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Title: Math 3121 Abstract Algebra I


1
Math 3121Abstract Algebra I
  • Lecture 9
  • Finish Section 10
  • Section 11

2
HW Due Today
  • Hand in
  • Due Tues, Oct 28Page 73 12, 14, 16Page 84
    18
  • Pages 94-95 10, 24, 36
  • Do not hand in
  • Pages 94-95 19, 39 

3
Section 10
  • Section 10 Cosets and the Theorem of Lagrange
  • Modular relations for a subgroup
  • Definition Coset
  • Theorem of Lagrange For finite groups, the order
    of subgroup divides the order of the group.
  • Theorem For finite groups, the order of any
    element divides the order of the group

4
Modulo a Subgroup
  • Definition Let H be a subgroup of a group G.
    Define relations L and R by
  • x L y ? x-1 y in H
  • x R y ? x y-1 in H
  • We will show that L and R are equivalence
    relations on G.
  • We call L left modulo H.
  • We call R right modulo H.
  • Note
  • x L y ? x-1 y h, for some h in H
  • ? y x h, for some h in H
  • x R y ? x y-1 h, for some h in H
  • ? x h y, for some h in H

5
Modulo a Subgroup is an Equivalence Relation
  • Theorem Let H be a subgroup of a group G. The
    relations L and R defined by
  • x L y ? x-1 y in H
  • x R y ? x y-1 in H
  • are equivalence relations on G.
  • Proof We show the three properties for
    equivalence relations
  • 1) Reflexive x-1 x e is in H. Thus x L x.
  • 2) Symmetric x L y ?x-1 y in H
  • ? (x-1 y) -1 in H
  • ? y-1 x in H
  • ? y L x
  • 3) Transitive x L y and y L z ? x-1 y in H
    and y-1 z in H
  • ? (x-1 y )( y-1 z) in H
  • ? (x-1 z) in H
  • ? x L z
  • Similarly, for x R y .

6
Cosets
  • The equivalence classes for these equivalence
    relations are called left and right cosets modulo
    the subgroup.
  • Recall x L y ? x-1 y h, for some h in H
  • ? y x h, for some h in H
  • Cosets are defined as follows
  • Definition Let H be a subgroup of a group G.
  • The subset
  • a H a h h in H
  • is called the left coset of H containing a, and
    the subset
  • H a a h h in H
  • is called the right coset of H containing a.

7
Examples
  • Cosets of nZ in Z are
  • nZ, nZ1, nZ2, , nZ (n-1)
  • For example 2Z in Z has two cosets.
  • Cosets of 0, 3 in Z6
  • Cosets of 0, 2, 4 in Z6

8
Abelian versus Nonabelian
  • Note In abelian groups the left cosets are the
    right cosets. In nonabelian case left and right
    dont always agree.
  • H e, µ1 in G S3 has different left and
    right cosets. For left cosets, make a
    multiplication table GH. For right cosets, make
    a multiplication table HG. (See slides after
    this one).
  • H ?0, ?1, ?2 in S3 has same left and right
    cosets.

9
Multiplication Table for S3 in Cycle Notation
e (1 2 3) (1 3 2) (2 3) (1 3) (1 2)
e e (1 2 3) ( 13 2) (2 3) (1 3) (1 2)
(1 2 3) (1 2 3) (1 3 2) e (1 2) (2 3) (1 3)
(1 3 2) (1 3 2) e (1 2 3) (1 3) (1 2) (2 3)
(2 3) (2 3) (1 3) (1 2) e (1 2 3) (1 3 2)
(1 3) (1 3) (1 2) (2 3) (1 3 2) e (1 2 3)
(1 2) (1 2) (2 3) (1 3) (1 2 3) (1 3 2) e
10
Left Cosets for S3 in Cycle Notation
e (2 3)
e e (2 3)
(1 2 3) (1 2 3) (1 2)
(1 3 2) (1 3 2) (1 3)
(2 3) (2 3) e
(1 3) (1 3) (1 3 2)
(1 2) (1 2) (1 2 3)
Distinct left cosets of H e, (2 3) e, (2
3) e H (2 3) H (1 2 3), (1 2) (1 2 3) H
(1 2) H (1 3 2), (1 3) (1 3 2) H (1 3) H
11
Right Cosets of lt(2 3)gt for S3 in Cycle Notation
e (1 2 3) (1 3 2) (2 3) (1 3) (1 2)
e e (1 2 3) ( 13 2) (2 3) (1 3) (1 2)
(2 3) (2 3) (1 3) (1 2) e (1 2 3) (1 3 2)
Distinct right cosets of H e, (2 3) e, (2
3) H H (2 3) (1 2 3), (1 3) H (1 2 3)
H (1 3) (1 3 2), (1 2) H (1 3 2) H (1 2)
12
Counting Cosets
  • Theorem For a given subgroup of a group, every
    coset has exactly the same number of elements,
    namely the order of the subgroup.
  • Proof Let H be a subgroup of a group G. Recall
    the definitions of the cosets aH and Ha.
  • a H a h h in H
  • H a a h h in H
  • Define a map La from H to aH by the formula
    La(g) a g. This is 1-1 and onto.
  • Define a map Ra from H to Ha by the formula
    Ra(g) g a. This is 1-1 and onto.

13
Lagrange
  • Theorem (Lagrange) Let H be a subgroup of a
    finite group G. Then the order of H divides the
    order of G.
  • Proof Let n number of left cosets of H, and
    let m the number of elements in H. Then m is
    the number of elements in any left coset. Thus
    n m the number of elements of G. Here m is the
    order of H, and n m is the order of G.

14
Orders of Cycles
  • The order of an element in a finite group is the
    order of the cyclic group it generates. Thus the
    order of any element divides the order of the
    group.

15
HW Section 10
  • Dont hand in
  • Pages 101 3, 6, 9, 15
  • Hand in Tues, Nov 4
  • Pages 101-102 6, 8, 10, 36, 40

16
Section 11 (as time permits)
  • Direct Products and Finitely Generated Abelian
    Groups
  • Cartesian Product of sets
  • Direct product of groups
  • Structure of Zn?Zm
  • Structure of products of cyclic groups
  • Next time Structure of Finitely Generated
    Abelian Groups

17
Cartesian Products
  • Definition The Cartesian product of a finite
    collection of sets Sk, for k 1 to n is the set
    of all n-tuples (s1, s2, , sn), with sk in Sk.
    The Cartesian product is denoted by
  • S1 S2 Sn
  • or by product notations such as

18
Projection Maps
  • The map pi S1 S2 Sn ? Si that takes the
    n-tuple s (s1, s2, , sn) to its ith component
    si is called the ith projection map.
  • In other words For any s in S1 S2 Sn the
    result of applying the ith projection map to s is
    called the ith component of s and denoted by si .
  • Note Two elements a and b of the product are
    equal if and only if ai bi, for all i 1,,n.

19
Direct Product of Groups
  • Theorem Let G1, G2, , Gn be groups with
    multiplicative notation. Define a binary
    operation on the Cartesian product G1G2 Gn by
    (a1, a2, , an)(b1, b2, , bn) (a1 b1, a2 b2,
    , an bn), then G1G2 Gn is a group with this
    binary operation. The set G1G2 Gn with this
    binary operation is called the direct product of
    the groups G1, G2, , Gn.
  • Proof Note that the binary operation is defined
    component-wise. That is, if a and b are in the
    product, then (a b)i ai bi. 1) Associativity
    follows because each component binary operation
    is associative. 2) The identity is e the n-tuple
    e (e1, e2, , en), where each ei is the
    identity of its own component group Gi. 3) For
    each a (a1, a2, , an) in the product, the
    n-tuple a-1 ((a1)-1, (a2)-1, , (an)-1) is the
    inverse of a.

20
Direct Sums of Groups
  • Sometimes we call the direct product a direct
    sum, especially if we use additive notation.
  • A direct product is characterized by its
    projection maps. These turn out to be
    homomorphisms.
  • On the other hand, the direct sum is
    characterized by injection maps ji Gi ? G1G2
    Gn that each take ai in Gi to the n-tuple (e1,
    e2, , ai, , en) that has identities in all
    components except for the ith, which has ai.
    These also turn out to be homomorphisms.

21
Examples
  • Z2Z3
  • Z3Z5
  • Z2Z2
  • Z3Z3
  • Z2Z6
  • Z9Z6

22
Internal Products
  • Each component group of a direct product can also
    be considered a subgroup by injection. All of
    these subgroups commute with each other. Thus
    any element g of the product G1G2 Gn can be
    uniquely written in the form g g1 g2 gn with
    gi in Gi. When this happens, G is said to be an
    internal product of the subgroups Gi.

23
LCM and GCD of two numbers
  • Let x and y be integers, then
  • LCM(x, y) is the least multiple of x and y
  • LCM(x, y) min m in Z m is a multiple of x
    and m is a multiple of y
  • GCD(x, y) is the greatest common divisor of x and
    y
  • GCD(x, y) max m in Z m divides x and m
    divides y
  • LCM(x, y) x y /GCD(x, y).

24
Methods of finding LCM and GCD
  • Euclidean Algorithm to find GCD in the form a x
    b y Start with positive integers x and y. Set
    r0 x, r1 y. Given rk-1 and rk, find qk and
    rk1 such that rk-1 qk rk rk1, with 0
    rk1 lt rk. Continue until rk1 0. Then GCD(x,
    y) rk. Then work backward to write GCD in the
    form a x b y.
  • Example Find GCD of 64 and 58.
  • For small numbers use prime factorization.
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