Algebra - Midterm II

Question 1

free group + normal closure + free abelian group

Answer 1

the free group generated by S is the set of all products of {a_i} and their inverses. Its elements are all products of a_{i1}^𝞮1…a_{ik}^𝞮k where 𝞮i∈{-1,1}.

the normal closure of R is the smallest normal subgroup containing R, denoted <>. Its elements are all products of the form (g_1^{-1})(r_1^𝞮1)(g_1) … (g_n^{-1})(r_n^𝞮n)(g_n) where 𝞮i∈{-1,1}

The free abelian group generated by two elements is written ≤a,b | ab=ba ≥

Question 2

action + faithful + kernel + stabilizer

Answer 2

A (left) action of a group G on a set X is a map GxX→X, (g,x)⥛gx satisfying:

• ex=x
• (g1g2)x = g1(g2x)

We call an action faithful if it is injective

the kernel of the action is the group of elements g acting identically (gx=x for all x∈X)

The stabilizer of x is the group
Gx = {g∈G : gx=x} < G

Question 3

equivalence classes of an orbit + transitive + free + Theorem + Theorem

Answer 3

We say x~y if there exists g∈G such that gx=y

The equivalence classes are called the orbits of the action

An action is called transitive if it only has one orbit

An action is called free if every stabilizer is trivial

ex. every group acts on itself by multiplication on the left. Here, the action is free

Theorem: Let G act on X freely and transitively. Then there exists a bijection φ:X→G such that φ(gx)=gφ(x)

Theorem: If an action of G on X is transitive then |X|=[G:Gx] for any x

ex. for every g, x maps to x^g = g^{-1}xg is an automorphism of G on itself

Question 4

center + stabilizer + centralizer + normalizer

Answer 4

Z(G) = {g : gx=xg for all x∈G}
this is the center of G

For h∈G, the stabilizer of h is {g: gh=hg} called the centralizer of h

If A⊆G then the centralizer of A is Z_G(A) = {g∈G : ga=ag for all a∈A}

If H≤G, then the normalizer of H is {g∈G : H= g^{-1}Hg }

Question 5

free abelian group generated by S

Answer 5

the free abelian group generated by S is {S | (si)(sj)=(sj)(si) for all si∈S }

All other abelian groups generated by S are natural homomorphism images of that

Question 6

basis stuff + gcd

Answer 6

look into the notes …

Question 7

Fundamental Theorem on finitely generated abelian groups

Answer 7

If G is a finitely generated abelian group, then there exists unique d1 | d2 | … | dk and r>=0 such that

G = C_{d1} \oplus C_{d2} \oplus … \oplus C_{dk} \oplus Z^r

where D_{di} is the cyclic group of order di and r=# zeros on the diagonal of the smith normal form

ALTERNATE FORM: If G is a finitely generated abelian group, then there exists unique r >=0 and q1,…,qn which are powers of (not necessarily distinct) prime numbers

G = C_{q1} \oplus … \oplus C_{qn} \oplus Z^r

Question 8

p-group + Theorem + Corollary + Proposition

Answer 8

A group is called a p-group if the order of every element is finite and is a power of p, where p is prime

Theorem: If p divides |G|, then there exists an element g of order p

Corollary: A finite group G is a p-group if and only if |G|=p^k for some k

Proposition: the center of a finite p-group is non-trivial

Question 9

1st Sylow Theorem

Answer 9

Suppose that p^k divides |G|. Then G contains a subgroup of order p^k

Question 10

Sylow p-subgroups

Answer 10

If p^k is the highest order of prime dividing |G|, then a subgroup of order p^k is called a Sylow p-subgroup

Question 11

2nd Sylow Theorem

Answer 11

Let P be a p-subgroup of G and let H be a Sylow p-subgroup.

Then there exists some g∈G such that g^{-1}Pg < H

In particular, all Sylow subgroups are conjugate to one another

Question 12

groups of each order

Answer 12

1: {e}
2: Z2 = C2
3: C3
4: C2xC2, C4
5: C5
6: C6 = C2xC3, S3=D3
7: C7
8: C8, C2xC2xC2, C2xC4, Q, D4

Question 13

3rd Sylow Theorem

Answer 13

the number of Sylow p-subgroups divides |G| and is congruent to 1 (mod p)

Question 14

ring

Answer 14

A ring R is a set with two binary operators: + and • s.t.

1) (R,+) is an abelian group
2) a(b+c)=ab+ac
3) a(bc)=(ab)c

also, may want it to have an identity or to be commutative

ex. R,C,Mn(R), C(X), Z[G]

Question 15

(left) zero divisor + left invertible

Answer 15

a is a left zero divisor if there exists some b≠0 such that ab=0

(similar for right)

a is a zero divisor if it is simultaneously a left and right zero divisor

ex. Z/nZ where n is not prime has zero divisors
ex. In Mn(R), the zero divisors are the matrices which are not divisible

If R has identity, a is left invertible if there exists some b such that ba=1.

Question 16

integral domain + division ring + field + homomorphism

Answer 16

A commutative ring with 0≠1 is called an integral domain if it has no zero divisors

If every non-zero element is invertible, then it is called a division ring

A commutative division ring is called a field

A homomorphism if a map f between rings R1 and R2 such that f(a+b)=f(a)+f(b), f(ab)=f(a)f(b)

Question 17

(left) ideal

Answer 17

A left ideal of R is a non-empty subset I which is a subgroup of the additive group such that ra∈I whenever r∈R, a∈I

Note: an ideal is a subring (without identity)

The trivial ideals are {0} and R

the smallest left ideal containing a1,a2,…,an is

{x1a1+x2a2+…+xnan + k1a1 + … + knan : xi∈R, ki∈Z}

the smallest (two-sided) ideal is much more complicated:
{ x11a1y11+ a12a1y12 + ... + (xnm)(an)(ynm) }

If R is a ring and I a (two-sided) ideal then R/I is an additive group

Question 18

principal ideal + principal ring + PID

Answer 18

An ideal is called principal if it is generated by one element.

A ring is called a principal ring if all its ideals are principal.

A principal ideal domain (PID) is a domain in which all ideals are principal.
Recall: integral domain is commutative + no zero divisors

Question 19

Theorems about ideals (3) + ex

Answer 19

Theorem 1: If f:R1 to R2 is a homomorphism, hten

f(R1) = R1 / ker f
f(r) from r+kerf

Theorem 2: If I,J are two ideals, then
I/(I⋂J) = (I+J)/J

Theorem 3: If I subset J are two ideals then
R/J = (R/I) / (J/I)

ex. if R=Z and (n) is the ideal generated by n, then
(n) + (m) = (gcd(n,m))
(n) ⋂ (m) = (lcm(n,m))

Question 20

prime ideal + Theorem + Theorem

Answer 20

An ideal P≠R is said to be prime if it cannot be written as the product of two ideals
i.e. for all ideals A,B if AB⊆P then A⊆P or B⊆P

Theorem: P is prime if and only if
for every a,b∈R if ab∈P then a∈P or b∈P

Theorem: If R is commutative with identity then an ideal P is prime if and only if R/P is an integral domain

Question 21

maximal + theorem + theorem

Answer 21

An ideal I is called maximal if I≠R and for every ideal J such that I⊆J either I=J or J=R

ex. In Z, (n) subset of (m) IFF m divides n

Theorem: Let R be a ring with identity. Then every proper ideal is contained in a maximal ideal.

Theorem: Let R be a commutative ring with identity. Then an ideal I is maximal if and only if R/I is a field

Question 22

Theorem (direct product, ideals) + Corollary + Corollary

Answer 22

Theorem:
If R is a ring and A1,…,An are ideals and

(1) A1+A2+…+An = R
(2) Ai ⋂ (A1+…+A_{i-1}+A_{i+1}+…+An) = {0}

Then R = A1 x … x An

Corollary 1: Under the same conditions,
R / (A1⋂…⋂An) = R/A1 x … x R/An

Corollary 2: if n=p1^{k1}…ps^{ks} for pairwise different primes pi then
Z/nZ = Z/p1^{k1}Z x … x Z/ps^{ks}Z

Question 23

divides + Proposition + Corollary + Proposition

Answer 23

R: commutative ring
a|b if there exists a x such that b=ax

We get an equivalence relation: a~b if and only if a|b and b|a. So a=b times a unit.

Proposition: a|b if and only if (a) contains (b)

Corollary: a~b if and only if (a)=(b)

Proposition: u is a unit (ie. an invertible element) if and only if (u)=R

Question 24

irreducible + prime

Answer 24

A non-unit element a is irreducible if whenever b|a either b is a unit or b~a
(the latter, if R is a domain means b=au for a unit u)

If R is a domain, this can be reformulated as: if whenever a=a1a2 either a1 or a2 is a unit.

a is irreducible iff (a) is maximal among proper principal ideals

A non unit element a is prime if a|b1b2 implies a|b1 or a|b2

Question 25

Chinese Remainder Theorem

Answer 25

A1,…,Ak ideals in R with identity such that Ai+Aj=R for i≠j

If b1,…,bn∈R then ∃b∈R s.t. b-bi∈Ai.

Moreover, b is unique modulo A1⋂A2⋂…⋂An