Math 120c - Abstract Algebra
Spring term 2012
Midterm Solutions
1. Solution:
a. Z/2Z Z/2Z
b. Z/9Z Z/3Z, where : Z/3Z Aut(Z/9Z) = Z/6Z via 1 2
c. D4 Z/3Z which has order 24
d. S3
e. S3
f. 8, D4 and Q8
g. 60, A5
h. 24, D4 Z/3Z and A4
2. Solution:
Write G as
Math 120c: Final Exam
Due on Wednesday June 6, 2012 at noon
Time Limit: 4 hours
The exam consists of 2 problems. Each problem is worth 50 points.
Directions:
You may work beyond the time limit. If you decide to do so,
you are required to clearly mark wha
Math 120c - Abstract Algebra
Spring term 2012
Homework 1 Solutions
1. Solution:
Since P is a p-subgroup of G, thus should be contained in some Sylow p-group
of G, say Q.
Now P H Q, while H Q is also a p-subgroup of H , thus the equality
holds, P = H Q.
2.
Math 120c - Abstract Algebra
Spring term 2012
Homework 2 Solutions
1. Solution:
Dn =< a, b|an = b2 = 1, (ab)2 = 1 >
For n = 2k + 1, Dn =< a >, Z (Dn ) = cfw_1;
For n = 2k , Dn =< a2 >, Z (Dn ) =< ak >;
So the derived series is:
For n = 2k + 1, Dn < a > cf
Math 120c - Abstract Algebra
Spring term 2012
Homework 3 Solutions
1. Solution:
a. Consider : A R R[G] A[G] generated by x g rg g g (f (rg )x)g ,
where f : R A is a ring homomorphism.
Clearly, this is an A-module epimorphism, and then straightforward to c
Math 120c - Abstract Algebra
Spring term 2012
Homework 4 Solutions
1. Solution:
If M is simple, then by Schurs lemma, EndR (M ) is a division algebra.
Conversely, if we have a decomposition of M into homogenous components:
M = iI Mi , then EndR (M ) = iI
Math 120c - Abstract Algebra
Spring term 2012
Homework 5 Solutions
1. Solution:
a. Choose a K -basis cfw_a1 , , am and cfw_b1 , , bn of A and B , we know that
A B has a K -basis cfw_ai bj , 1 i m, 1 j n.
If c = cij ai bj Z (A B ), then for any a A, we h
Math 120c - Abstract Algebra
Spring term 2012
Homework 6 Solutions
1. Solution:
By Maschkes theorem, its sucient to prove the case that V is simple.
Now, assume : G GL(V ) be the representation, consider H = CG (V )
be the kernel of the action of G on V ,
Math 120c: Midterm
Due on Monday May 7, 2012 at noon
Time Limit: 4 hours
The exam consists of 5 problems. Each problem is worth 20 points.
Directions:
You may work beyond the time limit. If you decide to do so,
you are required to clearly mark what part
MATH 120C SPRING 2011 HOMEWORK 2 SOLUTIONS
1. Solution:
Only needs to show that elements in N and K commute.
Pick g N, h K , since ghg 1 h1 N K = cfw_1, we have gh = hg .
2. Solution:
Note that |Aut(Z/4Z )| = (4) = 2, since S is non-abelian, the nontrivia