9
Abstract Algebra Notes – Version 04 Spring, 2012 1. PRELIMINARIES Consider the binary operation defined on the set as follows. Figure 1. Cayley table for a binary operation This Cayley table defines the product and the product . Definition 1. A set is closed under a binary operation iff for all . Definition 2. A binary operation defined set is associative iff for all . Note that the set is closed under a binary operation defined by the table since the interior of the table contains only symbols from . In addition, if we wanted to verify that the given operation is associative, then we would need to verify the identity given in Definition 2, where we could substitute any of the six elements in for , any of the six for , and any of the six for . So, by the fundamental counting principle, the number of identities we would need to verify would be . It’s likely intuitive that not all six by six tables would define associative operations. As a matter of fact, the number of ways a table could be completed is determined as follows. If we were to take the table given above, we can observe that there are 6

Abstract Algebra Notes V04

Embed Size (px)

DESCRIPTION

Abstract Algebra

Citation preview

Page 1: Abstract Algebra Notes V04

Abstract Algebra Notes – Version 04Spring, 2012

1. PRELIMINARIES

Consider the binary operation defined on the set as follows.

Figure 1. Cayley table for a binary operation

This Cayley table defines the product and the product .

Definition 1. A set is closed under a binary operation iff for all .

Definition 2. A binary operation defined set is associative iff for all .

Note that the set is closed under a binary operation defined by the table since the interior of the table contains only symbols from . In addition, if we wanted to verify that the given operation is associative, then we would need to verify the identity given in Definition 2, where we could substitute any of the six elements in for , any of the six for , and any of the six for . So, by the fundamental counting principle, the number of identities we would need to verify would be .

It’s likely intuitive that not all six by six tables would define associative operations. As a matter of fact, the number of ways a table could be completed is determined as follows. If we were to take the table given above, we can observe that there are 6 choices for the definition of the product . Similarly, there are six different choices for each of the other 35 products, or a

total of tables. Likewise the number of tables is which,

as it turns out is more than the estimated number of atoms in the universe. (Don’t believe me? Google it!)

Definition 3. A set that is closed under an associative binary operation is called a semigroup.The interested reader is encouraged to visit the On-line Encyclopedia of Integer Sequences at

http://www.research.att.com/~njas/sequences/?q=binary+operation&sort=0&fmt=0&language=english

Page 2: Abstract Algebra Notes V04

(sequences A002489 and A023814) where collections of important integer sequences have been studied, computed and recorded. At that site can be found more than the following information:

2 3 4 5 6Number of tables

16 19683 4294967296

Number of assoc tables

8 113 3492

Note that the Cayley table defined in Figure 1 is indeed a semigroup.

Definition 4. Suppose that is a semigroup. Then is commutative under iff for all .

Note that the semigroup defined in Figure 1 is not commutative since but .

Definition 5. Suppose that is a semigroup. An element is called a left zero of

iff for all . (Similarly we can define the term right zero.)

Definition 6. Suppose that is a semigroup. An element is called a left identity of

iff for all . (Similarly we can define the term right identity.)

We note that if an element is both a left identity and a right identity, then we simply call that element an identity. Similarly if an element is both a left zero and a right zero, then we will simply say that the element is a zero. By observation of the table defined in Figure 2, is an identity and is a zero.

Figure 2. Cayley table for noncommutative semigroup

The following semigroup has two left zeros, namely and .

Figure 3. Semigroup having two left zeros

2

Page 3: Abstract Algebra Notes V04

Notice also that the semigroup above has one right identity, namely but no left identity.

Proposition 1. If is a left identity of and is a right identity of , then .

Proof: Suppose the hypothesis is true. [WTS: ] But since is a left identity.

Likewise, since is a right identity. Therefore .

Proposition 2. If is a left zero of and is a right zero of , then .Proof: (The proof is left to the gentle reader.)

Exercise 1. For each of the semigroups provided in the handout, determine which elements (if any) are left (right) identities, and left (right) zeros, and which semigroups are commutative.

Note: There is a hierarchy (of sorts) applied as mathematicians conjecture about and prove results. For example, we have

Theorems – the main results of a research effort. It is common for a research paper or a thesis to have two or three main theorems.

Propositions – usually relatively minor results but still somewhat interesting. Research publications usually contain numerous propositions.

Corollaries – logical consequences of theorems or propositions. Usually corollaries occur when some part of a hypothesis is restricted to a special case, allowing the conclusion of the theorem or proposition to be reached without further proof.

Lemmas – an intermediate result, usually proved in advance of the proof of a main theorem or a proposition. In many cases a lemma will contain the crucial logic that provides the backbone of the proof of the theorem. Lemmas can be used to shorten otherwise long proofs of more important results since those proofs can just reference the lemma.

Exercise 2. Use the program CHECKER.exe to find a semigroup having 3 elements (i.e. a 3x3 Caley table) that has exactly 3 left zeros.

Solution to Exercise 2. The following table satisfies the requirements of the exercise.

Figure 4. Semigroup in which all elements are left zeros.

3

Page 4: Abstract Algebra Notes V04

Exercise 3. Show that the table above is associative directly (without using the checker program).

Solution to Exercise 3. The expression always simplifies to since is a left zero. Similarly, the expression simplifies to and then to again since x is a left zero.

Proposition 3. If is a semigroup in which all elements are left zeros, then all elements are also right identities.

Proof: Suppose the hypothesis is true. [WTS if then is a right identity, i.e., for all .] Suppose that x is an arbitrary element of . In order to show that x is a right

identity, let . Then since is a right zero. Hence is a right identity.

Exercise 4. Use the program CHECKER.exe to find a semigroup of order 3 (i.e. a 3x3 Caley table) that has exactly 2 left zeros.

4

Page 5: Abstract Algebra Notes V04

2. SUBSEMIGROUPS

Consider the following Caley table (Table #1 from the original handout).

Figure 6. Semigroup #1 on handout.

We can highlight a portion of that table as follows, showing that under the given operation, the set is itself a semigroup, since is closed under that “restricted” operation.

Figure 7. A subset of a semigroup that is also a semigroup.

Definition 7. Suppose that is a semigroup and that Then is a subsemigroup of iff the set is closed under the operation . (Frequently, we will simply say that is a subsemigroup of .)

Note that is associative under the operation since is associative and Note also that if is commutative then so is . It is customary to use the word order to describe the number of elements in any particular semigroup. So, for example, the semigroup given in Figure 7 above has order 4 and the subsemigroup has order 2.

Exercise 5. Use the homework handout to find an example of a noncommutative semigroup that has a commutative subsemigroup – the subsemigroup should contain at least 2 elements.

In what follows, we create the subsemigroup tree for the Caley table given in Figure 7.

Figure 9. The subsemigroup tree for semigroup in Figure 7.

5

Page 6: Abstract Algebra Notes V04

Definition 8. Suppose that is a semigroup and . Then is called an idempotent of iff .

Note that in the semigroup of Figure 8 (#18), the idempotents are and .

Exercise 6. Use the original handout to list all idempotents for each given semigroup.

Proposition 4. Suppose that is a semigroup and is the set of all idempotents of . If is commutative, then is a subsemigroup of .

Proposition 5. Suppose that is a semigroup with subsemigroups and . Then is also a subsemigroup of .

Definition 9. Suppose that is a semigroup and . Then .

Proposition 6. Suppose that is a semigroup and . Then is a subsemigroup of .

Definition 10. Suppose that is a semigroup and . Then is defined inductively as follows

If has been defined for some , then .

Proposition 7. Suppose that is a semigroup and . Then for all .

Proof (by induction on n): Since , the conclusion holds for .

Suppose now that for some , and consider . By associativity

and the definition of , , and by the induction

hypothesis, .

Proposition 8. Suppose that is a semigroup and . Then for all positive integers .

Proof: Suppose the hypothesis is true. [WTS if and are positive integers, then] Write . Then, we want to show that . Our proof is

by induction on . By Definition 9 and Proposition 4, , and the conclusion holds for

.

6

Page 7: Abstract Algebra Notes V04

Suppose now that for some , and consider . By

associativity and the definition of ,

, and by the

induction hypothesis, . It follows that

.

Corollary 1. for all positive integers such that .

Definition 11. Suppose that is a semigroup and . Then and we call the subsemigroup of generated by .Consider table #2 in the in-class handout. For that semigroup, we have

Exercise 7. Use the homework handout to list all subsemigroups generated by single elements for each semigroup.

Solution to Exercise 7. The semigroup #3, let’s call it S, has the property that

Definition 12. Suppose that is a semigroup and . If for some , then is called a cyclic semigroup.

Definition 13. Suppose that is a semigroup and . If is closed, then we say that is removable.

Exercise 8. Use the homework handout to list all subsemigroups having a removable element.

7