# MATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN

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1 NAME: MATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN 1. INSTRUCTIONS (1) Timing: You have 80 minutes for this midterm. (2) Partial Credit will be awarded. Please show your work and provide full solutions, NOT just answers. You may use results from class or the text, but they must be quoted. Cross out anything that the grader should ignore. (3) There are 4 questions in total, with the following scoring breakdown: Question 1 Question 2 Question 3 Question 4 Total Points (4) Good Luck! Date: November 26,

2 2 INSTRUCTOR: HAROLD SULTAN 2. QUESTIONS (1) Provide a definition of when two groups are isomorphic [2pts]. Are the groups Z 9 Z 4 and Z 6 Z 6 isomorphic [3pts]? justify your answer. ANSWER: Two groups G, G are isomorphic if there exists a bijective map φ : G G such that g, h G we have φ(gh) = φ(g)φ(h). Equivalently two groups are isomorphic if there exists a one to one and onto homomorphism from one group to the other. The groups Z 9 Z 4 and Z 6 Z 6 are NOT isomorphic. The element (1, 0) in Z 9 Z 4 is of order 9, while no element in Z 6 Z 6 has order greater than 6. To see this note that for any (a, b) Z 6 Z 6, (a, b) + (a, b) + (a, b) + (a, b) + (a, b) + (a, b) = (6a, 6b) = (0, 0) Z 6 Z 6

3 MATH 28A MIDTERM 2 3 (2) Define the kernel of a homomorphism [2pts]? Given a homomorphism φ : Z 4 Z 2 Z 6 defined by φ(j) = (0, 3j (mod 6)) Compute φ(3) [1pt]? Is φ injective [2pt]? ANSWER: Given a group homomorphism φ : G G, the kernel of φ is defined by ker(φ) = {g G φ(g) = e }. That is, the kernel consists of all elements in the domain that are sent by φ to the identity element in the range. φ(3) = (0, 3 3 mod (6)) = (0, 3) ker(φ) = {0, 2}. Since the kernel is non-trivial, φ is not injective.

4 4 INSTRUCTOR: HAROLD SULTAN (3) Consider the group G = Z 4 Z 6 and the subgroup H =< (2, 2) >. List the elements in the subgroup H [1pt]? Is the subgroup H normal, justify your answer [1pt]? What is the size of the factor group G/H [1pt]? What group is the factor group G/H isomorphic to [2pts]? ANSWER: The subgroup H contains the elements {(2, 2), (0, 4), (2, 0), (0, 2), (2, 4), (0, 0)} The subgroup H is normal because G is abelian and any subgroup of an abelian group is normal. The size of the factor group G/H is G / H = 24/6 = 4 The factor group G/H is of order 4 and is abelian (because a factor group of an abelian group is abelian). Hence there are two options for what the group G/H is isomorphic to as there are 2 abelian groups of order four up to isomorphism. Namely Z 4 and Z 2 Z 2. In order to differentiate between these two groups we must determine if G/H contains any elements of order 4 or not. If G/H contains an element of order 4 then G/H is isomorphic to Z 4. On the other hand, if G/H does not contain any elements of order 4 then it is isomorphic to Z 2 Z 2. The elements in G/H are the following four cosets H = {(2, 2), (0, 4), (2, 0), (0, 2), (2, 4), (0, 0)} (1, 0) + H = {(3, 2), (1, 4), (3, 0), (1, 2), (3, 4), (1, 0)} (0, 1) + H = {(2, 3), (0, 5), (2, 1), (0, 3), (2, 5), (0, 1)} (1, 1) + H = {(3, 3), (1, 5), (3, 1), (1, 3), (3, 5), (1, 1)} The element H in G/H has order 1, while the three elements (1, 0) + H; (0, 1) + H; (1, 1) + H all have order 2 in G/H. By the above, the group G/H is isomorphic to Z 2 Z 2.

5 MATH 28A MIDTERM 2 5 (4) State Lagrange s Theorem relating the size of a subgroup H to the size of a finite group G. [2pt] Define the center of a group [1pts]. Let G be a mystery group of order pq where p, q are any prime numbers. Assume that the center of G, Z(G), contains at least 2 distinct elements. Prove that G must be abelian [2pts]. [Hint: What is are possible sizes of Z(G)? Then use the fact that if a factor group G/Z is cyclic, then G is abelian. [1pt] extra credit if you can prove this fact.] ANSWER Lagrange s theorem states that if H is a subgroup of a finite group G, then H is a divisor of G. The center of a group G is the set of all elements in G which commute with all other elements in G. Equivalently, Z(G) = {g G gh = hg h G}. Since Z(G) is a subgroup of G, by lagrange s theorem the order of Z(G) must divide the order of G. Since G = pq, it follows that the possibilities for Z(G) are 1, p, q, pq. Since by assumption Z(G) contains at least two elements, the possibilities for Z(G) are p, q, pq. Since the center is always a normal subgroup, we can consider the factor group G/Z(G). The size of the factor group G/Z(G) is equal to pq/{p, q, pq} or equivalently, the possibilities for the size of the factor group are 1, p, q depending on whether the size of the center is pq, q, p, respectively. Since there is only one group of any prime order (namely the cyclic group of that order), the factor group G/Z(G) must be isomorphic to the trivial group, Z p or Z q, respectively. Since all three options for G/Z(G) all cyclic the proof is completed by the fact that if a factor group G/Z is cyclic, then G is abelian.

6 6 INSTRUCTOR: HAROLD SULTAN address:

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