quiz 2 (groups)
Name: DrSarah Greenwald
Start Time: Aug 13, 2000 19:08 Time Allowed: 12 min
Number of Questions: 5

Question 1  (8 points)

Match the mathematician with the example that illustrates their math.

1. Leopold Kronecker (Fundamental Theorem of Finite Abelian Groups)
  a. Z_6={0,1,2,...,5} under +mod6 has
a subgroup of order 2 Z_2={0,3} under +mod6
and a subgroup of order 3 Z_3={0,2,4} under +mod6.
2. Marjorie Lee Browne
  b. Z_6={0,1,2,...,5} under +mod6 is the direct product of
Z_2={0,1} under +mod2 and Z_3={0,1,2} under +mod3.
3. Johann Carl Friedrich Gauss (Fundamental Theorem of Algebra)
  c. When we adjoin any root r of f(x)=x^6+6x^5+17x^4+32x^3+37x^2+26x+6 to the complex numbers C, we still get C. I.e. C(r)=C for all roots r.
4. Peter Ludwig Mejdell Sylow (Sylow's First Theorem)
  d. 2x2 matrices with determinant 1 satisfying A times A transpose equals the identity form a group under matrix multiplication.
1 -->
2 -->
3 -->
4 -->

Question 2  (3 points)

What is the negation of the closure condition of groups:
For all a,b in G, a*b in G.

1. For all a,b in G, a*b is not in G.  
2. There exists a in G s.t. for all b in G, a*b is not in G.  
3. There exists a in G, there exists b in G s.t. a*b is not in G.  
4. There exists a not in G, there exists b not in G s.t. a*b is not in G.  
5. none of the above  

Question 3  (3 points)

What is the negation of the identity condition of groups:
There exists i in G s.t. for all a in G, i*a=a.

1. For all i not in G, there exists a not in G s.t. i*a is not equal to a.  
2. There exists i not in G s.t. for all a not in G i*a is not equal to a.  
3. For all i in G, there exists a in G s.t. i*a is not equal to a.  
4. There exists i in G s.t. for all a in G i*a is not equal to a.  
5. none of the above  

Question 4  (3 points)

How could one start the first line of a "disproof" of the associative property (#2 of a group definition)?

1. To disprove associativity, we will show that for any a, b, c in G, a*(b*c) is not (a*b)*c  
2. To disprove associativity, take a=5, b=6 and c=7. Notice that a,b and c are in G. We will show that a*(b*c) is not (a*b)*c  
3. To disprove associativity, let a, b, c in G be arbitrary. We will show that a*(b*c) is not (a*b)*c  
4. To disprove associativity, we will show that for any a, b, c not in G, a*(b*c) is not (a*b)*c  

Question 5  (3 points)

How could the first line of a proof of inverses (#4) in a group read?

1. To prove that #4 holds, we will look at every element of G by cases. For each element a, we will produce ainverse in G s.t. a*ainverse=1.  
2. To prove that #4 holds, take a=5. Then ainverse is 1/5.  
3. To prove that #4 holds, we will produce a in G and solve for ainverse to show that a*ainverse=1.  
4. To prove that #4 holds, let a in G be arbitrary. We will show that a*ainverse=1 for all ainverse in G..