CS Lecture 14 Powerful Tools     !

Build your toolbox of abstract structures and concepts. Know the capacities and limits of each tool.

Today’s Lecture: groups

What makes this calculation possible are abstract properties of integers and addition. Closure: the sum of two integers is an integer Associativity: (x + y) + z = x + (y + z) Identity: there is an integer 0 such that  x, 0 + x = x + 0 = x Inverse:  x  an integer –x s.t. x + (-x) = (-x) + x = 0 Associativity Identity Inverse Inverse, Closure

closure, identity, associativity, inverse integers / addition naturals / addition odd integers / addition even integers / addition rationals / addition reals / addition complex numbers / addition all four not inverse not closure, not identity all four

Let S be a non-empty set. Let  be a binary operator on S. Definition of a Group ( S,  ) is called a group if it has these properties: closure: associativity: identity: inverse:

Commutativity A group ( S,  ) is commutative if A commutative group is also called an Abelian group.

integers / + naturals / + odd integers / + even integers / + rationals / + integers /  rationals /  rationals – {0} /  group not inverse not closure, not identity group not inverse group

is a group closure associativity identity: 0 inverse:

Cancellation Theorem: inverse, closure associativity inverse identity Proof:

Identity? 1 Is this a group? But 0 has no inverse!

group  Not: closure inverse

 group Notice that each row and column is a permutation of the elements.

Theorem: Each row and column of the multiplication table is a permutation of the group elements. Proof: Suppose not. By closure, if a row is not a permutation, it must have repeated elements. ·  b  c  a x x  By cancelation:

Conjecture: S p is a group for prime p.  Not: closure inverse means that So  is not closed.

Theorem: Suppose (S,  ) has 1)closure 2)associativity 3)identity 4)cancelation And S is finite, Then (S,  ) is a group. Proof: a a This row is a permutation of the elements of S. Therefore, every a has an inverse, a -1, such that a  a -1 = 1. a -1 1

Cancelation modulo n Proof: Theorem: Corollary:

Let * denote multiplication modulo n. Suppose (a*b, n) > 1. Then there is a prime p such that p | n and p | a*b. p | a*b  p | ab-kn, for some k (note: ab is NOT modulo n)  p | ab (since p | n)  p | a or p | b (since p is prime)  (a,n)  p or (b,n)  p (a contradiction) closure: associativity. identity. cancelation: Theorem: with multiplication modulo n is a group. Proof:

A permutation  on [1..n] is a one-to-one function [1..n] mapping onto [1..n]

Composition of permutations Let  1 and  2 be permutations on [ 1.. n ]. The composition of  1 and  2, written  1   2 is given by  1   2 ( x ) =  1 (  2 ( x ) ) x  2  1  1   2 ( x )

Example: “  2 composed with  1 ” Notice: Composition of permutations is not always commutative.

A n = set of all permutations on [1.. n ]  = permutation composition Theorem: ( A n,  ) is a group Proof: closure a b c associativity identity a a -1 inverse

Subgroups Let ( S,  ) be a group. Then H is a subgroup of S if H  S and ( H,  ) is a group.

Example: (H,+) is a group.

Lagrange’s Theorem: If H is a subgroup of a finite group G, then the size of H divides the size of G. Example:

Proof of Lagrange’s Theorem (slide 1 of 3): Lemma: Proof: If aH were smaller it would mean but by cancelation, Definition:

Proof of Lagrange’s Theorem (slide 2 of 3): Definition: Lemma: Proof: Similarly Suppose Then

Lemma: Proof: Every element in G appears in at least one of these. Thus, this list contains a finite list of distinct sets, all of size | H |, that partition G. Proof of Lagrange’s Theorem (slide 3 of 3):

Example: