Monoids Definition: A binary operation on a set M is a function : M M M. If : M M M, we say that is well defined on M or equivalently, that M is closed under the operation. Examples:
Definition: A monoid is a pair (M, ) where M is a set M satisfying: (A0) M is closed under : : M M M. (A1) identity element e M: e a = a e = a, a M. (A2) Associativity: Note: Due to (A2), we can denote a b c = a (b c) = (a b) c. Examples: a (b c) = (a b) c, a, b, c
Definition: The elements of a subset A M are called generators of the monoid (M, ) if any element y M with the exception of the identity element can be obtained from them by successively applying the operation : y M \ {e}, a 1,..., a n A, y = a 1 a 2... a n. Examples: Definition: A set of generators A M is called free if any element y M can be written in terms of the elements of A in a unique way, i.e. there are no two different ways of writing m using and elements in A: a 1,..., a n, b 1,..., b m A, y = a 1... a n = b 1... b m m = n, a i = b i, i.
Example: Definition: A monoid with a free set of generators is called a free monoid. Definition: (N, +) is the free monoid generated by 1. Other examples: Definition: A commutative (or Abelian) monoid is a monoid (M, ) where also satisfies: Commutativity: a, b M, a b = b a.
Addition on N: 1) Consider the set N given by the Constructive Definition. Definition: We define + : N N N by A + B = A B for any sets A and B such that A B =. Note: (N, +) thus defined is a monoid: (A1) + has identity element 0 = : A + = A = A, A finite set. (A2) + is associative because the union of sets is: A (B C) = (A B) C = A + ( B + C ) = ( A + B ) + C for A, B, C disjoint. Exercise: Prove that the monoid (N, +) is commutative: (A3) a, b N, a + b = b + a.
Multiplication on N: 1) Consider the set N given by the Constructive Definition. Definition: We define : N N N by A B = A B. Note: (N, ) thus defined is a monoid: (A1) has identity element 1 = { } : A { } = A { } = A, A finite set. (A2) + is associative: A (B C) = (A B) C. Moreover, the new operation satisfies: (D) is distributive with respect to +, i.e. a, b, c N, a(b + c) = ab + ac. Similarly, (b + c)a = ba + ca. Indeed, for all finite sets A, B, C, A (B C) = A B A C, (B C) A = B A C A,
Finite monoids Definition:Let (M, ) be a monoid with M = finite set. The Cayley table of (M, ) is the table whose rows/columns correspond to the elements of M, and whose entry on the row a and column b is a b. * b a a b Examples: Z 2 = {[0], [1]}, where [0] = { all even numbers } and [1] = { all odd numbers }. + [0] [1] 1. The Cayley table for the monoid (Z 2, +): [0] [0] [1] [1] [1] [0] + [0] [1] 2. The Cayley table for the monoid (Z 2, ): [0] [0] [0] [1] [0] [1] 3. Z 5 = {[0], [1], [2], [3], [4]}, where [0] = {... 10, 5, 0, 5, 10, 15,...} = { all multiples of 5}, [1] = {... 9, 4, 1, 6, 11, 16,...} = = { numbers whose division by 5 yields remainder 1}, [2] = {... 8, 3, 2, 7, 12, 17,...} = = { numbers whose division by 5 yields remainder 2}... For convenience, when it is clear that we work modulo a certain number, we can write the equivalence classes without the square brackets, i.e. we ll write 2 and mean [2].
The Cayley table for the monoid (Z 5, +): + 0 1 2 3 4 0 0 1 2 3 4 1 + 4 = 2 + 3 = 5 0 ( mod 5), 1 1 2 3 4 0 2 + 4 = 3 + 3 = 6 1 ( mod 5), 2 2 3 4 0 1 because 3 + 4 = 4 + 3 = 7 2 ( mod 5), 3 3 4 0 1 2 4 + 4 = 8 3 ( mod 5). 4 4 0 1 2 3 1 is a generator for (Z 5, +) because n = 1 +... + 1. }{{} n times (Z 5, +) is not freely generated by 1 because. e.g., 2 can be written in two different ways using 1 and +: 2 = 1 + 1 = 1 +... + 1. }{{} 7 times 4. The Cayley table for the monoid (Z 5, ): + 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 because 3 0 3 1 4 2 4 0 4 3 2 1 (Z 5, +) is generated by {0, 2} because 2 3 = 6 1 ( mod 5), 2 4 = 8 3 ( mod 5), 3 3 = 9 4 ( mod 5), 3 4 = 12 2 ( mod 5) 4 4 = 16 1 ( mod 5). 0 = 0 in Z 5, 1 = identity elem. in Z 5, 2 = 2 in Z 5, 3 = 2 2 2 in Z 5 4 = 2 2 in Z 5. (Z 5, +) is not freely generated by {0, 2} because. e.g., 1 = 2 2 2 2 so 2 = 2 2 2 2 2. (Z 5, +) is generated by {0, 3} because 0 = 0 in Z 5, 1 = identity elem. in Z 5, 2 = 3 3 3 in Z 5, 3 = 3 in Z 5 4 = 3 3 in Z 5.
(Z 5, +) is not freely generated by {0, 3} because. e.g., 1 = 3 3 3 3 so 3 = 3 3 3 3 3. (Z 5, +) is not generated by {0, 4} because 4 4 = 1 in Z 5 so only 0, 1 and 4 can be written using {0, 4} and. Other examples of monoids 1. (M m n (N), +)= the monoid of m n matrices with entries in N, where + is the matrix addition. (M m n (N), +) is generated by {E ij ; i {1,..., m} and j {1,..., n}}, where E ij is the matrix having the entry on the i-th row and j-th column equal to 1, all other entries equal to 0. 2. For every set X, the set of functions {f : X X} together with, the composition of functions, forms a monoid. 3. The set of functions {f : R R; f continuous } together with, the composition of functions, forms a monoid. Monoid Homomorphisms Previously we claimed that (N, +) is the free monoid generated by one element, because for any n N nonzero, n = 1 } + {{... + 1 }. n times
On the other hand, consider the set M = {e, a, aa, aaa,...} of all words made out of the letter a, with the concatenation operation : a...a }{{} m times a...a }{{} n times = a...a }{{} m+n times Here e denotes the empty word, which is the identity element for. Then (M, ) has equal claims to the title of free monoid generated by one element. Although (N, +) and (M, ) are two different monoids, it is clear that one can be obtained from the other by relabeling: Thus relabel 0 by e, then n by a...a }{{} and n times + by, and (N, +) has become (M, ). If we could write the Cayley table of (N, +),. + n m m + n then relabel, we d get * a...a }{{} n times a...a }{{} m times a...a }{{} m+n times which is the Cayley table of (M, ). In fact we have constructed a map f : N M which sends the monoid operation + to the operation.
Definition: Let (X, ) and (Y, ) be two monoids with identity elements e X and e Y, respectively. A monoid homomorphism f : (X, ) (Y, ) is a function f : X Y such that f(e X ) = e Y for all m, n X. and f(m n) = f(m) f(n), Definition: A monoid homomorphism f : (X, ) (Y, ) which is bijective is called monoid isomorphism. Notation: In this case we say that (X, ) and (Y, ) are isomorphic, and write (X, ) = (Y, ). Example: 1. (N, +) = (M, ). We say that (N, +), the free monoid with one generator, uniquely defined up to an isomorphism (i.e. by a relabeling of its elements and its operation.) 2. f : (N, +) (N, ) given by f(x) = e x is a monoid homomorphism, injective but not surjective. 3. Let (M, +) be a monoid and let be another binary operation on M. For each element a M, define f : M M by f(x) = a x. Prove that f : (M, +) (M, +) is a monoid homomorphism iff satisfies the distributivity condition for all x, y M. a (x + y) = a x + a y, 4. Consider the map f : Z 3 Z 3 given by f(n) = n + 1. Define a new operation on Z 3 such that f : (Z 3, ) (Z 3, ) is a monoid isomorphism.
References An elementary introduction to sets, relations, functions: http://www.cosc.brocku.ca/ duentsch/archive/methprimer1.pdf The same, as taught by a philosophy professor: http://people.umass.edu/gmhwww/595t/text.htm Some more advanced texts on Set Theory, the Axiom of Choice and the Set of Natural Numbers: http://kaharris.org/teaching/582/ http://www.math.toronto.edu/weiss/set theory.pdf http://math.boisestate.edu/ holmes/holmes/head.pdf Further reading. Pick your own: http://www.e-booksdirectory.com/listing.php?category=55 http://www.e-booksdirectory.com/listing.php?category=540