henceψ is an epimorphism andB ∼= A+/ ker(ψ).

By aboveab = 1 is a relation inB. Let thenγ ∈ B be any element of the bicyclicmonoid,γ = γnγn−1 . . . γ1, whereγi = α or γi = β. Sinceαβ = ι, we may assumethat if γj = β for some indexj, thenγt = β for all t with j ≤ t ≤ n. This shows thatγ = βkαm for somek,m ≥ 0, and hence

B = {βkαm | k,m ≥ 0} .

Furthermore, these elements are all different from each other: if γ = βkαm andδ =βrαs, then

γ(0) = βkαm(0) = βk(0) = k and δ(0) = βrαs(0) = βr(0) = r ,

and forn ≥ max{m, s},

γ(n) = βkαm(n) = βk(n−m) = n+ k −m,

δ(n) = βrαs(n) = βr(n− s) = n+ r − s .

Therefore,γ = δ just in casek = r andm = s. This means that

B = 〈a, b | ab = 1〉

is a (monoid) presentation ofB.The bicyclic monoid has many (semigroup) presentations, ofwhich we mention

B = 〈a, b | aba = aab, a = aab, bab = abb, b = abb〉 .

⊓⊔

3.3 Embeddings into 2-generator semigroups

Evans’ embedding result

The following embedding result was proved by Evans in 1952. The proof given here isdue to Subbiah (1973). We say that a setX is denumerable, if it is finite or if it has thecardinality ofN. The elements of a denumerable setX can be listed:X = {x1, x2, . . . }.(This is not the case forR, which is not denumerable).

Theorem 3.13 (Evans).Let S be a semigroup generated by denumerably many ele-ments. ThenS can be embedded into a semigroup generated by two elements.

3.3 Embeddings into 2-generator semigroups 36

We shall use the fact that each semigroup is isomorphic to a subsemigroup of the fulltransformation semigroup, and we actually prove

Theorem 3.14 (Sierpinski).Letα1, α2, . . . : N+ → N+ be any transformations. Thereexist two transformationsβ1, β2 : N+ → N+ such that eachαi is a composition of thesetwo transformations:αi ∈ [β1, β2].

Proof. DefineXn = {2n(2m− 1)− 1 | m = 1, 2, . . . } .

Now,Xn∩Xm = ∅ for all n 6= m, becausek ∈ Xn if and only if n is the highest powerfor which2n dividesk + 1. Also,

⋃

n≥1

Xn = 2N + 1 ,

and hence the setsXn form a partition of the odd positive integers. Define

β1(n) = 2n (n ≥ 1) , β2(n) =

{n− 1 if n is even,

αk(2−(k+1)(n+ 1) + 2−1) if n ∈ Xk.

Now, for all k ≥ 1,αk = β22β

k1β2β1 ,

since

β22βk1β2β1(n) = β22β

k1β2(2n) = β22β

k1 (2n − 1) = β22(2

k(2n − 1)) =

β2(2k(2n− 1)− 1) = αk(2

−(k+1)(2k(2n − 1) + 2−1) =

αk(2−1(2n − 1 + 1) = αk(n) .

This proves the claim. ⊓⊔

Theorem 3.13 follows from Sierpinski’s result by using suitable isomorphisms. Weomit this straightforward reduction.

Embedding word monoids

LetA = {a1, a2, . . . , an} be a finite alphabet, and letB = {a, b} be a binary alphabet.The homomorphismα : A∗ → B∗ defined by

α(ai) = abi (i = 1, 2, . . . , n)

is injective as can be easily shown. Therefore

Theorem 3.15.Each finitely generated word monoidA∗ can be embedded in a wordmonoidB∗ generated by two letters.

4

Combinatorial Topics

4.1 The defect theorem

Submonoids of word monoids

For a subsetX ⊆ A∗ of words we shall write

X+ = [X]A∗

for the subsemigroup thatX generates in the word monoidA∗, that is,X+ consists ofall catenations of the words inX. Also, the corresponding monoid is denoted byX∗,that is,X∗ = X+ ∪ {1}. Notice that if1 ∈ X already, thenX∗ = X+.

If w ∈ A∗ is a word, then we write simplyw∗ = {w}∗.Letu, v ∈ A+ be two words. Henceforth we writeu = v (notu ≡ v) for the equality

in A∗.Thenu is

• a factor of v, if v = w1uw2 for some wordsw1, w2 ∈ A∗;• aprefix of v, if v = uw for somew ∈ A∗;• asuffix of v, if v = wu for somew ∈ A∗.

The length |w| of a wordw ∈ A∗ is the number of the occurrences of the letters init: if w = a1a2 . . . an with ai ∈ A, then|w| = n. The length of the empty word is zero.Clearly, ifw = uv, then|w| = |u|+ |v|, and so the length function is a homomorphism| | : A∗ → (N,+).

The next lemma is immediate.

Lemma 4.1.If u1u2 = v1v2 in A∗, then eitheru1 is a prefix ofv1 or v1 is a prefix ofu1,that is, there is a wordw ∈ A∗ such that eitheru1 = v1w or v1 = u1w.

We first prove another criterion for freeness of submonoids of a word monoid. Fora submonoidM of a word monoidA∗ we letM+ = M \ {1}. HereM+ is alwaysa subsemigroup ofA+, because the empty word cannot be a catenation of nonemptywords. ThebaseofM , Base(M) =M+ \M2

+, is the base of the semigroupM+. Recallthat the identity (the empty word) is automatically in a submonoid. The base of a freemonoid is called acode.

A submonoidM of a word monoidA∗ need not be free. As an example, consider themonoidM = {a, ab, ba}∗.

4.1 The defect theorem 38

Lemma 4.2.For all submonoidsM of A∗ the baseBase(M) is the unique minimalgenerating set ofM (as a monoid), i.e., if a subsetN generatesM , thenBase(M) ⊆ N .

Proof. In order to show thatBase(M) generatesM suppose to the contrary that thereexists a wordw ∈ M that cannot be written as a product of words fromBase(M), andlet w is a shortest word such thatw ∈ M+ butw /∈ Base(M)+. Sincew /∈ Base(M),there are two wordsu, v ∈ M+ such thatw = uv. By the minimality ofw, the shorterwordsu andv are inBase(M)+, but now alsow ∈ Base(M)+, which is a contradiction.

If a subsetN ⊆ M generatesM , then for allu ∈ Base(M), u ∈ N∗ (= M), butuis not a product of two or more words fromM , and hence necessarilyu ∈ N , too. ⊓⊔

Theorem 4.1.LetM be a submonoid of a word monoidA∗.M is free if and only if

u, v, uw, wv ∈M =⇒ w ∈M .

Proof. LetX = Base(M).Assume first thatM is free. Suppose that for a wordw, bothuw andwv are inM ,

whereu, v ∈M :

u = u1u2 . . . uk , wv = uk+1uk+2 . . . uk+r

uw = v1v2 . . . vt , v = vt+1vt+2 . . . vt+s ,

where eachui, vi ∈ X. Now,

uwv = u1u2 . . . ukuk+1 . . . uk+r = v1v2 . . . vtvt+1 . . . vt+s ,

and sinceM is freely generated byX, k+r = t+s andui = vi for all i = 1, 2, . . . , k+r.It follows thatw = uk+1 . . . ut (sincek ≤ t), and sow ∈M as required.

Assume then that the condition is valid forM . Suppose that there exists a word,which has two different factorizations overX,

u1u2 . . . un = v1v2 . . . vm (ui, vi ∈ X) .

We may suppose here thatu1 6= v1, for otherwise we have alreadyu2 . . . un = v2 . . . vm.Now, eitheru1 is a prefix ofv1 or v1 is a prefix ofu1. Suppose that the former holds,v1 = u1w (the other case is quite symmetric to this one). It follows that

u1w ∈M and u2u3 . . . un = wv2v3 . . . vm .

The latter holds, sinceA+ is cancellative. By assumption,w ∈ M , but this contradictsthe fact thatv1 = u1w ∈ Base(M). We conclude thatM is free. ⊓⊔

4.1 The defect theorem 39

The defect theorem

For the defect theorem we need the following result, which isinteresting on its own.

Theorem 4.2.If Mi (i ∈ I) are free submonoids ofA∗, then so isM = ∩i∈IMi.

Proof. Clearly, the intersectionM is a submonoid ofA∗. If now u, v, uw,wv ∈ M ,thenu, v, uw,wv ∈ Mi for all i ∈ I and hence by Theorem 4.1,w ∈ Mi for all i ∈ I,that is,w ∈M . The claim follows from Theorem 4.1. ⊓⊔

By above, ifX ⊆ A∗ is any subset then the intersection

⋂{M | X ⊆M, M is a free submonoid}

is a free submonoid. Clearly, it is the smallest free submonoid that contains the subsetX. The base of this intersection is denoted simply byHull(X), and it is called thefreehull of X. In particular,X∗ ⊆ Hull(X)∗, sinceX∗ is a monoid andX ⊆ Hull(X)∗.

Theorem 4.3 (Defect theorem).LetX ⊆ A+ be a finite set of words. IfX is not a code(a base of a free monoid), then|Hull(X)| ≤ |X| − 1.

Proof. SinceX ⊆ Hull(X)∗, each wordu ∈ X can be uniquely written in terms ofHull(X), u = x1x2 . . . xk with xi ∈ Hull(X). Let α : X → Hull(X) be the mappingsuch that

α(u) = x1 if u ∈ x1 ·Hull(X)∗.

Assume thatX∗ is not free, and hence that there exists a wordw ∈ X∗ having twodifferent factorizations overX:

w = u1u2 . . . un = v1v2 . . . vm (ui, vi ∈ X) ,

whereu1 6= v1. (Note that ifu1 = v1, then there is a shorter wordu2 . . . un withtwo different factorizations overX). We haveα(u1) = α(v1), and thereforeα is notinjective. (If α(u1) 6= α(v1), then the wordw would have two different factorizationsoverHull(X), butHull(X)∗ is free.)

We show next thatα is surjective. Suppose to the contrary that there exists a wordw ∈ Hull(X) such thatα(u) 6= w for all u ∈ X. Let

Y = (Hull(X) \ {w})w∗ .

HenceX ⊆ Y ∗. Now,Y ∗ is free, since if

w1w2 . . . wm = v1v2 . . . vr for somewi, vi ∈ Y ,

wherewi = yiwki andvi = ziw

ti for yi, zi ∈ Hull(X) \ {w} andki, ti ≥ 0. Then

y1wk1y2w

k2 . . . ymwkm = z1w

t1z2wt2 . . . zrw

tr ,

4.2 Ehrenfeucht’s conjecture 40

and, sinceHull(X) is free,y1 = z1, k1 = t1, . . . , yk = zk, km = tm andm = r. Butnow alsowi = vi for all i, which shows thatY ∗ is free.

We have, however, thatX ⊆ Y ∗, and also thatY ∗ ⊂ Hull(X)∗, which contra-dicts the minimality ofHull(X) as the smallest base of a free submonoid contain-ing X. This shows thatα is surjective. Becauseα is not injective, it follows that|Hull(X)| = |α(X)| < |X|, and the claim is proven. ⊓⊔

Example 4.1.Let A = {a, b} andX = {ab, aba, bab, ba}. The submonoidX∗ is notfree, sinceab·aba = aba·ba. Consider any free submonoidM such thatX∗ ⊆M . Now,for u = ab, v = ba andw = a, u, v, uw,wv ∈ X∗ ⊆ M , and hence by Theorem 4.1,alsow = a ∈ M . Similarly, b ∈ M , and henceM = A∗, which means thatHull(X) ={a, b}. Here|X| = 4 > 2 = |Hull(X)| as required by Theorem 4.3. ⊓⊔

Some corollaries

A wordw ∈ A+ is said to beprimitive , if it is not a proper power of another word, thatis,w = uk implies thatk = 1 andu = w.

Two wordsu, v ∈ A+ areconjugates(of each other), if there are wordsw1, w2 ∈ A∗

such thatu = w1w2 andv = w2w1. Hence, ifu is a conjugate ofv, thenu is obtainedfrom v by cyclically permuting the wordv.

Corollary 4.1. Each wordw ∈ A+ is a power of a unique primitive word.

Proof. Supposew = un = vm for some wordsu 6= v ∈ A+ and integersn,m ≥ 1.Now, the setX = {u, v} is not a code, sincew has two different factorizations over{u, v}. By Theorem 4.3,|Hull(X)| < |X|, and henceHull(X) = {z} for somez ∈ A∗,but this means thatu, v ∈ z∗, that is, they are powers of the wordz: u = zr andv = zs.If aboveu andv are primitive, thenr = 1 = s, and henceu = v. This shows the claim.

⊓⊔

Corollary 4.2. Two wordsu, v ∈ A∗ commute,uv = vu, if and only if they are powersof a common word.

Proof. If uv = vu, thenX = {u, v} is not a code, and hence again|Hull(X)| < |X| =2, and sou andv are powers of a common word. ⊓⊔

4.2 Ehrenfeucht’s conjecture

Equality sets

Let α, β : A∗ → B∗ be two homomorphisms between the word monoidsA∗ andB∗,where we assume that the alphabets are finite. Theirequality set is defined to be


E(α, β) = {w ∈ A∗ | α(w) = β(w)} .

Thus the equality set ofα andβ consists of all those words on which these homomor-phisms agree.

A word w ∈ E(α, β) is also called asolution of (α, β). A solutionw 6= 1 is anontrivial solution . Notice that we always have1 ∈ E(α, β). The empty word is thetrivial solution of (α, β).

Lemma 4.3.E(α, β) is a free monoid, for which

v, uv ∈ E(α, β) =⇒ u ∈ E(α, β) .

Proof. First of all the empty word is always inE = E(α, β), sinceα(1) = 1 = β(1).We use a restricted version of Theorem 4.1 to show the rest of the claim. Suppose

v, uv ∈ E. Thenα(uv) = α(u)α(v) = α(u)β(v) andα(uv) = β(uv) = β(u)β(v),and henceα(u) = β(u), becauseA∗ is cancellative. This means thatu ∈ E, too. Now,if u, v, uw,wv ∈ E, thenv,wv ∈ E and, by above,w ∈W . ThusE is free. ⊓⊔

The baseBase(E(α, β)) is the set of allminimal solutions of (α, β). By above, it isa code.

Example 4.2.(1) Consider the endomorphismsα, β : A∗ → A∗, whereA = {a, b}, andα andβ are defined by the images of the generatorsa andb:

α(a) = aba α(b) = baβ(a) = ab β(b) = aba.

Now, if w ∈ E = E(α, β) is nonempty, then it must begin with the lettera (becauseα(b) andβ(b) are ‘incomparable’). The next letter must beb, and, we obtain thus asolutionab ∈ E: α(ab) = aba · ba = uivab · aba = β(ab). It is easy to show thatE = (ab)∗ = {(ab)n | n ≥ 0}.

(2) Let thenα, β : A∗ → A∗ for A = {a, b, c} be defined by

α(a) = ab α(b) = bb α(c) = cβ(a) = a β(b) = bb β(c) = bc

If now a nonemptyw ∈ A∗ is in E = E(α, β), then eitherw begins witha or b, and itends withb or c. In this case we haveE(α, β) = b∗ ∪ ab∗c = {bn | n ≥ 0} ∪ {abnc |n ≥ 0}. ⊓⊔

Remark 4.1.It can be shown that the problem whetherE(α, β) contains a nonemptyword is algorithmically unsolvable, that is, there exists no algorithm which decideswhether or notE(α, β) = {1} on inputsα andβ. This is know as thePost correspon-dence problem. There are thus arbitrarily complex equality sets. ⊓⊔


Test sets

We are going to prove the following result, which was conjectured by Ehrenfeucht at thebeginning of 1970s and proved by Albert and Lawrence and independently by Guba in1985.

Theorem 4.4 (Ehrenfeucht’s conjecture).LetL ⊆ A∗ be any set of words. There existsa finite subsetT ⊆ L such that for all homomorphismsα, β : A∗ → B∗

α↾T = β↾T ⇐⇒ α↾L = β↾L .

The subsetT of L is called atest setof L, since in order to check whether any twohomomorphisms agree on the setL, it suffices to check whether they agree on the finitesubsetT .

Theorem 4.4 may be restated in the following form:

Theorem 4.5.LetL ⊆ A∗ be any set of words. There exists a finite subsetT ⊆ L suchthat for all homomorphismsα, β : A∗ → B∗

T ⊆ E(α, β) ⇐⇒ L ⊆ E(α, β) .

Systems of equations

LetX = {x1, x2, . . . , xn} be a special alphabet, the letters of which we callvariables.An equation overX is a pair of words(u, v) (for u, v ∈ X+). We shall usually writeu = v instead of(u, v).

We say that(w1, w2, . . . , wn) with wi ∈ A∗ is a solution of the equationu = v,if the substitutionxi 7→ wi in u andv results in the same word ofA∗. In other words,a solution is a (monoid) homomorphismα : X∗ → A∗ such thatα(u) = α(v) (hereα(xi) = wi), that is,(u, v) ∈ ker(α).

Example 4.3.Let X = {x, y, z}. Then a solution of the equationxyx = zyxz is ahomomorphismα : X∗ → A∗ (A is some alphabet) such thatα(xyx) = α(zyxz), thatis, w1w2w1 = w3w2w1w3, wherew1 = α(x), w2 = α(y) andw3 = α(z). Of course,w1 = w2 = w3 = 1 is a solution. This is thetrivial solution .

Let x 7→ w1, y 7→ w2, z 7→ w3 be any solution. By comparing the lengths weobtain |w1w2w1| = |w3w2w1w3|, that is,2|w1| + |w2| = |w1| + |w2| + 2|w3|, and so|w1| = 2|w3|, which means thatw1 = w3w for a wordw with |w| = |w3| (sincew1 andw3 begin the same word). When we substitutew1 byw3w, we obtain thatw3ww2w3w =w3w2w3ww3, from which we getww2w3w = w2w3ww3 by cancellation. Herew andw3 end this word, and hencew = w3, since |w| = |w3|. Therefore,w3w2w3w3 =w2w3w3w3, and sow3w2 = w2w3, which means thatw2 andw3 are powers of the sameword,w2 = ur andw3 = us, andw1 = w3w = w3w3 = u2s. Further, ifu is any word,thenw1 = u2s, w2 = ur, w3 = us is a solution for anyr, s ≥ 0. We have thus found allthe solutions of the equationxyx = zyxz. ⊓⊔


Any setE = {(ui, vi) | i = 1, 2, . . . } of equations (in variables fromX) is asystemof equation. A solution of E is a homomorphismα : X∗ → A∗ such thatα is a commonsolution to allui = vi.

Example 4.4.Let X = {x1, x2, y1, y2} and consider the system of equationsxi1yi1 =

xi2yi2 (i ≥ 1):

x1y1 = x2y2x1x1y1y1 = x2x2y2y2

...

.

This system of equations has the trivial solutionα(xi) = 1 = α(yi) (i = 1, 2), andthe solutionsα(x1) = w1 = α(x2), α(y1) = w2 = α(y2) (in any alphabet); alsoα(x1) = aa, α(x2) = a, α(y1) = a, α(y2) = aa is a solution. ⊓⊔

Two systemsE1 andE2 of equations areequivalent, if they have the same solutions.The following was proved by Culik and Karhumaki:

Theorem 4.6.Ehrenfeucht’s conjecture is equivalent to the statement: Every system ofequationsE is equivalent to a finite subsystemT ⊆ E.

Proof. For any alphabetB we letB = {a | a ∈ B} be a new alphabet, and write forall wordsu = a1a2 . . . ak ∈ B+, u = a1a2 . . . ak ∈ B

+. Hence the function is an

isomorphismB∗ → B∗.

Let us first assume that Ehrenfeucht’s conjecture is true, and let E be a system ofequations in variablesX = {x1, x2, . . . , xn}. LetX be a new set of variables defined asabove. To each equationu = v in E we attach a worduv ∈ X+ ·X

+. Let

L = {uv | u = v ∈ E} ⊆ (X ∪X)∗ .

By assumption,L has a finite test setT ⊆ L: for any homomorphismsα, β : (X∪X)∗ →A∗, if α(uv) = β(uv) for all uv ∈ T , thenα(uv) = β(uv) for all uv ∈ L.

For each homomorphismα : X∗ → A∗ define homomorphismsα1, α2 : (X∪X)∗ →A∗ as follows

α1(x) = α(x) α2(x) = 1

α1(x) = 1 α2(x) = α(x)

It follows that if α is a solution to

T = {u = v | uv ∈ T} ,

thenα(u) = α(v) for all u = v in T, andα1(uv) = α2(vv) for all uv in T , and soα1(uv) = α2(vv) for all uv in L, andα(u) = α(v) for all u = v in E. This proves thatE is equivalent to the finite subsystemT.


Assume then that each system of equations has an equivalent finite subsystem. LetL ⊆ B∗ be any set of words. LetB be a new alphabet as above. We form a system ofequationsE fromL as follows:

E = {u = u | u ∈ L} .

By assumption this system has an equivalent subsystemT. Let

T = {u | u = u ∈ T} .

HenceT is a finite subset ofX. Let α, β : B∗ → A∗ be homomorphisms such thatα(u) = β(u) for all u ∈ T . Let X = B ∪ B be our set of variables, and define ahomomorphismγ : X∗ → A∗: For allx ∈ B,

γ(x) = α(x) and γ(x) = β(x) .

Now, γ(u) = γ(u) for all u ∈ T , and henceγ is a solution to the finite systemT. Byassumption,γ is then a solution to the whole systemE, and, consequently,α(u) = β(u)for all u ∈ L. This shows thatT is a finite test set ofL. ⊓⊔

Hilbert’s basis theorem

In the proof of Ehrenfeucht’s conjecture we need the following original version ofHilbert’s basis theorem. LetZ[x1, x2, . . . , xn] denote the ring of polynomials with inte-ger coefficients in (commuting) variablesx1, x2, . . . , xn.

Theorem 4.7 (Hilbert’s basis theorem).LetPi (i ≥ 1) be any polynomials in the ringZ[x1, x2, . . . , xn]. There exists a finite subsetP1, P2, . . . , Pt of these polynomials suchthat everyPi can be expressed as a linear combination

Pi = P1Fi1 + P2Fi2 + · · · + PtFit ,

whereFij ∈ Z[x1, x2, . . . , xn].

The proof goes by showing first thatZ satisfies the above property, that is, for allintegerspi ∈ Z (i = 1, 2, . . . ) there are finitely many of these,p1, p2, . . . , pt, such thatall the integerspi can be expressed as a linear combination of these. In the second stepwe suppose that a ringR satisfies the finiteness (Noetherian) condition as stated intheclaim, and show that also the polynomial ringR[x] satisfies it. The claim then followsbecauseZ[x1, . . . , xn−1, xn] is isomorphicZ[x1, . . . , xn−1][xn]. We omit the proofs ofthese claims (and refer to a ‘standard text book in algebra’).

Corollary 4.3. Let Pi = Qi (i = 1, 2, . . . ) be a system of polynomial equations,wherePi, Qi ∈ Z[x1, x2, . . . , xn]. There exists a finite subsystemP1 = Q1, P2 =Q2, . . . , Pt = Qt of these equations such that if(m1, . . . ,mn) is a solution of thisfinite subsystem,

Pi(m1, . . . ,mn) = Qi(m1, . . . ,mn) (i = 1, 2, . . . , t) ,

then it is a solution the whole system of equations.


Proof. Consider the polynomialsRi = Pi −Qi ∈ Z[x1, x2, . . . , xn]. By Hilbert’s basistheorem there exists a finite subset of these,R1, . . . , Rt, such that

Ri = R1Fi1 +R2Fi2 + · · ·+RtFit ,

for eachRi (for someFij ). Now,

0 = Ri(m1, . . . ,mn) = Pi(m1, . . . ,mn)−Qi(m1, . . . ,mn)

for i = 1, 2, . . . , t impliesPi(m1, . . . ,mn) = Qi(m1, . . . ,mn) for all i. ⊓⊔

A proof of Ehrenfeucht’s conjecture

By Theorem 4.6 we need to prove that every infinite system of equations has an equiva-lent finite subsystem.

In order to make advantage of Hilbert’s Basis Theorem we transform the word equa-tionsu = v into polynomial equations inZ[x1, x2, . . . , xn]. To overcome the commu-tativity of the variables in the ring of polynomials, we use noncommuting matrices ofthese polynomial. For this we use of the free monoidSL(2,N) of the previous chapter.HereSL(2,N) consists of unimodular matrices, that is, those matrices with determinantequal to 1. Recall thatSL(2,N) is a free monoid generated by the matrices

A =

(1 10 1

)and B =

(1 01 1

).

The homomorphismµ : {a, b}∗ → SL(2,N) defined by

µ(a) =

(1 10 1

)and µ(b) =

(1 01 1

)(4.1)

is an isomorphism.LetX = {x1, x2, . . . , xk} be a set of word variables. We introduce for eachxi four

new integer variablesxi1, xi2, xi3 andxi4, and denote

X = {xij | 1 ≤ i ≤ k, 1 ≤ j ≤ 4} .

Further, for eachi = 1, 2, . . . , k define

Xi =

(xi1 xi2xi3 xi4

),

and letM(X) be the submonoid of the matrix monoidZ[X ]2×2 generated by the matri-cesX1,X2, . . . ,Xk.


Lemma 4.4.The monoidM(X) is freely generated byX1,X2, . . . ,Xk.

Proof. Consider two elementsM = Xr1Xr2 . . . Xrm andM ′ = Xs1Xs2 . . . Xst ofM(X). The upper right corner entryM12 of M contains the monomial

xr11xr21 . . . xrj−11xrj2xrj+14 . . . xrm−14xrm4 (4.2)

for j = 1, 2, . . . ,m. It is now easy to see that if these monomials exist in(M ′)12, thent = m andXrj = Xsj for j = 1, 2, . . . ,m. Therefore,(M)12 = (M ′)12 implies thatXr1 = Xs1,Xr2 = Xs2 , . . . ,Xrm = Xsm. The claim follows from this. ⊓⊔

⊓⊔

In particular, the monoid homomorphismϕ : X∗ → M(X) defined by

ϕ(xi) =

(xi1 xi2xi3 xi4

)(4.3)

is an isomorphism.The following lemma is now immediate.

Lemma 4.5.LetB be a binary alphabet. There exists a bijective correspondenceα↔αbetween the homomorphismsα : X∗ → B∗ and the monoid homomorphismsα : M(X) →SL(2,N) such that the following diagram commutes:

M(X)

X∗ B∗

SL(2,N)

-

? ?-

α

α

µϕ

We shall now prove the main result of this section, from whichTheorem 4.4.

Theorem 4.8.Each system of equations over a finite set of variables has an equivalentfinite subsystem.

Proof. Let X and X be as above. Since, by Theorem 3.15, each monoidB∗ can beembedded into a free monoid generated by two elements, we canrestrict our solutionsα : X∗ → B∗ to the case whereB is binary, sayB = {a, b}.

For a wordw ∈ X∗ we denote

ϕ(w) =

(Pw1 Pw2

Pw3 Pw4

),

wherePwj ∈ Z[X]. Consider any equationu = v, whereu, v ∈ X∗. Define a finitesystemP(u, v) of polynomial equations as follows:

P(u, v) =

{Puj = Pvj for 1 ≤ j ≤ 4,

xt1xt4 − xt2xt3 = 1 for 1 ≤ t ≤ k.


Let α : X∗ → B∗ be a homomorphism, and letα be the corresponding monoid homo-morphism from Lemma 4.5. Now, by Lemmas 3.10, 4.4 and 4.5,

α(u) = α(v) ⇐⇒ µα(u) = µα(v) ⇐⇒ αϕ(u) = αϕ(v) . (4.4)

Define then a mappingα′ : X → N by

α(Xi) =

(α′(xi1) α′(xi2)α′(xi3) α′(xi4)

).

Now, α′ is a solution of the equations on the second line of the definition of P(u, v).This is becauseα is into SL(2,N). Further,α′ extends in a unique way to a ring homo-morphismα′ : Z[X ] → Z, for which

αϕ(w) =

(α′(Pw1) α′(Pw2)α′(Pw3) α′(Pw4)

)

for all w ∈ X∗. Therefore by (4.4),α(u) = α(v) if and only if α′ is a solution of thewholeP(u, v).

Let nowE = {ui = vi | i ∈ I} be a system of equations, and letP = ∪i∈IP(ui, vi).By Hilbert’s Basis Theorem,P has an equivalent finite subsystemP0 = ∪i∈I0P(ui, vi).Consider the subsystemE0 = {ui = vi | i ∈ I0} of E. By above,

α is a solution toE0 ⇐⇒ α′ is a solution toP0

⇐⇒ α′ is a solution toP ⇐⇒ α is a solution toE ,

which proves thatE0 is equivalent toE, and this proves the theorem. ⊓⊔

Equations with Constants

Let X be a set of variables andA an alphabet ofconstants. An equationu = v withconstantsA is a pair of wordsu, v ∈ (X ∪A)∗. A solutionof such an equationu = v isa homomorphismα : (X ∪ A)∗ → B∗, whereA ⊆ B, α(u) = α(v) andα(a) = a forall a ∈ A. A homomorphismα is a solution to a systemE of equations with constantsA, if it is a common solution to the equations inE.

With each constanta ∈ A we associate a new variablexa and in this way eachequationu = v with constantsA can be presented as a finite system of equations:

{u′ = v′

xa = a for all a ∈ A,

whereu′ andv′ are obtained fromu andv, resp., by substitutingxa for a ∈ A. Thereforeeach system of equations with constantsA can be presented as a system of equationsover the variablesX ∪ {xa | a ∈ A} together with a finite number of equationsxa = acontaining a unique constant. Therefore Theorem 4.8 can be generalized:


Corollary 4.4. Each system of equation with constants is equivalent to a finite subsys-tem.

Let u1 = v1 andu2 = v2 be two equations, and leta, b be two different constants.It is easy to show thatα is a solution to the above pair of equations if and only ifα is asolution to the single equationu1av2u1bv2 = v1au2v1bu2. In particular, we have

Theorem 4.9.Each finite set of equations with constants is equivalent to asingle equa-tion with constants.

Note, however, that the single equation of Theorem 4.9 need not be among the origi-nal ones.

On generalizations to arbitrary monoids

We consider here examples of the equality sets and systems ofequations in a moregeneral setting of monoids.

Let againX = {x1, x2, . . . , xn} be a finite set of variables. Asolution of an equationu = v overX in a monoidM is a monoid homomorphismα : X∗ → M , for whichα(u) = α(v). We say that a monoidM satisfies thecompactness property(for systemsof equations), or CP for short, if for all sets of variablesX every systemE ⊆ X∗ ×X∗

is equivalent to one of itsfinite subsystemsT ⊆ E.Hence Ehrenfeucht’s Conjecture states that the finitely generated free monoids sat-

isfy CP. Below we mention two examples, where (the generalization of) Ehrenfeucht’sConjecture does not hold.

Example 4.5.(1) LetFin(A∗) denote the monoid of all nonemptyfinite subsetsof theword monoidA∗. It was shown by Lawrence that the monoidFin(A∗) does not satisfyCP even whenA is a binary alphabet. Indeed, in this case the systemE of equations

x1xi2x1 = x1x

i3x1 (i ≥ 1)

over three variables does not have an equivalent finite subsystem inFin(A∗).

(2) Recall that the bicyclic monoidB is a 2-generator and 1-relator semigroup witha presentation〈a, b | ab = 1〉. The monoidB is isomorphic to the monoid generated bythe functionsα, β : N → N:

α(n) = max{0, n − 1}, β(n) = n+ 1 .

Defineγi = βiαi, for i ≥ 0. Hence

γi(n) =

{i if n ≤ i,

n if n > i.

We observe thatγiγj = γmin{i,j}. Consider the systemE ⊆ {x1, x2, x3}∗ consisting of

the equations

4.3 Isbell’s Zig-Zag Theorem for Dominions 49

xi1xi2x3 = x3 (i = 1, 2, . . . ) .

As is easily seen the homomorphismδj defined byδj(x1) = β, δj(x2) = α andδj(x3) = γj, is a solution ofxi1x

i2x3 = x3 for all i ≤ j, but δj is not a solution of

xj+11 xj+1

2 x3 = x3. Hence the systemE does not have an equivalent finite subsystem,and therefore the bicyclic monoidB does not satisfy CP. ⊓⊔

However, as an example, we have

Theorem 4.10.Every commutative monoid satisfies CP.

4.3 Isbell’s Zig-Zag Theorem for Dominions

Dominions

Let nowS be any semigroup. An elementx ∈ S is said to bedominated by a subsetU ⊆ S, if α(x) = β(x) for all homomorphismsα, β : S → P (to any semigroupP )wheneverα andβ agree onU , α|U = β|U . Thedominion of a subsetU in a semigroupS is the set of all elements ofS that are dominated byU :

Dom(U,S) = {x ∈ S | α|U = β|U =⇒ α(x) = β(x)

for all homomorphismsα, β : S → T}

Lemma 4.6.LetU ⊆ S. The dominionDom(U,S) is a subsemigroup ofS, and[U ]S ⊆Dom(U,S).

Proof. Assumex, y ∈ Dom(U,S), and supposeα, β : S → P are homomorphisms, forwhich α|U = β|U . Now, α(xy) = α(x)α(y) = β(x)β(y) = β(xy) (sinceα andβare homomorphisms, andx andy are dominated byU ), and so alsoxy ∈ Dom(U,S).ThereforeDom(U,S) is a subsemigroup ofS. It is clear thatU ⊆ Dom(U,S), and thusthe second claim follows from the first. ⊓⊔

Zig-zag theorem

A rather neat condition forDom(U,S) was given by Isbell in 1966 using zig-zags: Letx ∈ S. A sequence

x = u0y1 = x1u1y1 = x1u2y2 = x2u3y2 = · · · = xmu2m−1ym = xmu2m (4.5)

is called azig-zag ofx (in S over U ), if ui ∈ U for 0 ≤ i ≤ 2m andxi, yi ∈ S for1 ≤ i ≤ m, and

u2i−1yi = u2iyi+1 and xiu2i = xi+1u2i+1 (i = 1, 2, . . . ,m− 1) (4.6)

u0 = x1u1 and u2m−1ym = u2m . (4.7)


The length of the zig-zag (4.5) ism. The sequenceu0, u1, . . . , u2m of elements ofU is called thespineof the zig-zag (4.5). Thevalue of the zig-zag (4.5) isx.

The zig-zag sequence becomes more appealing, when it is written in the followingform:

x =︷︸︸︷u0 y1 =

︷︸︸︷x1u1 y1 = x1

︷︸︸︷u1y1 =

x1︷︸︸︷u2y2 =

︷︸︸︷x1u2 y2 =

︷︸︸︷x2u3 y2 = x2

︷︸︸︷u3y2 =

...︷︸︸︷xm−1u2m−3 ym−1 = xm−1

︷︸︸︷u2m−3ym−1 =

xm−1︷︸︸︷u2m−2ym =

︷︸︸︷xm−1u2m−2 ym =

xm︷︸︸︷u2m−1ym = xm

︷︸︸︷u2m .

Lemma 4.7.If two elementsx and y in S have the same spineu0, u1, . . . , u2m, thenx = y.

Proof. If

x = u0y1 = x1u1y1 = · · · = xmu2m−1ym = xmu2m,

y = u0v1 = z1u1v1 = · · · = zmu2m−1vm = zmu2m ,

theny = u0v1 = x1u1v1, sinceu0 = x1u1; and therefore

y = x1u1v1 = · · · = xmu2m−1vm = xmu2m = x ,

by the conditions required in (4.5). ⊓⊔

We prove the following theorem only in one direction, the proof in the opposite direc-tion requires either new methods (topological as in the original Isbell’s proof (1966), ortensor products as in the proofs by Stenstrom (1971), Howie(1976) and Storrer (1976),or semigroup diagrams and HNN extensions as in Jackson’s proof (1991)), or exten-sive technics in combinatorics of presentations (as in Higgins’ proof (1991) based onJackson’s proof).

Theorem 4.11 (Isbell’s Zig-Zag Theorem).LetU be a subsemigroup of a semigroupS. Then

Dom(U,S) = U ∪ {x | there is a zig-zag ofx in S overU} .

Proof. (1) Assume first that there exists a zig-zag (4.5) ofx, and letα, β : S → P betwo homomorphisms that agree on the subsemigroupU : α|U = β|U . In particular,

α(ui) = β(ui) (for all 1 ≤ i ≤ 2m) . (4.8)


For notational convenience we extend the homomorphismsα andβ to S1 (where anidentity element1 is added, if it was not there) by settingα(1) = 1 = β(1). Letx0 = 1andym+1 = 1. Hence (4.5) becomes

x = x0u0y1, xi−1u2i−2yi = xiu2i−1yi, xiu2i−1yi = xiu2iyi+1 , (4.9)

for all 1 ≤ i ≤ m. We prove by induction that

α(x) = β(xiu2i) · α(yi+1) (for all 0 ≤ i ≤ m) , (4.10)

from which it follows thatα(x) = β(xmu2m)α(ym+1) = β(x) as required.To begin the induction, we obtain using (4.8) that

α(x) = α(u0y1) = α(u0)α(y1) = β(u0)α(y1) = β(x0u0)α(y1) ,

and hence the claim (4.10) holds fori = 0.Make an induction hypothesis that the claim (4.10) holds fori < m. We show that it

holds fori+ 1. Indeed,

α(x)IH= β(xiu2i)α(yi+1)

(4.6)= β(xi+1u2i+1)α(yi+1)

= β(xi+1)β(u2i+1)α(yi+1)(4.8)= β(xi+1)α(u2i+1)α(yi+1)

= β(xi+1)α(u2i+1yi+1)(4.6)= β(xi+1)α(u2i+2yi+2)

= β(xi+1)α(u2i+2)α(yi+2)(4.8)= β(xi+1)β(u2i+2)α(yi+2)

= β(xi+1u2i+2)α(yi+2) ,

and this completes the induction on (4.10), and proves that each zig-zag produces anelement of the dominion.

(2) In converse, we give only the very basic idea of the proof due to Jackson (1991)as modified by Higgins (1991).

Assume thatx ∈ Dom(U,S), but x /∈ U . (The claim is trivial forx ∈ U ). LetS = 〈A | R〉 be a presentation ofS, and assumet /∈ A is anew letter. LetP be asemigroup which is obtained fromS by adding a new generatort and the followingadditional relations:

t2 = 1, tu = ut and tut = u .

for all u ∈ U . Hence

P = 〈A, t | R, tt = 1, tu = ut (u ∈ U)〉 .

Defineα, β : S → P by

α(y) = y and β(y) = tyt (y ∈ S) .

Now, α(u) = u = tut = β(u) for all u ∈ U , and thereforeα|U = β|U , and, byassumption, alsoα(x) = β(x) holds, sincex ∈ Dom(U,S). Since nowx = txt and sotx = ttxt = xt, alsotx = xt. The proof goes on now by showing that the conditiontx = xt implies thatx has zig-zag inS overU . ⊓⊔

5

Green’s Relations

5.1 Definitions

Introducing the relations

Let S be a semigroup, and define the following relations onS,

xLy ⇐⇒ S1x = S1y

xRy ⇐⇒ xS1 = yS1

xJy ⇐⇒ S1xS1 = S1yS1 .

HereS1x, xS1 andS1xS1 are theprincipal left ideal , theprincipal right ideal and theprincipal ideal generated byx ∈ S. By the definitions,

xLy ⇐⇒ ∃s, s′ ∈ S1 : x = sy and y = s′x

xRy ⇐⇒ ∃r, r′ ∈ S1 : x = yr and y = xr′ .

As an exercise we have

Lemma 5.1.The relationsL, R and J are equivalence relations onS. In fact,L is aright congruence andR is a left congruence ofS.

We denote the corresponding equivalence classes containing x by

Lx = {y | xLy}, Rx = {y | xRy}, Jx = {y | xJy} .

Example 5.1.Consider the semigroupS from the table. Then

S1a = {a, d}, S1b = {a, b, c, d}, S1c = {a, b, c, d}, S1d = {a, d}

aS1 = {a}, bS1 = {a, b}, cS1 = {a, c}, dS1 = {d} .

· a b c d

a a a a ab a b b ac a c c ad d d d d

The equivalence classes with respect toL areLa = {a, d} = Ld andLb = {b, c} = Lc

and those with respect toR,Ra = {a}, Rb = {b}, Rc = {c} andRd = {d}. ⊓⊔

Example 5.2.Let TX be the full transformation semigroup onX. Now, forα, β ∈ TX ,

αRβ ⇐⇒ ∃γ, γ′ ∈ TX : α = βγ and β = αγ′ .

5.1 Definitions 53

Therefore,αRβ implies thatα(X) = β(X). On the other hand, ifα(X) = β(X), thendefineγ ∈ TX as a function, for which

γ(x) = somex with β(x) = α(x) .

Then, in the above notations,βγ(x) = β(x) = α(x), and henceα ∈ βT 1X . Similarly,

β ∈ αT 1X , and hence,αRβ ⇐⇒ α(X) = β(X). ⊓⊔

Theorem 5.1.The relationsL andR commute:L ◦R = R ◦ L.

Proof. Suppose that(x, y) ∈ L ◦R. This means that there exists an elementz ∈ S suchthatxLz andzRy. Therefore there are elementss, s′, r, r′ ∈ S such that

x = sz, z = s′x, z = yr, y = zr′ .

Denotet = szr′. Now,

t = sz · r′ = xr′,

x = sz = syr = szr′r = tr ,

which means thatxRt. On the other hand,

t = s · zr′ = sy,

y = zr′ = s′xr′ = s′szr′ = s′t ,

which implies thatyLt. SincexRt andtLy, also(x, y) ∈ R ◦ L. We have shown thatL ◦R ⊆ R ◦ L. The inclusion in the other direction is proved in the same way. ⊓⊔

D- and H-relations

The previous result is important because it implies that theproduct

D = L ◦R

is an equivalence relation.

Lemma 5.2.D is the smallest equivalence relation that contains bothL andR.

Proof. Clearly,L ⊆ D andR ⊆ D, sincexLx andxRx hold for allx ∈ S1.In order to show thatD is an equivalence relation one needs to show that it is transi-

tive: if xDz andzDy, then alsoxDy. This is an exercise.In order to show thatD is the smallest equivalence relation containingL ∪ R one

assumes an equivalence relationC with L ∪ R ⊆ C, and shows thatD ⊆ C. This is anexercise. ⊓⊔

5.1 Definitions 54

H

L R

D

J

Fig. 5.1. The inclusion diagram of the Green’s relations

LetH = L ∩ R .

HenceH is an equivalence relation, and it is the largest equivalence relation ofS that iscontained in bothL andR.

The inclusion diagram of theGreen’s relations is given in Fig. 5.1. Here one needsto observe thatD ⊆ J by Lemma 5.2.

We denote byDx andHx the corresponding equivalence classes that contain theelementx ∈ S. Clearly, for allx ∈ S

Hx = Lx ∩Rx .

On D-classes

Lemma 5.3.For each semigroupS we have

xDy ⇐⇒ Lx ∩Ry 6= ∅ ⇐⇒ Ly ∩Rx 6= ∅ .

Moreover,Dx =

⋃

y∈Dx

Ly =⋃

y∈Dx

Ry .

Proof. By the definition ofD,

xDy ⇐⇒ ∃z ∈ S : xLz and zRy ⇐⇒ ∃s ∈ S : xRs and sLy .

The first claim follows from this.For the second claim is trivial, sinceL ⊆ D andR ⊆ D. ⊓⊔

The situation can be visualized by the followingeggboxpicture:Here the rows areR-classes and the columns areL-classes. Their intersections are

H-classes, if nonempty (the intersectionLy ∩ Rz is nonempty, ifyDz). Indeed, ifu ∈Ly ∩Rz, thenLy = Lu andRz = Ru, and soLy ∩Rz = Hu.

5.1 Definitions 55

Ly ∩ Rz

Ly

Rz

Fig. 5.2. The eggbox visualization of aD-classDx

Example 5.3.LetX = {1, 2, . . . , 7}, and consider the subsemigroupS of the full trans-formation semigroupTX generated byα andβ as defined in the table.

1 2 3 4 5 6 7α 2 2 5 6 5 6 5β 3 4 3 4 7 4 7

This semigroup consists of six elementsα, β, αβ, βα, αβα andβαβ.By computing we getβαβα = βα, and henceβαRβαβ. Also,αβRαβα holds, as

doβαLαβα andαβLβαβ.

αβ

βαβ

αβα

βα

Fig. 5.3. Eggbox forDβα

We conclude (after some calculations) that the eggbox for the classDβα looks likein the figure. ⊓⊔

5.2 Green’s Lemma and Its Corollaries 56

5.2 Green’s Lemma and Its Corollaries

Green’s lemma(s)

Let S be a semigroup. For eachs ∈ S we define (as before) the mappingρs : S1 → S1

by∀z ∈ S1 : ρs(z) = sz .

The usefulness of the Green’s relations are due to the following version ofGreen’slemma.

Lemma 5.4.Let S be a semigroup,xLy, and lets, s′ ∈ S1 be such thatsx = y ands′y = x. Then

1. ρs : Rx → Ry is a bijection, andρs′ : Ry → Rx is a bijection,2. ρs′ = ρ−1

s is the inverse function ofρs restricted toRx ,3. ρs fixes theL-classes, that is,zLρs(z) for all z ∈ Rx ,4. ρs preservesH-classes, that is, for allu, v ∈ Rx: uHv ⇐⇒ ρs(u)Hρs(v) .

x z

Lx

Rx

y ρs(z)Ry �

Fig. 5.4. Green’s lemma

Proof. First we have to prove thatρs mapsRx into Ry. For this, letz ∈ Rx, that is,zS1 = xS1. We have now

szS1 = sxS1 = yS1 ,

which shows thatρs(z) = sz ∈ Ry as required. A symmetric argument shows thatρs′

mapsRy intoRx.If z ∈ Rx, thenzRx and therefore there are elementsu, u′ ∈ S1 such thatz = xu

andx = zu′. Now,

s′sz = s′s · xu = s′ · sx · u = s′yu = xu = z ,

and hence∀z ∈ Rx : s′sz = z . (5.1)


Henceρs′ρs(z) = ρs′(sz) = s′sz = z ,

and soρs′ρs is the identity mapping ofRx. Similarly, ρsρs′ is the idenitity mapping ofRy, from which we conclude Claims (1) and (2).

For Case (3) assumez ∈ Rx. By (5.1), the elementsz andρs(z) (= sz) are in thesameL-class.

For Case (4) we notice thatuHv if and only if uLv anduRv; and so

uHv =⇒ ρs(u)Lρs(v) and ρs(u)Rρs(v) =⇒ ρs(u)Hρs(v)

by Case (3) and sinceR is a left congruence (andρs(u) = su, ρs(v) = sv). On the otherhand, ifρs(u)Hρs(v), that is,suHsv, thenuHv, since, by equation (5.1),u = s′su andv = s′sv (and sinceR is a left congruence andρs′ fixes theL-classes). ⊓⊔

In particular,ρs maps eachH-classHz (with z ∈ Rx) bijectively onto theH-classHρs(z).

The next dual version of Lemma 5.4 is proved similarly. Hereλr : S1 → S1 is the

opposite version ofρr:∀z ∈ S : λr(z) = zr .

Lemma 5.5.Let S be a semigroup,yRz, and letr, r′ ∈ S1 be such thatyr = z andzr′ = y. Then

1. λr : Ly → Lz is a bijection, andλr′ : Lz → Ly is a bijection,2. λr′ = λ−1

r is the inverse function ofλr restricted toLy, and3. λr preserves theR-classes, that is,wRλr(w) for all w ∈ Ly.

?

x u

y z �

6

-

λr

ρs

λr′

ρs′

Fig. 5.5. Combination of Green’s lemmas

Lemma 5.6.Lete ∈ ES be an idempotent. IfxLe, thenxe = x. If xRe, thenex = x.


Proof. If xLe, thenx ∈ Le, that is,x = se for somes ∈ S1. Therefore, sincee = ee,x = se · e = xe. The other case is similar. ⊓⊔

Corollary 5.1. EachH-class contains at most one idempotent.

Sizes of the H-classes

In the above notations,ρs is a bijectionRx → Ry andλr is a bijectionLy → Lz andtherefore

Corollary 5.2. TheH-classes inside aD-class have the same cardinality, that is, ifxDy,then there exists a bijection betweenHx andHy.

Proof. Indeed, ifx, z ∈ S are in the sameD-class such thatxLy, yRz with

sx = y, s′y = x, yr = z, zr′ = y ,

then by the above lemmasρs : Hx → Hy andλr : Hy → Hz are bijections. Thereforeλrρs : Hx → Hz is a bijection. ⊓⊔

Green’s theorem for H-classes

The following rather crusial result is thelocation theoremof Miller and Clifford (1956).

Theorem 5.2.Letx, y ∈ S. Then

xy ∈ Rx ∩ Ly ⇐⇒ Ry ∩ Lx contains a unique idempotent.

LyLx

Ry

Rx xyx

ye

Fig. 5.6. Location theorem

Proof. Suppose first thatxy ∈ Rx ∩ Ly. SinceyLxy, we may chooses = x inLemma 5.4, and soρx : Ry → Rxy is a bijection. AlsoxyRx, and henceRxy = Rx,which means thatρx : Ry → Rx is a bijection. The mappingρx preservesL-classes andsoρx mapsRy ∩ Lx ontoRx ∩ Lx = Hx. Therefore there exists az ∈ Ry ∩ Lx suchthatρx(z) = x, that is,xz = x.


BecausezLx, there exists au ∈ S1 with z = ux. We have thus obtainedxux =xz = x, and thuszz = uxux = ux = z, which means thatz ∈ ES .

In the other direction, suppose there exists an idempotente ∈ Ry ∩ Lx. Now, byLemma 5.6,ey = y andxe = x. From eRy we obtain thatxeRxy, and thusxRxy.FromeLx we obtain thateyLxy, and thusyLxy. Soxy ∈ Rx ∩ Ly as required.

Finally, if Ry ∩Lx contains an idempotent, it must be unique by Corollary 5.1, sinceRy ∩ Lx is anH-class. ⊓⊔

In the following Green’s theorem, G is a subgroup of a semigroupS, if G is asubsemigroup, which is a group.

We start with a small lemma.

Lemma 5.7.Lete, f ∈ ES . Then for allx ∈ Re∩Lf there existsy ∈ Rf ∩Le such thatxy = e andyx = f .

Proof. Let x ∈ Re ∩ Lf . Hence by Lemma 5.6,x = ex = xf , and there areu, v ∈ S1,for which e = xu andf = vx. Now,y = fu is the required element of the claim:

First of allf = vx = vex = vxux = fux = yx (5.2)

ande = xu = xfu = xy . (5.3)

Hence,y ∈ Rf by (5.2) and the fact thaty = f · u; andy ∈ Le, by (5.3) and sincey = fu = vxu = ve. Soy ∈ Rf ∩ Le. ⊓⊔

Theorem 5.3.LetH be anH-class ofS. Then the following are equivalent:

1.H contains an idempotent.2. There existsx, y ∈ H such thatxy ∈ H.3.H is a subgroup ofS.

Proof. That Case (1) implies Case (2) is clear, since we can choosex = e = y for theidempotente ∈ H.

Assume Case (2). Then Case (1) follows by Theorem 5.2, since now Rx ∩ Ly =H = Ry ∩ Lx. ThereforeH contains an idempotente, and hence by the converse claimof Theorem 5.2, we know thatH is a subsemigroup ofS. Further,H is a monoid withidentity e by Lemma 5.6. We apply then Lemma 5.7 fore = f . This shows thatH is agroup. Hence Case (2) implies Case (3).

Case (3) implies Case (1), since the identity element of the groupH is an idempotentof S. ⊓⊔

Example: periodic semigroups

We say that a semigroupS is periodic, if each of its elements has a finite order, that is,the monogenic subsemigroup[x] is finite for allx ∈ S.



Lemma 5.8.If S is periodic, then each[x] contains an idempotent.

Theorem 5.4.For periodic semigroupsJ = D.

Proof. The inclusionD ⊆ J is valid for all semigroups. Thus we have to show thatJ ⊆ D.

Suppose thatxJy. Therefore there existu, v, r, s ∈ S1 such that

x = uyv and y = rxs .

Now,∀i ≥ 0 : x = (ur)ix(sv)i andy = (ru)iy(vs)i .

By Lemma 5.8, there exists a nonnegativen such that(ur)n ∈ ES . We may supposehere thatur 6= 1, for otherwise alreadyxRy and thusxDy, sinceR ⊆ D.

Denotez = rx, for short. We have then

x = (ur)nx(sv)n = (ur)n(ur)nx(sv)n = (ur)nx = (ur)n−1uz ,

and soxLz, sincez = r · x. Further,y = rxs = zs and if (vs)k ∈ ES , then

z = rx = r(ur)k+1x(sv)k+1 = (ru)k+1rxs(vs)kv = (ru)k+1y(vs)kv =

(ru)k+1y(vs)k(vs)kv = (ru)k+1y(vs)k+1(vs)k−1v = y(vs)k−1v ,

which shows thatzRy, and soxDy. This proves thatJ = D. ⊓⊔

Corollary 5.3. For all finite semigroupsS, J = D.

6

Inverse Semigroups

6.1 Regular Semigroups

Basic properties of regular elements

We say that an elementx of a semigroupS is regular, if there exists an elementy ∈ Ssuch that

x = xyx .

Also,S is regular, if each of its elements is regular.

Example 6.1.(1) All groups are regular:x = xx−1x, wherex−1 is the group inverse ofx. A regular semigroup is a group if and only if it has exactly one idempotent (Exercise).

(2) If e ∈ ES is an idempotent, thene = eee, and hence all idempotents of a semi-group are regular elements ofS.

(3) The full transformation semigroupTX is regular (Exercise). ⊓⊔

We obtain immediately

Lemma 6.1.Letx be a regular element inS: x = xyx. Thenxy ∈ ES andyx ∈ ES .

Lemma 6.2.Letx ∈ S. Then the following are equivalent:

1. x is regular.2. ∃e ∈ ES : xRe.3. ∃f ∈ ES : xLf .

Proof. Assume first thatx is regular, and letx = xyx. Thene = xy ∈ ES andf = yxsatisfyxRe andxLf .

Suppose then thatxRe. Now,x = ex by Lemma 5.6, and there exists ay ∈ S1 suchthate = xy. Thereforex = ex = xyx, andx is regular. The case forL is similar. ⊓⊔

Note that, by Lemma 5.7, ife, f ∈ ES , then for allx ∈ Re ∩ Lf there exists a(unique)y ∈ Rf ∩ Le such thatxy = e andyx = f .

6.1 Regular Semigroups 62

x e = xy

f = yx y

Rx = Re

Rf

Lx = Lf Le

Fig. 6.1. The eggbox ofDx for a regularx

Regular D-classes

We say that a classDx is regular, if it contains only regular elements.

Lemma 6.3.If x ∈ S is regular element, thenDx is regular.

Proof. Suppose thatx is regular. Hence, by Lemma 6.2, the classDx contains an idem-potente, that is,Dx = De.

Let thenz ∈ Dx. Now, zDe, and thus there existsu ∈ Dx with eRu anduLz. Sothere are elementsr, s, s′ ∈ S such that

e = ur, u = eu, u = sz, z = s′u .

Here,u = eu by Lemma 5.6. Now,

z = s′u = s′eu = s′esz = s′ursz = z · rs · z ,

and soz is regular. ⊓⊔

In particular,De is regular for each idempotente ∈ ES .

Lemma 6.4.If a D-classD is regular, then eachLx ⊆ D and eachRx ⊆ D containsan idempotent.

Proof. Whenx ∈ D, thenx = xyx for somey, and hencexLyx andxRxy, wherexyandyx are idempotents. ⊓⊔

From these we obtain

Theorem 6.1.A D-class is regular if and only if it contains an idempotent.

6.2 Inverse Semigroups 63

Inverse elements in semigroups

We say thaty ∈ S is aninverse elementof x ∈ S, if

x = xyx and y = yxy .

Note that an inverse element ofx, if such an element exists, need not be unique.

Lemma 6.5.Each regular elementx ∈ S has an inverse element.

Proof. If x ∈ S is regular, then for somey ∈ S, x = xyx. Now, yxy = yxy · x · yxy,and soyxy is also regular. Also,x = x · yxy · x and consequentlyyxy is an inverseelement ofx. ⊓⊔

Lallement’s lemma

Lemma 6.6.LetS be a regular semigroup, and letα : S ։ P be an epimorphism ontoa semigroupP . If e ∈ EP , thenα−1(e) ∩ ES 6= ∅, that is, there exists an idempotentf ∈ ES such thatα(f) = e.

Proof. Let x ∈ S be such thatα(x) = e, and lety be an inverse element ofx2 in S:x2 = x2yx2 andy = yx2y. Then forf = xyx,

α(f) = α(x)α(y)α(x) = α(x)2α(y)α(x)2 = α(x2yx2) = α(x2) = α(x)2 = e2 = e ,

that is,α(f) = e. Heref is an idempotent:xyx · xyx = x · yx2y · x = xyx. ⊓⊔

As a consequence we have

Theorem 6.2.Letρ be a congruence of a regularS, then

xρ ∈ ES/ρ =⇒ ∃e ∈ ES : xρ = eρ .

As an exercise

Theorem 6.3.If α : S → P is a homomorphism from a regular semigroupS, thenα(S)is regular. In particular, ifα is an epimorphism, thenP is regular.

6.2 Inverse Semigroups

Example

A semigroupS is called aninverse semigroup, if eachx ∈ S has aunique inverseelementx−1:

x = xx−1x and x−1 = x−1xx−1 .


Example 6.2.(1) If S is a group, then it is an inverse semigroup, andx−1 is the groupinverse of the elementx.

(2) ConsiderZ × Z as a drawing board so that you can draw a unit lineupu, downd, right r and left ℓ. A wordw ∈ A∗ = {u, d, r, ℓ}∗ produces a figure onZ × Z, whenyou start from(0, 0) and follow the instructions given by the wordw.

◦

·

· ·

•

◦

· •

·

w = uurd v = ruℓdur

Fig. 6.2. The figures drawn by the wordsw andv

We identify the wordsw1 andw2, and denotew1 ⊲⊳ w2, if they draw the same figurewith the same endpoint. This means that

u ⊲⊳ udu, d ⊲⊳ dud, r ⊲⊳ rℓr, andℓ ⊲⊳ ℓrℓ . (6.1)

It should be clear that ifw1 ⊲⊳ w2 andv ∈ A∗, then alsovw1 ⊲⊳ vw2 andw1v ⊲⊳ w2v,and hence⊲⊳ is a congruence on the word semigroupA+. The quotientT2 = A+/ ⊲⊳might be called the (two-dimensional)turtle semigroup. It is an inverse semigroup withu−1 = d, d−1 = u, r−1 = ℓ and ℓ−1 = r, and the inverse of an elementw ⊲⊳ fora wordw = x1x2 . . . xn ∈ A+ is obtained by reversing the order of the instructions:w−1 = x−1

n x−1n−1 . . . x

−11 . ⊓⊔

The semilattice of idempotents

If e ∈ ES for an inverse semigroup, theneee = e, and hence for all idempotentse,e−1 = e.

Theorem 6.4.Let S be an inverse semigroup. Then the idempotentsES form a sub-semigroup ofS. Moreover,ES is a semilattice, that is, the idempotents of an inversesemigroup commute.

Proof. Let e, f ∈ ES and consider the (unique) inverse elementx = (ef)−1 of ef .Now,

ef = ef · x · ef =

{ef · xe · ef

ef · fx · ef

and

xe · ef · xe = xefx · e = xe ,

fx · ef · fx = f · xefx = fx .


This means thatx = (ef)−1 = xe = fx. Herex ∈ ES, since

x2 = xe · fx = x · ef · x = x ,

and soef ∈ ES for all e, f ∈ ES , that is,ES is a subsemigroup ofS.Further,ES is commutative: Fore, f ∈ ES , alsoef, fe ∈ ES, and

ef · fe · ef = efef = (ef)2 = ef and fe · ef · fe = fefe = (fe)2 = fe ,

meaning thatfe = (ef)−1 = ef . ⊓⊔

Corollary 6.1. AssumeS = [X]S . If each generatorx ∈ X has a unique inverse ele-ment, thenS is an inverse semigroup:

(x1x2 . . . xn)−1 = x−1

n x−1n−1 . . . x

−11

for all xi ∈ X.

Proof. Assumex, y ∈ S have unique inverse elementsx−1 andy−1, respectively. Then

xy · y−1x−1 · xy = x · yy−1 · x−1x · y = x · x−1x · yy−1 · y = xy .

The rest of the claim follows by induction. ⊓⊔

Corollary 6.2. In an inverse semigroupS, for all x ∈ S, x = (x−1)−1.

A characterization

Theorem 6.5.LetS be a semigroup. The following are equivalent:

1. S is an inverse semigroup.2. S is regular and its idempotents commute.3. EachL-class andR-class contains an idempotent.

Proof. Case (1) implies Case (2) by Theorem 6.4.Suppose Case (2). By Lemma 6.4, eachL-class andR-class contains a unique idem-

potent. For the uniqueness letf ∈ Le, wheree, f ∈ ES . HenceeLf , and therefore therearex, y ∈ S1 such thate = xf andf = ye. From here we obtain

e = xf = xff = ef = fe = yee = ye = f .

Similarly, eRf implies thate = f . So Case (2) implies Case (3).Suppose Case (3). Now eachD-class contains an idempotent, and hence, by Theo-

rem 6.1, eachx ∈ S has an inverse element. Suppose an elementx has two inverse ele-mentsy andz. Now,yx, zx ∈ ES with yxLx andzxLx. Then, by assumption,yx = zx.A similar reasoning usingR shows thatxy = xz. Thereforey = yxy = zxz = z, andCase (1) follows. ⊓⊔


As exercises we have

Corollary 6.3. LetS be an inverse semigroup. Then

∀x ∈ S : x−1ESx ⊆ ES .

Theorem 6.6.LetS be an inverse semigroup, and letx, y ∈ S ande, f ∈ ES . Then

1. xLy ⇐⇒ x−1x = y−1y.2. xRy ⇐⇒ xx−1 = yy−1.3. eDf ⇐⇒ ∃z ∈ S : e = zz−1 and f = z−1z.

Partial ordering inverse semigroups

Recall that in any semigroupS the idempotents can be partially ordered by the relation:

e ≤ f ⇐⇒ ef = e = fe .

This partial order generalizes in an inverse semigroupS to all elements ofS as follows,

x ≤ y ⇐⇒ ∃e ∈ ES : x = ey .

Indeed, here≤ is

• reflexive, sincex = (xx−1) · x, wherexx−1 ∈ ES ;• antisymmetric, since ifx = ey andy = fx, thenx = ey = eey = ex, and so

x = ey = efx = fex = fx = y;• transitive, since ifx = ey andy = fz, then alsox = ey = efz, whereef ∈ ES .

If you restrict≤ onto ES you get the above partial order of idempotents. Indeed, ife ≤ f , then there existsg ∈ ES such thate = gf , and heree = gff = ef = fe asrequired.

As an exercise we state

Lemma 6.7.In an inverse semigroupS we have

x ≤ y ⇐⇒ ∃e ∈ ES : x = ye ⇐⇒ xx−1 = yx−1 ⇐⇒ x = xy−1x

⇐⇒ xx−1 = xy−1 ⇐⇒ x−1x = y−1x ⇐⇒ x−1x = x−1y ⇐⇒ x = xx−1y .

Groups as inverse semigroups

All groups are inverse semigroups, and moreover

Theorem 6.7.LetS be a semigroup. Then the following are equivalent:

1. S is a group.2. ∀x ∈ S ∃!x′ ∈ S : x = xx′x.

6.3 Representations by Injective Partial Mappings 67

3. ∀x ∈ S ∃!x′ ∈ S : xx′ ∈ ES .4. S is an inverse semigroup that satisfies:x = xyx =⇒ y = yxy.

Proof. Case (1) implies Case (2) by choosingx′ = x−1, the group inverse ofx.Case (2) implies Case (3): In one direction, the claim is clear, since ifx = xx′x, then

xx′ ∈ ES . Conversely assume thatx = xx′x for a uniquex′ (in which casexx′ ∈ ES),andxy ∈ ES . Now,

xyx = xy · xy · x = xyx · x′ · xyx ,

andxyx = xyx · y · xyx ,

wherefrom we obtain thaty = x′, when (2) is applied toxyx.Case (3) implies Case (4), since ifx = xyx, thenxy ∈ ES, and fromxy = xyxy ∈

ES we gety = yxy by the uniqueness assumption, and herey = x′ = x−1.Case (4) implies Case (1), since

xyy−1 = xx−1x · yy−1yy−1 = xyy−1 · x−1 · xyy−1

=⇒ x−1 = x−1 · xyy−1 · x−1

=⇒ x = x · yy−1x−1 · x = xx−1x · ·yy−1 = xyy−1 .

Similarly, x = yy−1x, and soy−1 is a group inverse ofy. ⊓⊔

Theorem 6.8.If an inverse semigroupS is right cancellative, then it is a group.

Proof. Supposexy ∈ ES . We havexy = xyxy, and hence by cancellation,x = xyxandy = yxy, which means thaty = x−1, sinceS is inversive. ⊓⊔

6.3 Representations by Injective Partial Mappings

Partial mappings

Let X 6= ∅ be a set. Apartial mapping α : X → X is a function from a subsetY =dom(α) of X ontoran(α) = α(Y ) ⊆ X. A partial mappingα : X → X is undefinedon allx /∈ dom(α).

Example 6.3.LetX = {1, 2, . . . , 5} and letdom(α) = {2, 4, 5} with α(2) = 1, α(4) =5 andα(5) = 1. We can representα as follows

α =

(2 4 51 5 1

).

Hereran(α) = {1, 5}. ⊓⊔


We say that a partial mappingα : X → X is injective, if α(x) 6= α(y) for all x 6= ywith x, y ∈ dom(α). The injective partial mappings form a semigroup, denotedIX ,under the usual composition:

(βα)(x) = β(α(x)) if x ∈ dom(α) and α(x) ∈ dom(β) .

We observe that

dom(βα) = α−1(ran(α) ∩ dom(β)) and ran(βα) = β(ran(α) ∩ dom(β)) .(6.2)

We denote byιY : X → X the partial function such thatdom(ιY ) = Y = ran(ι)andιY (y) = y for all y ∈ Y .

Theorem 6.9.IX is an inverse semigroup.

Proof. Let α ∈ IX , and letα−1 : X → X be defined by

α−1(y) = x if α(x) = y for x ∈ dom(α) ,

that is,dom(α−1) = ran(α) andran(α−1) = dom(α). Moreover,α−1(α(x)) = x forall x ∈ dom(α). Clearly,α−1 ∈ IX . Furthermore,

α−1α = ιdom(α) and αα−1 = ιran(α) .

Now,αα−1α = α andα−1αα−1 = α−1, and soα−1 is an inverse element ofα, andIXis a regular semigroup.

By Theorem 6.4, we need to show that the idempotents ofIX commute. For this weprove that

ε ∈ EIX ⇐⇒ ε = ιY for someY ⊆ X .

Indeed, supposeε is an idempotent ofIX . Then by (6.2),

dom(ε2) = ε−1(ran(ε) ∩ dom(ε)) = dom(ε) ,

and so by injectivity,

ran(ε) ∩ dom(ε) = ε(dom(ε)) = ran(ε) ,

which implies thatdom(ε) ⊆ ran(ε). Similarly,ran(ε2) = ran(ε) implies thatran(ε) ⊆dom(ε), and soran(ε) = dom(ε). Further,ε2(x) = ε(x) for all x ∈ dom(ε), and thus,by injectivity, ε(x) = x for all x ∈ dom(ε).

Finally, the idempotents commute, since for allY,Z ⊆ X,

ιZιY = ιY ∩Z = ιY ιZ ,

and so the theorem is proved. ⊓⊔


The Vagner-Preston representation

Theorem 6.10.Each inverse semigroupS has a faithful representation as a semigroupof injective partial mappings, that is, there exists an embeddingϕ : S → IX for somesetX.

Proof. We chooseX = S itself, and define for eachx ∈ S the mapping

τx : x−1S → S by τx(y) = xy (y ∈ x−1S) .

First of allx−1S = x−1xS, (6.3)

becausex−1xS ⊆ x−1S, and x−1S = x−1xx−1S ⊆ x−1xS. In particular, sincex−1x ∈ ES ,

y ∈ dom(τx) = x−1S ⇐⇒ y = x−1xy . (6.4)

To show thatτx ∈ IX we need to show that it is injective. For this lety, z ∈ x−1S besuch thatτx(y) = τx(z). Thusxy = xz, and so by (6.4),y = x−1xy = x−1xz = z.

We need also:dom(τyx) = dom(τyτx) . (6.5)

To prove this we notice that

z ∈ dom(τyτx) ⇐⇒ z ∈ dom(τx) and τx(z) ∈ dom(τy) ,

that isz ∈ dom(τyτx) ⇐⇒ z = x−1xz and xz = y−1yxz ,

by (6.4). Therefore ifz ∈ dom(τyτx), then

z = x−1xz = x−1y−1yxz = (yx)−1 · yx · z

and hencez ∈ dom(τyx) by (6.4). On the other hand, ifz ∈ dom(τyx), then z =(yx)−1 · yx · z, and soz = x−1x · x−1y−1yxz = x−1xz, from which it follows thatz ∈ dom(τx). Also, in this case, by commuting the idempotents,

xz = xx−1y−1yxz = y−1yxx−1xz = y−1yxz ,

and henceτx(z) = xz ∈ dom(τy). This proves (6.5).Define thenϕ : S → IX by

ϕ(x) = τx (x ∈ S) .

Now,ϕ is a homomorphism: Whenz ∈ dom(τyτx) = dom(τyx), then

τyτx(z) = τy(xz) = yxz = τyx(z) ,

and soτyτx = τyx, that is,

6.4 Congruences of Inverse semigroups 70

ϕ(yx) = τyx = τyτx = ϕ(y)ϕ(x) .

Finally, ϕ is injective: If ϕ(x) = ϕ(y), thenτx = τy, and so by (6.3),x−1xS =y−1yS, from which it follows thatx−1xRy−1y for the idempotentsx−1x andy−1y. ByTheorem 6.6,x−1x = y−1y. Now, sincex−1x ∈ x−1S = dom(τx) = dom(τy), itfollows thatτx(x−1x) = τy(x

−1x). Hereτx(x−1x) = xx−1x = x andτy(x−1x) =yx−1x = yy−1y = y. Thereforex = y as required. ⊓⊔

6.4 Congruences of Inverse semigroups

Heritage of images

Lemma 6.8.Let S be an inverse semigroup andα : S → P a homomorphism. Thenα(S) is an inverse subsemigroup ofP .

Proof. We have for allx ∈ S, α(x) = α(xx−1x) = α(x) · α(x−1) · α(x), and soα(S)is a regular subsemigroup ofP . Again, it sufficies to show that the idempotents ofα(S)commute.

Let g, h ∈ Eα(S). By Lemma 6.6, there existe, f ∈ ES so thatg = α(e) andh = α(f). Now,gf = α(ef) = α(fe) = fg, from which the claim follows. ⊓⊔

In particular,

Corollary 6.4. If ρ is a congruence of an inverse semigroupS, thenS/ρ is an inversesemigroup.

Therefore,

Lemma 6.9.LetS be an inverse semigroup, andρ its congruence. Then

xρy ⇐⇒ x−1ρy−1 .

We obtain also that for each homomorphismα : S → P for an inverse semigroupS,

∀x ∈ S : α(x−1) = α(x)−1 .

A subsemigroupT of an inverse semigroupS is called ainverse subsemigroup, iffor all x ∈ T alsox−1 ∈ T , wherex−1 is the inverse element ofx in S. Notice that notall subsemigroups of an inverse semigroup are inverse subsemigroups.

The following lemma is an exercise.

Lemma 6.10.LetS be an inverse semigroup, and letA be a subsemigroup ofS. ThenA is an inverse subsemigroup ofS if and only ifx−1 ∈ A for all x ∈ A.

Lemma 6.11.LetS be an inverse semigroup,α : S → P an epimorphism, and lete ∈EP . Thenα−1(e) is an inverse subsemigroup ofS.


Proof. First of all, if α(x) = e = α(y), thenα(xy) = e2 = e, and soxy ∈ α−1(e).This means thatα−1(e) is a sunsemigroup ofS.

The claim follows when we show thatx ∈ α−1(e) impliesx−1 ∈ α−1(e). For thiswe just notice that ifα(x) = e, thenα(x−1) = α(x)−1 = e−1 = e. ⊓⊔

For ideals we have the following rather strong closure property (which will be anexercise). In general, ifI is an ideal of a semigroupS, thenS is called anideal extensionof I by S/I, whereS/I is the Rees quotient.

Theorem 6.11.Let I be an ideal of a semigroupS. ThenS is an inverse semigroup ifand only ifI andS/I are inverse semigroups

Kernels and traces

RES?Let ρ be a congruence of a semigroupS. We define itskernel ker(ρ) andtrace tr(ρ)

as follows:

ker(ρ) = {x ∈ S | xρe for some e ∈ ES} =⋃

e∈ES

eρ ,

tr(ρ) = ρ(res)E = {(e, f) | e, f ∈ ES} .

Theorem 6.12.LetS be an inverse semigroup. Then for all congruencesρ andδ,

ρ ⊆ δ ⇐⇒ ∀e ∈ ES : eρ ⊆ eδ .

Proof. In one direction the claim is trivial. Suppose then thateρ ⊆ eδ for all e ∈ ES .Now,

xρy =⇒ xx−1ρyx−1 =⇒ xx−1δyx−1 =⇒ xδyx−1x ,

xρy =⇒ x−1xρx−1y =⇒ x−1xδx−1y =⇒ yx−1xδyx−1y ,

and thusxρy =⇒ xδyx−1y . (6.6)

On the other hand,

xρy =⇒ x−1ρy−1 =⇒ x−1yρy−1y =⇒ x−1yδy−1y =⇒ yx−1yδy ,

and, by combining this with (6.6), we obtain thatxδy holds. This shows thatρ ⊆ δ. ⊓⊔

Corollary 6.5. For an inverse semigroupS,

ρ = δ ⇐⇒ ∀e ∈ ES : eρ = eδ

for all congruencesρ andδ.


We have thenVagner’s theorem:

Theorem 6.13.LetS be an inverse semigroup, and letρ andδ be its congruences. Then

ρ = δ ⇐⇒ ker(ρ) = ker(δ) and tr(ρ) = tr(δ) .

In other words, Ifα : S ։ P andβ : S ։ T are epimorphisms from an inversesemigroupS, thenker(α) = ker(β) if and only if for all x ∈ S ande, f ∈ ES ,

α(x) ∈ EP ⇐⇒ β(x) ∈ ET ,

α(e) = α(f) ⇐⇒ β(e) = β(f) .

Proof. In one direction the claim is again trivial. Suppose then that ker(ρ) = ker(δ) andtr(ρ) = tr(δ). If e ∈ ES , then

eρx =⇒ ∃f ∈ ES : fδx =⇒ ff−1 = fδxx−1 ,

eρx =⇒ ee−1 = eρxx−1 =⇒ eδxx−1 ,

and thuseδf andfδx hold, from which we obtaineδx, that is,eρ ⊆ eδ. The case foreδ ⊆ eρ is similar, and hence we can conclude thateρ = eδ, and finallyρ = δ byCorollary 6.5. ⊓⊔

Classifications according to traces

Congruences of an inverse semigroup are classified according to their traces.

Lemma 6.12.For all x ∈ S ande ∈ ES , x−1ex ∈ ES .

Proof. Indeed, by commuting idempotents,x−1ex · x−1ex = x−1xx−1eex = x−1ex.⊓⊔

For a congruenceρ of an inverse semigroupS, we obtain a congruenceρmin bydefining

xρminy ⇐⇒ ∃e ∈ ES : xe = ye, x−1xρe andy−1yρe .

The next theorem states thatρmin identifies as few elements as possible under therestriction that it should identify exactly the same idempotents as the originalρ. In thisway the quotientS/ρmin is as large as possible.

Theorem 6.14.For a congruenceρ of an inverse semigroupS, ρmin is the smallest con-gruence whose trace equalstr(ρ).


Proof. To show thatρmin is an equivalence relation demands some calculations, whichwe leave as an exercise.

We show thatρmin is a congruence. For this supposexρminy and lete ∈ ES be suchthatxe = ye. Let z ∈ S. Now,f = z−1ez ∈ ES , and

xz · f = xzz−1 · ez = xe · zz−1z = ye · zz−1z = yzz−1 · ez = yz · f ,

and so there exists an idempotentf ∈ ES such thatxz · f = yz · f . Further,

(xz)−1 · xz = z−1x−1xz ρ z−1ez ρ z−1y−1yz = (yz)−1 · yz , (6.7)

and thereforexzρminyz. On the other hand,xe = ye implies thatex−1 = ey−1 bytaking inverses of both sides, and using this we obtain

(zx)−1 · zx = x−1 · z−1z · x = x−1x · x−1z−1z · x ρ ex−1z−1zx ,

(zy)−1 · zy = y−1 · z−1z · y = y−1y · y−1z−1zy · y−1y ρ ey−1z−1zye

= ex−1 · z−1z · xe ρ ex−1z−1zx · x−1x = ex−1z−1zx ,

which shows thatzxρminzy, sinceex−1z−1zx is an idempotent. Thusρmin is a congru-ence.

Next we show thattr(ρmin) = tr(ρ). Forf, g ∈ ES ,

fρming =⇒ ∃e ∈ E : fe = ge, fρe and gρe =⇒ fρg .

On the other hand, supposefρg. Now, for e = fg we havefe = ffg = fg = fgg =gfg = ge, and, clearly,fρe andgρe. Hence alsofρming and the traces are the same.

Finally, ρmin is the smallest of such congruences: For any congruenceδ with tr(δ) =tr(ρ),

xρminy =⇒ ∃e ∈ ES : xe = ye, x−1xρe, y−1yρe

=⇒ x−1xδe and y−1yδe =⇒ xδxe and yδye =⇒ xδy ,

and thereforeρmin ⊆ δ as required. ⊓⊔

In particular, we have thatρmin ⊆ ρ for all congruencesρ of an inverse semigroupS.For a congruenceρ of an inverse semigroupS defineρmax by

xρmaxy ⇐⇒ ∀e ∈ ES : x−1exρy−1ey .

Theorem 6.15.Let S be an inverse semigroup andρ its congruence. Thenρmax is thelargest congruence ofS whose trace equalstr(ρ).

Proof. First of all we show thatρmax is a congruence. That it is an equivalence relationis an easy exercise. Let thenxρmaxy and letz ∈ S. We have

zxρmaxzy ⇐⇒ ∀e ∈ ES : (zx)−1ezxρ(zy)−1ezy ⇐⇒ x−1z−1ezxρy−1z−1ezy .


Sincez−1ez ∈ ES, we have

xρmaxy =⇒ x−1 · z−1ez · xρy−1 · z−1ez · y ,

and thuszxρmaxzy holds if xρmaxy holds. Similarly, forxzρmaxyz. We conclude thatρmax is a congruence.

Next we show thattr(ρmax) = tr(ρ). For this letf, g ∈ ES . Then

fρmaxg =⇒ f−1ffρg−1fg =⇒ fρfg ,

and in the same way,fρmaxg =⇒ gρfg, from which we get thatfρg, and sotr(ρmax) ⊆ tr(ρ). In the other direction, for alle ∈ ES , if fρg, thenf−1efρg−1egand so we deduce thatfρmaxg, that is, the traces are the same.

Finally ρmax is the largest of such congruences. For this assumeδ isa congruencesuch thattr(δ) = tr(ρ). Now, for all e ∈ ES,

xδy =⇒ x−1eaδy−1ey =⇒ x−1eaρy−1ey =⇒ xρmaxy ,

sincex−1ex, y−1ey ∈ ES ; and soδ ⊆ ρmax as required. ⊓⊔

The above theorem states thatρmax identifies as many elements ofS as possiblewith the restriction that it does not identify any idempotents unlessρ does so. Certainly,ρ ⊆ ρmax, and so the quotientS/ρmax is an epimorphic image ofS/ρ.

Group congruences

We say that a congruenceρ of a semigroupS is agroup congruence, if S/ρ is a group.The following lemma holds already for regular semigroups.

Lemma 6.13.An inverse semigroup is a group if and only if it has a unique idempotent.

For a congruenceρ of an inverse semigroup, we have by Theorem 6.2 that

xρ ∈ ES/ρ =⇒ ∃e ∈ ES : eρ = xρ ,

and hence

Theorem 6.16.A congruenceρ of an inverse semigroupS is a group congruence if andonly if tr(ρ) = ES × ES.

Proof. If tr(ρ) = ES × ES, thenS/ρ has exactly one idempotent by Theorem 6.2. Inthe other direction the claim is equally clear. ⊓⊔

If ρ is a group congruence of an inverse semigroupS, then so isρmin, because nowtr(ρ) = tr(ρmin) = ES × ES. In particular,ρmin is the smallest group congruence ofS,and for all group congruencesδ of S, δmin = ρmin.

Thesmallest group congruenceof an inverse semigroupS is denoted byσS .Let thenρ be a group congruence. ThenσS = ρmin ⊆ ρ, and hence, by the ho-

momorphism theorem, there exists a unique homomorphismβ such that the followingdiagram commutes:


S S/ρ-ρ♯

R

�

S/σS

σ♯S β

This means that every groupG, which is a homomorphic image ofS, is a homomor-phic image of the groupS/σS , and in this senseS/σS is a maximal homomorphic imageof S.

Remark 6.1.If S is not an inverse semigroup, it need not have the smallest group con-gruence. As an example consider(N+,+). The group congruences of this semigroup areexactlyρn = {(p, q) | p ≡ p( mod n)}. ⊓⊔

Theorem 6.17 (Munn).In an inverse semigroupS,

xσSy ⇐⇒ ∃e ∈ ES : xe = ye .

Proof. Now,σS = σmin, and so

xσSy ⇐⇒ ∃e ∈ ES : xe = ye, x−1xσSe, y−1yσSe ,

wherex−1x, y−1y ∈ ES , and so the condition reduces to the claim. ⊓⊔


Theorem 6.18.In an inverse semigroupS,

1. xσSy ⇐⇒ ∃e ∈ ES : ex = ey.2. ker(σS) = {x ∈ S | ∃y ∈ S : xy = x}.

Idempotent separating congruences

A group congruence puts all idempotents in the same congruence class. An idempo-tent separating congruence does the opposite, it puts different idempotents to differentclasses.

A congruenceρ of a semigroupS is idempotent separating, if

∀e, f ∈ ES : eρf =⇒ e = f .

From this definition we have immediately

Lemma 6.14.If ρ is an idempotent separating congruence of an inverse semigroupS,thentr(ρ) = ιE = {(e, e) | e ∈ ES}.

By Theorem 6.15, for each inverse semigroupS there exists thegreatest idempotentseparating congruence, which will be denoted byµS.

6.5 Free Inverse Semigroups 76

Theorem 6.19.For all inverse semigroupsS,

xµSy ⇐⇒ ∃e ∈ ES : x−1ex = y−1ey .

Proof. This is clear, since from Theorem 6.15, sincetr(µS) = ιE, and so the claimfollows. ⊓⊔

As an exercise, using Theorem 6.6, we have

Theorem 6.20.For an inverse semigroupS,

1. µS ⊆ H;2. if ρ ⊆ L, thenρ is idempotent separating.

6.5 Free Inverse Semigroups

Generalized notion

Free inverse semigroups (and monoids) are defined in a similar way as free semigroupsexcept now the homomorphic extensions of mappings must be homomorphisms of in-verse semigroups.

It is somewhat more convenient to consider a generalized (but equivalent) notion offreeness. This is as follows.

Let L be a class of semigroups. LetF ∈ L be a semigroup in this class,X anonempty set, andϕ : X → F an injective mapping intoF such thatϕ(X) generatesF . We say that the pair(F,ϕ) is a free L-semigroup onX, if for any P ∈ L and anymappingα0 : X → P there exists a homomorphismα : F → P such that the followingdiagram commutes, that is,α0 = αϕ.

X

P

F-

R

ϕ

αα0

In this case, we say also thatϕ(X) generatesF freely. The homomorphismα iscalled aϕ-extensionof the mappingα0. The mappingϕ is often omitted from the nota-tion, and then we talk about freeL-semigroupsF (on a setX).

Note that here we do not assume thatX is a subset ofF . Nor do we claim that suchaL-free semigroup exists for a given classL.

Note also that we require thatϕ is a bijection fromX onto a generating setϕ(X) ofF . In literature one usually omits this requirement, and derives it from a more generalcondition in certain classes (like ‘varieties of semigroups’).

The following is immediate.


Lemma 6.15.If F is a freeL-semigroupF on X and α0 : X → P a mapping forP ∈ L, then theϕ-extensionα : F → P is unique.

Free semigroups with involution

Let S be a semigroup. A mappingδ : S → S is called aninvolution of S, if it satisfies

∀x, y ∈ S : δ(xy) = δ(y)δ(x) and δ(δ(x)) = x .

The pair(S, δ) or simplyS is then asemigroup with involution.Usually we write

δ(x) = xδ ,

sinceδ behaves in much the same fashion as the inverse function on a inverse semigroupor on a group. Hence the above conditions become

(xy)δ = yδxδ and (xδ)δ = x .

Example 6.4.(1) If S is an inverse semigroup, then the mappingδ(x) = x−1, wherex−1

is the inverse element ofx in S, is an involution.This is the involution we are mainlyinterested in.

(2) If G is the special linear group ofn-by-n-matrices (with entries inZ and determi-nant equal to 1), then the mappingδ(A) = AT , which maps a matrixA to its transpose,is an involution ofG. ⊓⊔

Let (S, δ) be a semigroup with involution. A subsemigroupP of S is a subsemi-group with involution , if P is closed under the involutionδ, that is, if for allx ∈ P ,alsoδ(x) ∈ P .

Let (S, δ) and(P, ν) be two semigroups with involution. Then a mappingα : S → Pis ahomomorphismα : (S, δ) → (P, ν), if

∀x, y ∈ S : α(xy) = α(x)α(y) and α(xδ) = α(x)ν ,

that is, ifα is a homomorphism that respects (or iscompatible with) the involutions.If in the above bothS andP are inverse semigroups and their involutions are the

inverse functions,x 7→ x−1, then the latter condition becomes:

∀x ∈ S : α(x−1) = α(x)−1 .

LetX andX ′ be two nonempty sets of the same cardinality such thatX ∩X ′ = ∅,and letψ : X → X ′ be a bijection. WriteY = X∪X ′, and define the mappingw 7→ w−1

of wordsw ∈ Y + by

x−1 =

{ψ(x) if x ∈ X ,

ψ−1(x) if x ∈ X ′ .

and


w = x1x2 . . . xn =⇒ w−1 = x−1n x−1

n−1 . . . x−11 .

Now, (Y +,−1 ) is a semigroup with involution, and the following theorem states that itis afree semigroup with involution. Note that hereY + itself is just the free semigroupon the setY of letters – it is not an inverse semigroup.

Later we shall usually write

X ′ = X−1 = {x−1 | x ∈ X} and Y = X ∪X−1 .

Theorem 6.21.(Y +,−1 ) is a free semigroup with involution, that is, if(P, δ) is a semi-group with involution andα0 : X → P is any mapping, then there exists a (unique)homomorphismα : (Y +,−1 ) → (P, δ) such thatα0(x) = α(x) andα0(x

−1) = α(x)δ

for all x ∈ X.

Proof. Let α0 : X → P be a s stated. We extendα0 as follows:

α(x) =

{α0(x) if x ∈ X ,

α0(x)δ if x ∈ X ′ ,

and for allxi ∈ Y ,

α(x1x2 . . . xn) = α(x1)α(x2) . . . α(xn) .

By the second condition,α is a homomorphism, and by the first condition it respects theinvolutions:

α((x1x2 . . . xn)−1) = α(x−1

n x−1n−1 . . . x

−11 ) = α(x−1

n )α(x−1n−1) . . . α(x

−11 )

= α(xn)δα(xn−1)

δ . . . α(x1)δ = (α(x1)α(x2) . . . α(xn))

δ

= α(x1x2 . . . xn)δ .

⊓⊔

Freeness of inverse semigroups

Let F be an inverse semigroup,X be a nonempty set andϕ : X → F an injective map-ping onto a generating setϕ(X) of F . We say that(F,ϕ) is a free inverse semigrouponX, if for all mappingsα0 : X → P , whereP is an inverse semigroup, there exists ahomomorphismα : F → P such thatα(x−1) = α(x)−1 andα0 = αϕ.

Herex−1 is the inverse element ofx in the inverse semigroup.

Existence of free inverse semigroups

That a free inverse semigroup on a setX exists follows from general results on varietiesof algebras. These are classes of algebras (say, semigroupswith involution) that areclosed under homomorphic images, taking subalgebras (subsemigroups with involution)


and arbitrary direct products. It was shown by Birkhoff some60 years ago that if aclassV of algebras forms a variety, then it has the free algebras on every nonempty setX. Birkhoff also gave a characterization of the varieties byidentities that are specialrelations on words.

We prove the existence of a free inverse semigroup by taking the approach of Vagner.Let (Y +,−1 ) be the free semigroup with involution as constructed in the above. De-

fine a relationρX on (Y +,−1 ) as follows,

ρX = {(uu−1u, u) | u ∈ Y +} ∪ {(uu−1vv−1, vv−1uu−1) | u, v ∈ Y +} .

Further, letρcX be the congruence generated byρX , that is,ρcX is the smallest congruencethat contains the relationρX . Let us denote by

ZX = (Y +,−1 )/ρcX

the quotient semigroup. For readability, we write

uρcX = [u] ,

so thatZX = {[u] | u ∈ Y +} and the product[u][v] = [uv] satisfies the followingproperties (ofρX ):

[uu−1u] = [u] and [uu−1vv−1] = [vv−1uu−1] .

Theorem 6.22.ZX is a free inverse semigroup onX.

Proof. First, we show thatZX is a regular semigroup. Indeed, for allu ∈ Y +,

[u][u−1][u] = [uu−1u] = [u] ,

and thus[u−1] = [u]−1 is an inverse element of[u].In order to show thatZX is an inverse semigroup, we need to prove that its idempo-

tents commute. For this it is clearly enough to show that the idempotents are exactly theelements[uu−1], since then

[uu−1][vv−1] = [uu−1vv−1] = [vv−1uu−1] = [vv−1][uu−1] .

First of all each[uu−1] is an idempotent, since[uu−1] = [u][u]−1 andZX is regular.Suppose then that[w] is an idempotent ofZX . Then[u−1u][uu−1] = [uu−1][u−1u] bythe definition ofρX , and consequently,

[w] = [w][w] = [ww−1w][ww−1w] = [w][w−1w][ww−1][w] = [w][ww−1][w−1w][w]

= [w]2[w−1]2[w]2 = [w][w−1]2[w] = [ww−1][w−1w] = [(ww−1)(ww−1)−1] ,

which is of the required form. Therefore,

EZX= {[uu−1] | u ∈ Y +} . (6.8)

6.6 Concrete Free Inverse Semigroups 80

In particular,ZX is an inverse semigroup. Evidently, the set{[x] | x ∈ X} generatesZX as an inverse semigroup.

Denote simplyϕ : (Y +,−1 ) → ZX the natural epimorphism ontoZX , that is,

∀u ∈ Y + : ϕ(u) = [u] .

Here the restriction ofϕ : X → ZX is clearly a bijection.Let P be any inverse semigroup, and letα0 : X → P be a mapping. The mapping

α0 can be extended immediately toα0 : Y → P by setting

∀x ∈ X : α0(x−1) = α0(x)

−1 .

SinceY + is a free semigroup, the mappingα0 extends in a unique way to a homo-morphismα : Y + → P . We just have to show that this homomorphism respects theinvolutions. First of all,ρcX ⊆ ker(α), sinceP is an inverse semigroup and thus for allu, v ∈ Y +, α(uu−1u) = u andα(uu−1vv−1) = α(vv−1uu−1). Now, according to thehomomorphism theorem there exists a homomorphismβ : ZX → P for whichα = βϕ.In particular, for allx ∈ Y , we haveα0(x) = α(x) = β(ϕ(x)), and the claim follows.(The uniqueness ofβ is also immediate). ⊓⊔

If we add an identity (the empty word) toZX we obtain afree inverse monoid onX, which is denotedZ1

X . This monoid satisfies the property that ifα0 : X → P is amapping into a inverse monoidP , then there exists a unique monoid homomorphismα : Z1

X → P that respects the involutions.

Theorem 6.23.For a nonempty setX, Z1X is a free inverse monoid onX.

6.6 Concrete Free Inverse Semigroups

Free groups

The free inverse semigroupZX as constructed in the above is abstract in the sense thatit is defined as a quotient semigroup. We shall have now a quicklook at a more concreterepresentation ofZX due to Scheiblich (1973). This construction is more difficult thanVagner’s approach, but using Scheiblich’s approach McAlister has contributed to thetheory of semigroups some of its most beautiful results.

In this section we concentrate on the construction a free inversemonoid.Recall that agroup homomorphismα : G → H is a mapping between the groups

G andH such that

α(xy) = α(x)α(y) and α(x−1) = α(x)−1 ,

wherex−1 denotes the group inverse ofx ∈ G, that is,xx−1 = 1G = x−1x. In particu-lar,α(1G) = 1H .

We say that a groupF is freely generatedby its subsetX (and thatF is a freegroup), if all mappingsα0 : X → G for a groupG can be extended to a group homo-morphismα : F → G.


Free groups on words

LetX be an alphabet and let

X−1 = {x−1 | x ∈ X}

be a new alphabet withX ∩ X−1 = ∅. DenoteY = X ∪ X−1. Consider the monoidpresentation

GX = 〈Y | ∀x ∈ X : xx−1 = 1, x−1x = 1〉 .

It is immediate thatGX is a presentation of a group, where two wordsu andv presentthe same element if and only if one getsv from u by a finite number of

(I) insertions of factorsxx−1 or x−1x for x ∈ X, and(D) deletions of factorsxx−1 or x−1x for x ∈ X.Write u ∼ v, if there exists a sequencew1, w2, . . . , wk of words such thatu = w1

andwk = v, andwi+1 is obtained fromwi by using a rule (I) or (D). This relationis clearly an equivalence relation, and it is a congruence onY ∗, since if u ∼ v andw ∈ Y ∗, thenwu ∼ wv anduw ∼ vw hold. Let us denote the congruence classes by[u] = {v | v ∼ u}. Now,Y ∗/ ∼ is a group. Indeed, this quotient has[1] as its identityand the inverse elements are determined by the rule[u]−1 = [u−1], whereu−1 comesfrom the obvious involution:u 7→ u−1 and(u−1)−1 7→ u.

We say that a wordw ∈ Y ∗ is reduced, if it contains no factorxx−1 or x−1x forany lettersx ∈ X. Clearly, from any wordw ∈ Y ∗ one gets a reduced word by deletingall the factorsxx−1 andx−1x iteratively. Therefore each congruence class[u] containsa reduced word.

Example 6.5.For X = {a, b}, we have the following reduction of the wordw =ab−1abaa−1b−1ba−1b,

w ∼ ab−1abb−1ba−1b ∼ ab−1aba−1b ,

where the last word is reduced, and presents the same wordw in GX . ⊓⊔

Lemma 6.16.For each wordw ∈ Y ∗ there exists a unique reduced wordr(w) such thatw = r(w) in GX . Therefore each congruence class[u] contains a unique reduced word.

Proof. Assume there are two different reduced wordsu andv in the same class, andlet u = w1, w2, . . . , wk = v be a sequence of words, wherewi+1 ∼ wi using oncea rule (I) or (D). DenoteN =

∑ki=1 |wi| (the sum of the lengths of the words in this

sequence). We can assume that the sequencew1, w2, . . . , wk is chosen such thatN isminimum. Sinceu andv are both reduced (and different), we must have|u| < |w2| and|wk−1| > |v|, that is, rule (I) is used inw1 ∼ w2 and (D) is used inwk−1 ∼ wk. Itfollows that the sequence|w1|, |w2|, . . . , |wk| has a maximal value (a turning point): forsomei, |wi−1| < |wi| > |wi+1|. This means thatwi is obtained fromwi−1 by addinga factorxx−1, andwi+1 by deleting a factoryy−1. These factors cannot coincide, forotherwisewi−1 = wi+1 contradicts the minimality ofN . If the factors overlap with


each other, thenx−1 = y andwi contains a factorxx−1x or x−1xx−1. In this case,againwi−1 = wi+1. Hence the factorsxx−1 andyy−1 should be disjoint. However, nowwe get a smaller value ofN by replacingwi by the word whereyy−1 is removed fromwi−1 and replacingwi+1 by the word where the factorxx−1 is then added at its place.

⊓⊔

Hence two wordsu, v ∈ Y ∗ are equal inGX if and only if r(u) = r(v).This allows us to presentGX more concretely as follows. Let

FX = {w ∈ Y ∗ | w = r(w)}

be the set of all reduced words inY ∗ (for Y = X ∪X−1), and define a product by

u · v = r(uv)

for all u, v ∈ FX .The relations ofGX are just the group axioms, and therefore

Theorem 6.24.FX is a free group, freely generated byX.

Proof. Let α0 : X → G be a mapping into a groupG. Eachw ∈ FX is a reduced wordin the letters ofY = X ∪X−1, and it can be uniquely written as a product

w = xε11 xε22 . . . xεkk (6.9)

wherexi ∈ X andεi ∈ {−1,+1} (so thatx+1 = x), andxi 6= xi+1. Extendα0 to FX

by settingα(w) = α0(x1)

ε1α0(x2)ε2 . . . α0(xk)

εk .

By the uniqueness in (6.9), this is a well defined mapping, andit is rather immediate thatα is a group homomorphism. ⊓⊔

Idempotents: The idea behind Scheiblich’s construction

Consider a free inverse monoidM = Z1X . If e is an idempotent ofM , then by (6.8)

there exists a wordu ∈ Y ∗ such thate = [uu−1]. If here the wordu is not reduced, sayu = v1xx

−1v2, then

uu−1 = v1xx−1v2 · v

−12 xx−1v−1

1 = v1 · v−11 v1xx

−1v2 · v−12 xx−1v−1

1

= v1xx−1v−1

1 v1v2v−12 v−1

1

= (v1x)(v1x)−1(v1v2)(v1v2)

−1 ,

and thereforee = [(v1x)(v1x)−1][(v1v2)(v1v2)

−1], where the factorsv1x andv1v2 areshorter thanu and they more reduced. From this observation we obtain


Lemma 6.17.If e ∈ EM (whereM = Z1X), then there are reduced wordsu1, u2, . . . un

such thate = [u1u

−11 ][u2u

−12 ] . . . [unu

−1n ] (ui ∈ FX) . (6.10)

In (6.10) the idempotents[uiu−1i ] and [uju

−1j ] commute in all places, and so for

each finite subsetA = {u1, u2, . . . , un} of reduced words there corresponds a uniqueidempotente as in (6.10):

A 7→ e . (6.11)

Lemma 6.18.If A 7→ e andB 7→ f , thenA ∪B 7→ ef .

Note that the mapping (6.11) is not injective: fore = xyy−1x−1 = xyy−1x−1 ·xx−1,we have both{xy} 7→ e and{xy, x} 7→ e. To overcome the noninjectivity we do thefollowing. LetA = {u1, u2, . . . , un} 7→ e, so that (6.10) holds. If hereui = v1v2, then

u1u−11 . . . uiu

−1i . . . unu

−1n = u1u

−11 . . . v1v2v

−12 v−1

1 . . . unu−1n

= u1u−11 . . . v1v2v

−12 v−1

1 · v1v−11 . . . unu

−1n

= u1u−11 . . . unu

−1n · v1v

−11

and so alsoA ∪ {v1} 7→ e. This means that we can safely add all the prefixes of thewordsui intoA without changing its valuee. Denote

A = {v | v a prefix of au ∈ A}

the set of all words that are prefixes of words fromA, including the empty word1 andthe wordsu ∈ A, themselves. By the above,A 7→ e, and this mapping is injective.

In conclusion, letK = {A | A = A}

be the set of all finite nonempty subsets of reduced words thatare closed under takingprefixes,

A ∈ K if and only if w = uv ∈ A =⇒ u ∈ A .

Lemma 6.19.The mappingK → EM (whereM = Z1X) given byA 7→ e is a bijection.

WriteAe 7→ e for the uniqueAe ∈ K that corresponds to the idempotente. Hence

Ae = {u1, u2, . . . , un} =⇒ e = [u1u−11 ][u2u

−12 ] . . . [unu

−1n ]

andAe ∪Af = Aef .

Therefore the idempoyents ofM can be identified with the closed setsA of reducedwords, and the product of the idempotents becomes defined as the union of such sets.


An overview of Scheiblich’s construction

We take now a look at the free inverse monoids without proofs.Let againY = X ∪X−1, and letFX be the free group of reduced words onX. For

eachw ∈ Y ∗, sayw = x1x2 . . . xn with xi ∈ Y , denote

w = {1, r(x1), r(x1x2), . . . , r(x1x2 . . . xn)} ,

where eachr(u) is the reduced word corresponding tou. Hencew consists of the reducedwords of the prefixes ofw. Define also

K = {A ⊆ FX | 0 < |A| <∞ and w ∈ A =⇒ w ⊆ A} .

Note that ifw ∈ FX , then alsow ⊆ FX . In general,K consists of those nonempty andfinite sets of words such that the reduced prefixes of the wordsin A are also inA.

From now on we letg andh denote the elements (reduced words) of the free groupFX , and we writeg · h for r(gh).

DenoteMX = {(A, g) ∈ K × FX | g ∈ A} ,

and define a product inMX as follows,

(A, g)(B,h) = (A ∪ g · B, g · h) ,

whereg · B = {g · h′ | h′ ∈ B} consists of reduced words only.Here the componentg · B need not be inK, butA ∪ g · B is always inK (because

g ∈ A), and hence the product is well defined,(A, g)(B,h) ∈ MX . (Already the proofof this requires some effort).

Lemma 6.20.For all g ∈ FX andA ∈ K, g−1 · A ∈ K ⇐⇒ g ∈ A.

Theorem 6.25.MX is a free inverse monoid onX.

In the following we sketch out some of the highlights of the proofs.If (A, g) ∈MX , then by the definitiong ∈ A, sayA = {u1, u2, . . . , un} andg = un.

Ifw = u1u

−11 u2u

−12 . . . un−1u

−1n−1un ∈ Y ∗ ,

then w = A andr(w) = g. This gives us a simpler representation of the elements ofMX :

(A, g) = (w, r(w))

and thusMX = {(w, r(w)) | w ∈ Y ∗} .

The following result bindsMX with Z1X (that is, with the congruenceρcX):

u = v and r(w) = r(u) ⇐⇒ uρcX .

It follows from this that the mappingψ : Z1X →MX defined by

uρcX 7→ (w, r(w))

is a bijection. In fact, one more proof shows thatψ is an isomorphism.


On Reynolds construction

Reynolds (1986) proved that we get the free inverse semigroups also from the partialmappings as follows.

Recall thatIFXconsists of the injective partial mappings of the free groupFX into

itself. Letψ : X → IFXbe defined by

ψ(g) = ψg where ψg(h) = g · h (h 6= 1, h 6= g−1) .

Then the inverse subsemigroupI ′X generated byψ(X) is a free inverse semigroup (onX).

On Munn’s construction

Munn presented the free inverse semigroups using (orientedword) trees. His variant isprobably the most intuitive, but it is also (probably) less useful than Scheiblich’s con-struction.

lecture notes on semigroups - semantic scholar · lecture notes on semigroups tero harju department...

Documents