Abstract Algebra: Midterm I Review

Definitions

Set

A set is a collection of elements

Map

A mapfromtois a rule which assigns a unique element ofto an elements of.

1. surjective: iff
2. injective: iff if, thenif, then
   3.bijective: iffand

Product

A product ofandis a set

Equivalence Relations

An equivalence relation(or) onis a subset
1.reflexivity:
2.symmetry: if, then
3. transitivity: if, then

Equivalence Class

An equivalence class ofis

Quotient

A quotient ofbyis

Z/nZ

iff

Group

A groupis a set G and a binary operations.t.,
Associativity:
Identity:
Inverses:

Homomorphism

A homomorphism fromtois a map (of sets),s.t.

Symmetric Groups

Letbe a set, we define(cardinality of X=number of elements in X) is a group of all bijective maps(bijections)

Subgroup

is a group, Subgroupofis a groupwith the group operation restricted from.

Subgroup Diagram

Subgroup diagram of a group,trival subgroup(),is contained as a set. They are all subgroups of

Proper Subgroup

Proper subgroup ofis subgroup which is not trivial oritself.

Cyclic Subgroup

$Ga{a^k | k \in Z}\star = \star_G$

Order

,Order of $$ is minimal,s.t.

Isomorphism

Isomorphism is bijective homomorphism.(which is bijective：injective,subjective)

Kernel

Kernel of, Kernel is the preimage of identity:

Coset

,is the coset of.

Direct Product

are groups, direct product of&is a setwith the following operation

Normal

,id normal iff
iff

Kernel And Image

ifis a homomorphism
image of
kernel of

Inner Direct Product

is a group,,is the internal direct product ofandiff
1.(warning:)
2.(i.e. as samll as possible)
3.for any(warning: G is not abelian)

Propositions

• Proposition1
• Proof

has exactly n elements

if, then
Let, then


then,

ifand, then[i] and [j] do not intersect
We prove by contradiction:
assume
then,

According to, we have
Since a and b are integers, we have
and since we have,


which caused a contradiction, and it should be. Proved.
Everybelongs to one of the classes
Proof:
By,has at least n elements.
By,has at most n elements.

• Proposition2
• Proof

Addtion is correctly define on,the classis the same for diff. elements ofand.

• Proposition3
• Proof

Identity element e is laways unique in any group G.

By contradiction


which caused a contradiction

• Proposition3.18
• Proof

is a group. The inverseis unique

By contradiction


which caused a contradiction that

• Proposition3.19
• Proof

For any

• Proposition3.20
• Proof

• Proposition3.21
• Proof

For any

• Proposition3.22
• Proof

• Proposition

1. Operation of composition (of maps) is associative
2. Identity:
3. Inverses: Just flip it.

• Proposition3.30
• Proof

is a subgroup ofiff:

1.(identity of subgroup)

2.(closure of the group operation)if, then,

3.(closure of inverses)if, then.

is Problem #2 on HW#3
is a subgroupthat all of three is true

• Proposition3.31
• Proof

is a subgroup iff () and ()

• Lemma
• Proof

$={a^k|Z}$ is a subgroup of G

if, then

• Lemma
• Proof

$={a^k|Z}$ is a subgroup of G

if, then

• Proposition

is cyclic iff $\exists g \in G, G=$

• Proposition
• Proof

are groups and they are homomorphism, then:
1.
2.
3.if, then(image ofunder)
3’. as, then(image of:)
4. if, thenis a subgroup of G (we callis called pre-image of)

1.
2.
3.
4. Let us consideris a subgroup?
Prop 3.31:

Lemma6.3

The following are equivalent(1-3 therioticallt useful, 4 griginal warm-up, 5 How to use this lemma for actual)

• Corollary
• Proof

ifis abelian, then every subgroup is normal.

We have, as
Q: Why do we need the definition of normality
Idea:,, we want group operation on(factor/quotient group of):


such that
Problem: if
How to deal with such sits? Whether

or not.

(Remark:, it is really good!)

Theorem

• Theorem
• Proof

$H \leq G \quad s.t. \quad a \in H$

part is already proved.Let us prove the second part.Any subgroup, then
$$\forall k \in Z, a^k \in H, \text{then} \leq H$$
Therefore $HGa \in H$

• Theorem

Every cyclic group is abelian,

Theorem 6.4

H-cosets partition G: we can finda set of elements ofsuch thatthat

Theorem 6.10

Lagrange’s Theorem
= finite group,,(Note: all three are natural numbers)

Cor 6.11

,(Remember,= order of=minimals.t.)

Theorem 9.7+9.8

Ifis a cyclic group, then:
if it is finite order(n), then it is isomorphic to.
if it is of infinite order, then it is isomorphic to.

Thm/Prop

ifare isomorphic, then,then other is.
ifare finite, then,then other is.
ifare cyclic, then,then other is.

• Theorem
• Proof

ifis a homomorphism, then
(means isomorphic,is bijective)

1. is normal,
   
   2.To find

• Theorem
• Proof

ifis internal direct product of, then

Let’s consider


Q: is it well-defined?
A:, by 2, we know that

Q:homomorphism?




+surj+inj

• Proposition
• Proof

内容

内容

Z/nZ

iff