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c 2001-2012 T. Uyar, A. Yayımlı, E. Harmancı Discrete Mathematics Algebraic Structures You are free: I to Share – to copy, distribute and transmit the work I to Remix – to adapt the work Under the following conditions: H. Turgut Uyar Ay¸segul Gen¸cata Yayımlı Emre Harmancı I Attribution – You must attribute the work in the manner specified by the author or licensor (but not in any ¨ way that suggests that they endorse you or your use of the work). I Noncommercial – You may not use this work for commercial purposes. I Share Alike – If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one. 2001-2012

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Topics Algebraic Structure

Algebraic Structures Introduction Groups I algebraic structure: Rings I carrier set I operations: binary, unary I constants: identity, zero Lattices Partially Ordered Sets Lattices Boolean Algebra

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Operation Closed Operation Examples

I every operation is a function

I binary operation: Example ◦ : S × S → T I subtraction is closed on Z + I unary operation: I subtraction is not closed on Z ∆ : S → T

I closed: T ⊆ S

5 / 70 6 / 70 Binary Operation Properties Binary Operation Example

Example ◦ : Z × Z → Z Definition a ◦ b = a + b − 3ab commutativity: commutative: ∀a, b ∈ S a ◦ b = b ◦ a I a ◦ b = a + b − 3ab = b + a − 3ba = b ◦ a Definition I associative: associativity: (a ◦ b) ◦ c = (a + b − 3ab) + c − 3(a + b − 3ab)c ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c) = a + b − 3ab + c − 3ac − 3bc + 9abc = a + b + c − 3ab − 3ac − 3bc + 9abc = a + (b + c − 3bc) − 3a(b + c − 3bc) = a ◦ (b ◦ c)

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Constants Examples of Constants

Example

I identity for < N, max > is 0 Definition Definition zero for < , min > is 0 identity: zero: I N + x ◦ 1 = 1 ◦ x = x x ◦ 0 = 0 ◦ x = 0 I zero for < Z , min > is 1 left identity: 1 ◦ x = x left zero: 0 ◦ x = 0 I l I l Example I right identity: x ◦ 1r = x I right zero: x ◦ 0r = 0 ◦ a b c

a a b b I b is a left identity b a b c I a and b are right zeros c a b a

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Constants Inverse

Theorem Theorem I if x ◦ y = 1: I x is a left inverse of y ∃1l ∧ ∃1r ⇒ 1l = 1r ∃0l ∧ ∃0r ⇒ 0l = 0r I y is a right inverse of x Proof. Proof. if x ◦ y = y ◦ x = 1 x and y are inverse 1l ◦ 1r = 1l = 1r 0l ◦ 0r = 0l = 0r I

11 / 70 12 / 70 Inverse Algebraic Families

Theorem if the operation ◦ is associative: w ◦ x = x ◦ y = 1 ⇒ w = y I algebraic family: algebraic structure, axioms Proof. I commutativity, associativity w = w ◦ 1 I inverse elements = w ◦ (x ◦ y) = (w ◦ x) ◦ y = 1 ◦ y = y

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Algebraic Family Examples Subalgebra

Example Definition

I axioms: subalgebra: 0 0 0 0 0 I x ◦ y = y ◦ x let A =< S, ◦, ∆, k > ∧ A =< S , ◦ , ∆ , k > I (x ◦ y) ◦ z = x ◦ (y ◦ z) 0 I x ◦ 1 = x I A is a subalgebra of A if: 0 structures for which these axioms hold: I S ⊆ S I 0 0 0 I ∀a, b ∈ S a ◦ b = a ◦ b ∈ S I < Z, +, 0 > 0 0 0 I ∀a ∈ S ∆ a = ∆a ∈ S I < Z, ·, 1 > 0 I k = k I < P(S), ∪, ∅ >

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Subalgebra Examples Semigroups

Example Definition + semigroup: < S, ◦ > I < Z , +, 0 > is a subalgebra of < Z, +, 0 >. I ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c) I < N, −, 0 > is not a subalgebra of < Z, −, 0 >.

17 / 70 18 / 70 Semigroup Examples Monoids

Example Definition < Σ+, & > monoid: < S, ◦, 1 > + I Σ: alphabet, Σ : strings of length at least 1 I ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c)

I &: string concatenation I ∀a ∈ S a ◦ 1 = 1 ◦ a = a

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Monoid Examples Groups

Definition Example group: < S, ◦, 1 > < Σ∗, &,  > I ∀a, b, c ∈ S (a ◦ b) ◦ c = a ◦ (b ◦ c) ∗ I Σ: alphabet, Σ : strings of any length I ∀a ∈ S a ◦ 1 = 1 ◦ a = a I &: string concatenation −1 −1 −1 I ∀a ∈ S ∃a ∈ S a ◦ a = a ◦ a = 1 I : empty string I Abelian group: ∀a, b ∈ S a ◦ b = b ◦ a

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Group Examples Group Example: Permutation Composition

Example I permutation: a bijective function on a set

I < Z, +, 0 > is a group. I representation:   I < Q, ·, 1 > is not a group. a1 a2 ... an I < Q − {0}, ·, 1 > is a group. p(a1) p(a2) ... p(an)

23 / 70 24 / 70 Permutation Examples Group Example: Permutation Composition

Example I permutation composition is associative

A = {1, 2, 3} I identity permutasyon: 1A  1 2 3   1 2 3    p = p = a1 a2 ... an 1 1 2 3 2 1 3 2 a1 a2 ... an  1 2 3   1 2 3  p = p = 3 2 1 3 4 2 3 1 I the set of permutations of the elements of a set,  1 2 3   1 2 3  the permutation composition operation p = p = 5 3 1 2 6 3 2 1 and the identity permutation constitute a group

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Group Example: Permutation Composition Group Example: Permutation Composition

Example (permutations on {1, 2, 3, 4}) A 1 p p p p p p p p p p p A 1 2 3 4 5 6 7 8 9 10 11 Example 1 111111222222

2 223344113344 I p8  p12 = p12  p8 = 1A: −1 −1 3 342423341413 p12 = p8 , p8 = p12 4 434232434131 I p14  p14 = 1A: −1 p14 = p14 p12 p13 p14 p15 p16 p17 p18 p19 p20 p21 p22 p23 1 333333444444 I G1 =< {1A, p1,..., p23}, , 1A > is a group 2 112244112233 3 241412231312 4 424121323121

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Group Example: Permutation Composition Left and Right Cancellation

Example Theorem  1A p2 p6 p8 p12 p14 a ◦ c = b ◦ c ⇒ a = b

1A 1A p2 p6 p8 p12 p14 c ◦ a = c ◦ b ⇒ a = b p2 p2 1A p8 p6 p14 p12 Proof. p6 p6 p12 1A p14 p2 p8 a ◦ c = b ◦ c p8 p8 p14 p2 p12 1A p6 ⇒ (a ◦ c) ◦ c−1 = (b ◦ c) ◦ c−1 −1 −1 p12 p12 p6 p14 1A p8 p2 ⇒ a ◦ (c ◦ c ) = b ◦ (c ◦ c ) p14 p14 p8 p12 p2 p6 1A ⇒ a ◦ 1 = b ◦ 1 ⇒ a = b

I < {1A, p2, p6, p8, p12, p14}, , 1A > is a subgroup of G1

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Definition ring: < S, +, ·, 0 > Theorem The unique solution of the equation a ◦ x = b is: x = a−1 ◦ b. I ∀a, b, c ∈ S (a + b) + c = a + (b + c) I ∀a ∈ S a + 0 = 0 + a = a Proof. a ◦ c = b I ∀a ∈ S ∃(−a) ∈ S a + (−a) = (−a) + a = 0 −1 −1 ⇒ a ◦ (a ◦ c) = a ◦ b I ∀a, b ∈ S a + b = b + a −1 ⇒ 1 ◦ c = a ◦ b I ∀a, b, c ∈ S (a · b) · c = a · (b · c) ⇒ c = a−1 ◦ b I ∀a, b, c ∈ S

I a · (b + c) = a · b + a · c I (b + c) · a = b · a + c · a

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Field References

Grimaldi Definition Chapter 5: Relations and Functions field: < S, +, ·, 0, 1 > I I 5.4. Special Functions I all properties of a ring I Chapter 16: Groups, Coding Theory, I ∀a, b ∈ S a · b = b · a and Polya’s Method of Enumeration

I ∀a ∈ S a · 1 = 1 · a = a I 16.1. Definitions, Examples, and Elementary Properties −1 −1 −1 I ∀a ∈ S ∃a ∈ S a · a = a · a = 1 I Chapter 14: Rings and Modular Arithmetic I 14.1. The Ring Structure: Definition and Examples

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Partially Ordered Set Partial Order Examples

Definition partial order relation: Example (set of sets, ⊆) I reflexive I A ⊆ A I anti-symmetric I A ⊆ B ∧ B ⊆ A ⇒ A = B I transitive I A ⊆ B ∧ B ⊆ C ⇒ A ⊆ C I partially ordered set (poset): a set with a partial order relation defined on its elements

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+ Example (Z, ≤) Example (Z , |)

I x ≤ x I x|x

I x ≤ y ∧ y ≤ x ⇒ x = y I x|y ∧ y|x ⇒ x = y

I x ≤ y ∧ y ≤ z ⇒ x ≤ z I x|y ∧ y|z ⇒ x|z

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Comparability Comparability Examples

a  b: a precedes b I Example

I a  b ∨ b  a: a and b are comparable + I Z , |: 3 and 5 are not comparable

I total order (linear order): I Z, ≤: total order all elements are comparable with each other

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Hasse Diagrams Hasse Diagram Examples

Example

I a  b: a immediately precedes b ¬∃x a  x  b

I Hasse diagram: {1, 2, 3, 4, 6, 8, 9, 12, 18, 24} I draw a line between a and b if a  b the relation | I preceding element is below

41 / 70 42 / 70 Consistent Enumeration Consistent Enumeration Examples

Example

I consistent enumeration: f : S → N a  b ⇒ f (a) ≤ f (b)

I there can be more than one consistent enumeration

I {a 7−→ 5, b 7−→ 3, c 7−→ 4, d 7−→ 1, e 7−→ 2}

I {a 7−→ 5, b 7−→ 4, c 7−→ 3, d 7−→ 2, e 7−→ 1}

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Maximal - Minimal Elements Maximal - Minimal Element Examples

Example

Definition maximal element: max ∀x ∈ S max  x ⇒ x = max Definition max : 18, 24 minimal element: min min : 1 ∀x ∈ S x  min ⇒ x = min

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Bounds Bound Example

Example (factors of 36)

Definition Definition A ⊆ S A ⊆ S M is an upper bound of A: m is a lower bound of A: ∀x ∈ A x  M ∀x ∈ A m  x inf = greatest common divisor sup = least common multiple M(A): set of upper bounds of A m(A): set of lower bound of A sup(A) is the supremum of A: inf (A) is the infimum of A: ∀M ∈ M(A) sup(A)  M ∀m ∈ m(A) m  inf (A)

47 / 70 48 / 70 Lattice Poset - Lattice Relationship

Definition lattice: < L, ∧, ∨ >

∧: meet, ∨: join I If P is a poset, then < P, inf , sup > is a lattice.

I a ∧ b = inf (a, b) I a ∧ b = b ∧ a a ∨ b = b ∨ a I a ∨ b = sup(a, b)

I (a ∧ b) ∧ c = a ∧ (b ∧ c) I Every lattice is a poset where these definitions hold. (a ∨ b) ∨ c = a ∨ (b ∨ c)

I a ∧ (a ∨ b) = a a ∨ (a ∧ b) = a

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Duality Lattice Theorems

Definition Theorem dual: a ∧ a = a ∧ instead of ∨, ∨ instead of ∧ Proof. Theorem (Duality Theorem) a ∧ a = a ∧ (a ∨ (a ∧ b)) Every theorem has a dual theorem in lattices.

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Lattice Theorems Lattice Examples

Example

Theorem a  b ⇔ a ∧ b = a ⇔ a ∨ b = b < P{a, b, c}, ∩, ∪ >

⊆ relation

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Definition Definition distributive lattice: lower bound of lattice L: 0 upper bound of lattice L: I I ∀a, b, c ∈ L a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) ∀x ∈ L 0  x ∀x ∈ L x  I I I ∀a, b, c ∈ L a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) Theorem Every finite lattice is bounded.

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Counterexamples Counterexamples

Example Example

a ∨ (b ∧ c) = a ∨ 0 = a a ∨ (b ∧ c) = a ∨ 0 = a (a ∨ b) ∧ (a ∨ c) = I ∧ c = c (a ∨ b) ∧ (a ∨ c) = I ∧ I = I

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Distributive Lattice Join Irreducible

Definition Theorem join irreducible element: A lattice is nondistributive if and only if it has a sublattice a = x ∨ y ⇒ a = x ∨ a = y isomorphic to any of these two structures. I atom: a join irreducible element which immediately succeeds the minimum

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Example (divisibility relation) Theorem I prime numbers and 1 are join irreducible Every element in a lattice can be written as the join of join irreducible elements. I 1 is the minimum, the prime numbers are the atoms

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Complement Complemented Lattice

Theorem In a bounded, distributive lattice the complement is unique, if it exists. Proof. Definition a ∧ x = 0, a ∨ x = I , a ∧ y = 0, a ∨ y = I a and x are complements: a ∧ x = 0 and a ∨ x = I x = x ∨ 0 = x ∨ (a ∧ y) = (x ∨ a) ∧ (x ∨ y) = I ∧ (x ∨ y) = x ∨ y = y ∨ x = I ∧ (y ∨ x) = (y ∨ a) ∧ (y ∨ x) = y ∨ (a ∧ x) = y ∨ 0 = y

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Boolean Algebra Boolean Algebra - Lattice Relationship

Definition Boolean algebra: < B, +, ·, x, 1, 0 > Definition a + b = b + a a · b = b · a A Boolean algebra is a finite, distributive, complemented lattice. (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) a + 0 = a a · 1 = a a + a = 1 a · a = 0

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Definition dual: Example + instead of ·, · instead of + 0 instead of 1, 1 instead of 0 B = {0, 1}, + = ∨, · = ∧ Example Example (1 + a) · (b + 0) = b B = { factors of 70 }, + = lcm, · = gcd dual of the theorem: (0 · a) + (b · 1) = b

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Boolean Algebra Theorems References

a + a = a a · a = a Required Reading: Grimaldi a + 1 = 1 a · 0 = 0 a + (a · b) = a a · (a + b) = a I Chapter 7: Relations: The Second Time Around (a + b) + c = a + (b + c) (a · b) · c = a · (b · c) I 7.3. Partial Orders: Hasse Diagrams I Chapter 15: Boolean Algebra and Switching Functions

a = a I 15.4. The Structure of a Boolean Algebra a + b = a · b a · b = a + b

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