301 Abstract Algebra
2:00 - 2:50 MWF
Hentzel 432 Carver
hentzel@iastate.edu
Web Page   http://www.math.iastate.edu/hentzel/class.301

Text: A First Course in Abstract Algebra, sixth Edition 
by John B. Fraleigh

Course: Chapter 1,2,3,4,5 
  
Friday, September 21, 2001 

**********************************
*   TEST MONDAY SEPTEMBER 24     * 
*    pages  51 through 135       *
*    pages 197 through 208.      *
**********************************

Main Idea:  Study these problems for a good review
            for the test. 
Key Words:  What-ever-else you know, know this. 
Objective:  Focus your attention from the myriad of
            stuff we covered to the critical mass for
            Monday. 

1. (a) Write Pi in orbit notation for the 16 puzzle 
   (b) Square Pi
   (c) Find the inverse of Pi.
   (d) Tell if Pi is even or odd.
   (e) Find the order of Pi.

2. Prove that if G is a group and H is a subgroup of
   G, then |H| divides |G|.  This means that I want you
   to write down the proof of Lagrange's theorem.

3. Find integers x and y such that  19x + 17y = 1.

4. How many different bracelets with 6 beads can you  
you make with 3 red, two blue and 1 green bead.

5. If a group G acts on a set S
(a)  If s in S is given, let H = {g in G | gs = s} 
     is a subgroup of G.
(b)  Prove that all the elements in gH map
     s to the same element in the orbit of s.
(c)  Prove that different cosets map s to different
     elements in the orbit of s.

(d)  Prove that   Sum   F(g)  =    Sum  | Stabilizer(s) |
                 g in G           s in S.
     F(g) is the number of elements of S such that gs = s.
     | Stabilizer(s) | is the number of elements in g
     such that gs = s. 

6.  Suppose you have a commutative group satisfying these
    relations.
       
         27 a + 18 b + 12 c = 0
         30 a + 12 b +  6 c = 0

    Write G = Zr x Zs x Zt and give the elements a,b,c in 
    Zr x Zs x Zt. 







Previous Assignment:  
1.  Given the generators and relations
 3a + 6 b + 12 c = 0
 8a + 2 b +  6 c = 0
 2a + 3 b + 15 c = 0
Decompose the group into cyclic summands and
give the generators of each cyclic summand.

Z1 x Z1 x Z372
Essentially the group is cyclic of order 372.

a' = a+3b-3c = 0
b' =  b+117c = 0
c' =       c

| a'|    | 1  3  -3  || a |
| b'| =  | 0  1  117 || b |
| c'|    | 0  0   1  || c |

| a |    | 1 -3  354 || a'|
| b | =  | 0  1 -117 || b'|
| c |    | 0  0   1  || c'|

           0 x 0 x Z372
      a = (0,  0,   354)
      b = (0,  0,  -117)
      c = (0,  0,     1)

 3a + 6 b + 12 c =   (0,0,3*354-6*117+12) = (0,0, 372) = (0,0,0)  
 8a + 2 b +  6 c =   (0,0,8*354-2*117 +6) = (0,0,2604) = (0,0,0)
 2a + 3 b + 15 c =   (0,0,2*354-3*117+15) = (0,0, 372) = (0,0,0)

2. Given the generators and relations
 8a + 5 b + 12 c = 0
 6a + 3 b +  2 c = 0
 6a + 9 b +  6 c = 0
Decompose the group into cyclic summands and
give the generators of each cyclic summand.

    Z1 x Z2  x Z156
| a'|    |  6   3   2 || a |
| b'| =  |-37 -13 -13 || b |
| c'|    |-20  -7  -7 || c |

| a |    |  0   7 -13 || a'|
| b | =  |  1  -2   4 || b'|
| c |    | -1 -18  33 || c'|

3. Given the generators and relations
12a + 9 b +  2 c = 0
 8a + 3 b +  4 c = 0
 4a + 4 b +  6 c = 0
Decompose the group into cyclic summands and
give the generators of each cyclic summand.

       Z1 x Z2 x Z112
|a'| | 12   9   2 | |a |
|b'|=|-82 -60 -15 | |b | 
|c'| |-11  -8  -2 | |c | 

|a | |  0   2  -15| |a'|
|b |=|  1  -2   16| |b'| 
|c | | -4  -3   18| |c'| 

12a + 9 b +  2 c =  (0,  0,  0) = (0,0,0)         
 8a + 3 b +  4 c =  (0, -2,  0) = (0,0,0)
 4a + 4 b +  6 c =  (0,-18,112) = (0,0,0)