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 --> Choose Match abcd 2 --> Choose Match abcd 3 --> Choose Match abcd 4 --> Choose Match abcd

### 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..