Exam 1 Solutions. Solution: The 16 contributes 5 to the total and contributes 2. All totaled, there are 5 ˆ 2 10 abelian groups.

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1 Math 5372 Spring 2014 Exam 1 Solutions 1. (15 points) How many abelian groups are there of order: (a) 16 For any prime p, there are as many groups of order p k as there are partitions of k. The number of partitions of k is the number of ways to write k as a non-increasing sum of positive integers. In this case, k 4 and we have 4 = 4, 4 = 3 + 1, 4 = 2 + 2, 4 = , 4 = Thus, there are five abelian groups of order 16 (and five abelian groups of order p 4 for any prime p.) (b) ˆ 25 The 16 contributes 5 to the total and contributes 2. All totaled, there are 5 ˆ 2 10 abelian groups. (c) ˆ 9 ˆ 625 Since 9 3 2, and , we know there will be 2 ˆ 5ˆ the number of partitions of 6 groups. Here are the partitions of 6: 6, 5 + 1, 4 + 2, , 3 + 3, , , , , , All totaled, there are 11 of them so the total number of groups is 11 ˆ (10 points) If G is a group with normal subgroup N and rg : Ns k, show that g k P N for every g P G. Hint: Make use of the factor group G{N. We make use of the fact that the order of an element always divides the order of the group, in the form: If H has order n, then h n e for any h P H. In this particular case, H G{N, a group of order k. Given any g P G, gn will be an element of H so pgnq k will be the identity, N, in H. That is, N pgnq k g k N. This means that g k must be in N.

2 Page 2 3. (15 points) If G and H are groups, prove that G teu is normal in G H and that pg Hq{pG teuq H. Hint: Both parts are done easiest by using the right homomorphism φ : G H Ñ H. The right homomorphism is φpg, hq h, of course. This is obviously an onto map from G H to H. It actually IS a homomorphism: φppg 1, h 1 qpg 2, h 2 qq φpg 1 g 2, h 1 h 2 q h 1 h 2 φpg 1, h 1 qφpg 1, h 1 q. The kernel of this map is the set of all pg, hq that map to the identity. That is, we need h e so the kernel consists of all ordered pairs pg, eq where g P G. That is, kerpφq G teu. Since kernels of homomorphisms are normal, G teu is normal in G H. Finally, since φ is onto, by the First Isomorphism Theorem, G H{kerpφq H, or pg Hq{pG teuq H. 4. (30 points) Define a homomorphism α : Z Ñ Z 40 by αpxq 5x mod 40. (a) Find αpzq. Write your answer in the form ă n ą for some integer n. αpzq t0, 5, 10, 15, 20, 25, 30, 35u ă 5 ą. (b) Find Kerpαq. Write your answer in the form ă n ą for some integer n. We need all k in Z for which αpkq 0 in Z 40. That is, we need k with 5k 0 ` 40j for some integer j. That is, we need k 8j so Kerpαq ă 8 ą. (c) If H ă 6 ą, a subgroup of Z, find αphq. Elements of H all have the form 6n for some integer n, so all elements in αphq have the form 30n modulo 40. That is, αphq t0, 30, 20, 10u ă 10 ą. (d) If K ă 4 ą t0, 4, 8, 12, 16, 20, 24, 28, 32, 36u, a subgroup of Z 40, find α 1 pkq. The formula we learned in the book was that if αpxq y, then α 1 pyq x Kerpαq. On the other hand, if y is not in the range of α then α 1 pyq is empty. For a set like K, α 1 pkq is the set of all things in Z that map into K which is the union of α 1 pkq over all k P K. This gives the union of a bunch of empty sets along with the inverse images of the elements of K that are in the range. This means that α 1 pkq α 1 p0q Y α 1 p20q Kerpαq Y 4 Kerpαq ă 4 ą.

3 5. (20 points) Let G Z 400 Z 500 Z 600. Page 3 (a) Express G as a direct product of cyclic groups of prime power order. Put your answer in the standard order: powers of 2 first, in increasing order, then increasing powers of 3, etc. We have G Z 400 Z 500 Z 600 Z 16 Z 25 Z 4 Z 125 Z 8 Z 3 Z 25 Z 4 Z 8 Z 16 Z 3 Z 25 Z 25 Z 125 (b) Express G as a direct product of the form Z m1 Z m2 Z mk m 1 m 2,..., m k 1 m k. where G Z 4 Z 8 Z 16 Z 3 Z 25 Z 25 Z 125 pz 16 Z 3 Z 125 q pz 8 Z 25 q pz 4 Z 25 q Z 6000 Z 200 Z 100 Z 100 Z 200 Z (10 points) Find the Smith-Normal Form for each of the following matrices. Possible hint: You can use facts I mentioned in class to avoid some calculations (a) A By problem 5(b) we know the invariant factors are 100, 200, 6000 so we have A

4 (b) B ˆ Page 4 ˆ ˆ B (twice top row subtracted from bottom row) ˆ 0 80 (five times bottom added to top) ˆ 0 80 (three times first column subtracted from second) 10 0 ˆ10 0 (interchange and negate rows) 0 80 Problems 7-10 are purely extra credit. You can work on them if you have time. 7. (8 points) Related to problem 6, in each case, find the matrices P and Q so that P MQ is in Smith-Normal Form. (Obviously (b) is much easier than (a)!) (a) A

5 Page 5 (b) B If we interchange the top two rows and first two columns of the main matrix, we get P MQ with P and Q ˆ ˆ10 0 ˆ2 1 ˆ1 3 so P MQ with P and Q. The P and Q are not unique for either part a or part b. 8. (2 points) Related to problem 4, identify Z{Kerpαq as some well-known group. This is, of course, too easy. We had Kerpαq ă 8 ą 8Z so Z{Kerpαq Z{8Z Z 8.

6 Page 6 9. (4 points) If G is a group with normal subgroup H, if H is cyclic, prove that every subgroup of H is also normal in G. We have that H ă a ą is normal in G. Thus, given any g P G, gag 1 P H or gag 1 a k for some integer k. Now let N be a subgroup of H. By the Fundamental Theorem of Cyclic groups, N ă a m ą for some integer m. Now every element of N has the form a mj, and given g P G, ga mj g 1 pgag 1 q mj pa k q mj a kmj pa m q kj, so ga mj g 1 P N, or gng 1 Ď N. 10. (6 points) Suppose that H ă v 1, v 2, v 3 ą is a subgroup of Z 3 and A pv 1 v 2 v 3 q is the matrix with the v s as columns. If A is invertible, show that Z 3 {H has size equal to detpaq, the absolute value of the determinant of A. What happens if A is not invertible? The key here is that if P AQ B where B is in Smith-Normal form, then detpp q 1 and detpqq 1. Consequently, detpbq detpaq, the determinant of 0 0 B always being nonnegative. Next, we know that B will have the form n1 0 n n 3 which tells us that detpbq n 1 n 2 n 3. Also, Z 3 {H Z 3 {pn 1 Z n 2 Z n 3 Zq Z n1 Z n2 Z n3, and this group also has size n 1 n 2 n 3. If A is not invertible, then the Smith-Normal Form for A will have one of the following forms: n 1 0 or or even (if A 0.) In these cases, 0 0 n n we would have Z 3 {H Z Z n1 Z n2, or Z 2 Z n1, or Z 3, respectively. In each case, this group would be infinite.

Exam 2 Solutions. In class questions

Exam 2 Solutions. In class questions Math 5330 Spring 2018 Exam 2 Solutions In class questions 1. (15 points) Solve the following congruences. Put your answer in the form of a congruence. I usually find it easier to go from largest to smallest