Algebra - Midterm I

Question 1

group + 5 properties

Answer 1

(1) associativity
(ab) c = a(bc)
(2) identity element
(3) inverses

A monoid satisfies (1) and (2)
A semigroup satisfies (1)

Properties:

• identity is unique
• inverse is unique
• (a^{-1})^{-1} = a
• (ab)^{-1} = b^{-1} a^{-1}
• a_1 a_2 … a_n is well-defined

Question 2

commutative / abelian

Answer 2

If ab=ba for all a,b in G

Question 3

examples of groups

Answer 3

• (Z,+)
• (Q,+), (Q*,•)
• vector spaces wrt addition
• GLn(R) = {A \in Mn(R) with det(A)≠0 }
• SLn(R) = { A \in Mn(R) with det(A) = 1 }
• SLn(Z)
• Z / nZ = Z_n

Question 4

Sn, An, Dn

Answer 4

Sn = group of permutations on {1,2,…,n}

ie. bijections from {1,2,…,n} to itself

An = even permutations in Sn

Dn = group of symmetries on a regular n-gon

|Sn|=n!
|An| = n!/2
|Dn| = 2n

Question 5

subgroup

Answer 5

A non-empty subset H in G such that

• g,h \in H then gh \in H
• g \in H then g^{-1} \in H

subgroup criterion: for g,h \in H then gh^{-1} \in H

Question 6

subgroup generated by S

Answer 6

For S subset of G, the subgroup generated by S is denoted and is the smallest subgroup containing S (ie. intersection of all subgroups containing S)

It consists of all products g_1^{a_1} … g_m^{a_m} where g_i \in S and a_i \in \bZ

Question 7

homomorphism

Answer 7

a map f:G⟹H between groups is a homomorphism if f(ab)=f(a)f(b)

f(g)=e is the trivial homomorphism

A homomorphism is called a

• monomorphism if it is injective
• epimorphism if it is onto
• isomorphism if it is bijective (ie. it has an inverse)

If ∃f:G⟹H isomorphism, we write G≅H

Question 8

equivalence relation

Answer 8

~ is an equivalence relation if

• a~a
• a~b then b~a
• a~b and b~c then a~c

ex. for H≤G define g~h if gh^{-1} \in H

Question 9

cosets

Answer 9

We call gH a left coset

The set of left cosets is denoted G/H

Note: the map aH↔︎Ha is not well-defined
ex. G=S_3, H=≤(12)>, then H(13) = H(132) but (13)H≠(123)H

Question 10

Lagrange’s Theorem (+ 2 corollaries)

Answer 10

If H≤G then |H| divides |G|

Corollary 1: If |G| is prime, the only subgroups are the trivial ones

Corollary 2: If |G| is prime, then G is cyclic

Question 11

index

Answer 11

|G/H| = [G:H]

If G is finite, [G:H] = |G| / |H|

If G>H>K
[G:K] = [G:H][H:K]

Question 12

completely classify ≤g>

Answer 12

If all g^n are different, then ≤g> ≅≈\bZ

If not, then ker is a non-zero subgroup of \bZ. d=min(H⋂\bZ) so dZ ≈ ker(n \mapsto g^n)

Then ≤g>≈Z/dZ

Question 13

Theorem (subgroups of Z)

Answer 13

Every subgroup of Z is of the form dZ for some d \in \bN

Question 14

normal subgroup + Theorem

Answer 14

H≤G is normal if gH = Hg

Equivalently, g^{-1}Hg = g

We write H◁G

Theorem: If [G:H] = 2 then H◁G

Question 15

conjugation

Answer 15

Conjugation by g∈G is the map x \mapsto g^{-1}xg

This is an isomorphism

(an isomorphism with itself is called an automorphism)

we write it as x^g = g^{-1}xg

Question 16

simple

Answer 16

A group is simple if the only normal subgroups of G are G and {e}

ex. Z_p for prime p are simple
ex. A_n for n≥5 is simple

Question 17

Theorem (coset representations)

Answer 17

The multiplication (gH)(hH)=(gh)H is well defined ⇔ H◁G

G/H is then a group.

Note: the map g \mapsto gH is a homomorphism and is called the “canonical epimorphism”

Question 18

First Isomorphism Theorem

Answer 18

Let φ:G→H be a homomorphism.

Then φ(G) ≈≅ G/ker(φ)

Moreover, the isomorphism is the map which sends φ(g) to g(kerφ)

Proof: well-defined, onto, one-to-one

Question 19

center

Answer 19

Z(G) = { h∈G : gh=hg for all g∈G}

This is normal in G

Question 20

5 propositions for normal

Answer 20

Proposition 1: If G>H and G▷N then H⋂N◁H

Proposition 2: If G>H and G▷N then N ◁ ≤H⋃N>

Proposition 3: If G>H and G▷N then ≤H⋃N>=HN=NH

Proposition 4: If N and K are normal in G then N⋂K ◁ G

Proposition 5: G / (N⋂K) is isomorphic to a subgroup of G/N x G/K

Question 21

2nd and 3rd isomorphism Theorems

Answer 21

2nd Isomorphism Theorem:
For G>H, G▷N then HN/N ≈≅ H / (N⋂H)

3rd Isomorphism Theorem:
If G▷N and G▷K and N≥K (so N▷K) then G/K ▷ N/K

and (G/K) / (N/K) ≈≅ G/N

(should also probably know proofs of these two)

Question 22

Direct product

Answer 22

Let G_i be a collection of groups

Then \Pi_{i \in I} G_i = { (g_i)_{i \in I} \mid g_i \in G_i } is the direct product of the groups

We have multiplication (g_i)(h_i) = (g_ih_i)

π_j : \Pi G_i → G_j is the projection (this is an epimorphism)

If H is a group and φ_i : H → G_i are homomorphisms then we have a homomorphism ψ:H→ \Pi G_i which sends h to (φ_i(h))

This is uniquely determined by π_i ○ ψ = φ_i

Question 23

Category Theory

Answer 23

LOL

Question 24

actions

Answer 24

and all that shite