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Math 3121 Abstract Algebra I. Lecture 12 Finish Section 14 Review. Next Midterm. Midterm 2 is Nov 13. Covers sections: 7-14 (not 12) Review on Thursday. Cosets of a Homomorphism.

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math 3121 abstract algebra i

Math 3121Abstract Algebra I

Lecture 12

Finish Section 14

Review

next midterm
Next Midterm
  • Midterm 2 is Nov 13.

Covers sections: 7-14 (not 12)

Review on Thursday

cosets of a homomorphism
Cosets of a Homomorphism

Theorem: Let h: G  G’ be a group homomorphism with kernel K. Then the cosets of K form a group with binary operation given by (a K)(b K) = (a b) K. This group is called the factor group G/K. Additionally, the map μ that takes any element x of G to is coset xH is a homomorphism. This is called the canonical homomorphism.

coset multiplication is equivalent to normality
Coset Multiplication is equivalent to Normality

Theorem: Let H be a subgroup of a group G. Then H is normal if and only if

(a H )( b H) = (a b) H, for all a, b in G

canonical homomorphism theorem
Canonical Homomorphism Theorem

Theorem: Let H be a normal subgroup of a group G. Then the canonical map : G  G/H given by (x) = x H is a homomorphism with kernel H.

Proof: If H is normal, then by the previous theorem, multiplication of cosets is defined and  is a homomorphism.

fundamental homomorphism theorem
Fundamental Homomorphism Theorem

Theorem: Let h: G  G’ be a group homomorrphism with kernel K. Then h[G] is a group, and the map μ: G/K  h[G] given by μ(a K) = h(a) is an isomorphism. Let : G  G/H be the canonical map given by (x) = x H. Then h = μ.

h

G

h[G]

μ

G/Ker(h)

proof of fundamental thoerem
Proof of Fundamental Thoerem
  • Proof: This theorem just gathers together what we have already shown. We have already shown that h[G] is a group. We have h(a) = h(b) iff aK = bK. Thus μ exists.μ((x)) = μ(x H) = h(x).

x

h(x)

h

G

h[G]

μ

x Ker(h)

G/Ker(h)

properties of normal subgroups
Properties of Normal Subgroups

Theorem: Let H be a subgroup of a group G. The following conditions are equivalent:

1) g h g-1 H, for all g in G and h in H

2) g H g-1= H, for all g in G

3) g H = H g, for all g in G

Proof:

1) ⇒ 2): H  g H g-1

1) ⇒ g H g-1 H ⇒ g H g-1 H and

g H g-1 H ⇒ 2)

2) ⇒ 3):

Assume 2). Then x in g H ⇒ x g-1 in H ⇒ x in H g

and x in H g ⇒ x g-1 in H ⇒ g x g-1 in g H

3) ⇒ 1):

Assume 3). Then h  H ⇒ g h  g H ⇒ g h  H g ⇒ g h g-1 H

automorphism
Automorphism

Definition: An isomorphism of a group with itself is called an automorhism

Definition: The automorphism ig: G  G given by ig (x) = g x g-1 is the inner automorphism of G by g. This sometimes called conjugation of x by g.

Note: ig is an automorphism.

more terminology
More Terminology
  • Invariant subgroups
  • Congugate subgroup. – examples in S3
hw section 14
HW: Section 14
  • Don’t hand in

Pages 142-143: 1, 3, 5, 9, 11, 25, 29, 31

  • Hand in:

Pages 142-143: 24, 37