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13.1: Monoids

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Recall that in Section 11.2 we introduced systems called monoids. Here is the formal definition.

Definition 13.1.1: Monoid

A monoid is a set M together with a binary operation with the properties

  • is associative: a,b,cM, (ab)c=a(bc) and
  • has an identity in M: eM such that aM, ae=ea=a

Note 13.1.1

Since the requirements for a group contain the requirements for a monoid, every group is a monoid.

Example 13.1.1: Some Monoids

  1. The power set of any set together with any one of the operations intersection, union, or symmetric difference is a monoid.
  2. The set of integers, Z, with multiplication, is a monoid. With addition, Z is also a monoid.
  3. The set of n×n matrices over the integers, Mn(Z), n2, with matrix multiplication, is a monoid. This follows from the fact that matrix multiplication is associative and has an identity, In. This is an example of a noncommutative monoid since there are matrices, A and B, for which ABBA.
  4. [Zn;×n],n2, is a monoid with identity 1.
  5. Let X be a nonempty set. The set of all functions from X into X, often denoted XX, is a monoid over function composition. In Chapter 7, we saw that function composition is associative. The function i:XX defined by i(a)=a is the identity element for this system. If |X| is greater than 1 then it is a noncommutative monoid. If X is finite, |XX|=|X||X| . For example, if B={0,1},|BB|=4. The functions z,u,i, and t, defined by the graphs in Figure 13.1.1, are the elements of BB . This monoid is not a group. Do you know why?
    One reason why BB is noncommutative is that tzzt because (tz)(0)=t(z(0))=t(0)=1 while (zt)(0)=z(t(0))=z(1)=0.
clipboard_e0e4ecd170c14b92b407fcf9e51eeb978.pngFigure 13.1.1: The functions on B2

Virtually all of the group concepts that were discussed in Chapter 11 are applicable to monoids. When we introduced subsystems, we saw that a submonoid of monoid M is a subset of M; that is, it is a monoid with the operation of M. To prove that a subset is a submonoid, you can apply the following theorem.

Theorem 13.1.1: Submonoid Test

Assume [M;] is a monoid and K is a nonempty subset of M. Then K is a submonoid of M if and only if the following two conditions are met.

  • If a,bK, then. abK; i. e., K is closed with under .
  • The identity of M belongs to K.

Often we will want to discuss the smallest submonoid that includes a certain subset S of a monoid M. This submonoid can be defined recursively by the following definition.

Definition 13.1.2: Submonoid Generated by a Set

If S is a subset of monoid [M;], the submonoid generated by S, S, is defined by:.

  1. (Basis) The identity of M belongs to S; and aSaS.
  2. (Recursion) a,bSabS.

Note 13.1.2

If S={a1,a2,,an}, we write a1,a2,,an in place of {a1,a2,,an}.

Example 13.1.2: Some Submonoids

  1. One example of a submonoid of [Z;+] is 2={0,2,4,6,8,}.
  2. The power set of Z, P(Z), over union is a monoid with identity . If S={{1},{2},{3}}, then S is the power set of {1,2,3}. If S={{n}:nZ}, then S is the set of finite subsets of the integers.

As you might expect, two monoids are isomorphic if and only if there exists a translation rule between them so that any true proposition in one monoid is translated to a true proposition in the other.

Example 13.1.3

M=[P{1,2,3};] is isomorphic to M2=[Z32;], where the operation in M2 is componentwise mod 2 multiplication. A translation rule is that if A{1,2,3}, then it is translated to (d1,d2,d3) where

di={1 if iA0 if iA

Two cases of how this translation rule works are:

{1,2,3} is the identity for M1(1,1,1) is the identity for M2{1,2}{2,3}={2}(1,1,0)(0,1,1)=(0,1,0).

A more precise definition of a monoid isomorphism is identical to the definition of a group isomorphism, Definition 11.7.2.

Exercises

Exercise 13.1.1

For each of the subsets of the indicated monoid, determine whether the subset is a submonoid.

  1. S1={0,2,4,6} and S2={1,3,5,7} in [Z8;×8].
  2. {fNN:f(n)n,nN} and {fNN:f(1)=2} in the monoid [NN;].
  3. {AZA is finite} and{AZAc is finite} in [P(Z);].
Answer
  1. S1 is not a submonoid since the identity of [Z8;×8], which is 1, is not in S1. S2 is a submonoid since 1S2 and S2 is closed under multiplication; that is, for all a,bS2, a×8b is in S2.
  2. The identity of NN is the identity function i:NN defined by i(a)=a, aN. If aN, i(a)=aa, thus the identity of NN is in S1. However, the image of 1 under any function in S2 is 2, and thus the identity of NN is not in S2, so S2 is not a submonoid. The composition of any two functions in S1, f and g, will be a function in S1:
    (fg)(n)=f(g(n))g(n) since f is in S1n since g is in S1fgS1
    and the two conditions of a submonoid are satisfied and S1 is a submonoid of NN.
  3. The first set is a submonoid, but the second is not since the null set has a non-finite complement.

Exercise 13.1.2

For each subset, describe the submonoid that it generates.

  1. {3} in [Z12;×12]
  2. {5} in [Z25;×25]
  3. the set of prime numbers in [P;]
  4. {3,5} in [N;+]

Exercise 13.1.3

n×n matrix of real numbers is called stochastic if and only if each entry is nonnegative and the sum of entries in each column is 1. Prove that the set of stochastic matrices is a monoid over matrix multiplication.

Answer

The set of n×n real matrices is a monoid under matrix multiplication. This follows from the laws of matrix algebra in Chapter 5. To prove that the set of stochastic matrices is a monoid over matrix multiplication, we need only show that the identity matrix is stochastic (this is obvious) and that the set of stochastic matrices is closed under matrix multiplication. Let A and B be n×n stochastic matrices.

(AB)ij=nk=1aikbkj

The sum of the jth column is

nj=1(AB)ij=nk=1a1kbkj+nk=1a1kbkj++nk=1ankbkj=nk=1(a1kbkj+a1kbkj++ankbkj)=nk=1bkj(a1k+a1k++ank)=nk=1bkj since A is stochastic=1 since B is stochastic

Exercise 13.1.4

A semigroup is an algebraic system [S;] with the only axiom that be associative on S. Prove that if S is a finite set, then there must exist an idempotent element, that is, an aS such that aa=a.

Exercise 13.1.5

Let B be a Boolean algebra and M the set of all Boolean functions on B. Let be defined on M by (fg)(a)=f(a)g(a). Prove that [M;] is a monoid. Construct the operation table of [M;] for the case of B=B2.

Answer

Let f,g,hM, and aB.

((fg)h)(a)=(fg)(a)h(a)=(f(a)g(a))h(a)=f(a)(g(a)h(a))=f(a)(gh)(a)=(f(gh))(a)

Therefore (fg)h=f(gh) and is associative.

The identity for  is the function uM where u(a)=1= the “one” of B. If aB, (fu)(a)=f(a)u(a)=f(a)1=f(a). Therefore fu=f. Similarly, uf=f.

There are 22=4 functions in M for B=B2. These four functions are named in the text. See Figure 13.1.1. The table for is

zituzzzzzizizitzzttuzitu


This page titled 13.1: Monoids is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Al Doerr & Ken Levasseur via source content that was edited to the style and standards of the LibreTexts platform.

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