Title: Math 3121 Abstract Algebra I
1Math 3121Abstract Algebra I
- Lecture 9
- Finish Section 10
- Section 11
2HW 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
3Section 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
4Modulo 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
5Modulo 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 .
-
6Cosets
- 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.
7Examples
- 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
8Abelian 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.
9Multiplication 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
10Left 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
11Right 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)
12Counting 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.
13Lagrange
- 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.
14Orders 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.
15HW Section 10
- Dont hand in
- Pages 101 3, 6, 9, 15
- Hand in Tues, Nov 4
- Pages 101-102 6, 8, 10, 36, 40
16Section 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
17Cartesian 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
-
18Projection 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.
19Direct 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.
20Direct 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.
21Examples
- Z2Z3
- Z3Z5
- Z2Z2
- Z3Z3
- Z2Z6
- Z9Z6
22Internal 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.
23LCM 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).
24Methods 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.