Math 236H May 6, 2008 Be sure to write your name on your bluebook. Use a separate page (or pages) for each problem. Show all of your work. 1. (15 points) Prove that the symmetric group S 4 is generated by the three transpositions, and. Solution: We know that S 4 is generated by 6 transpositions: the three listed above along with (13), (14) and (24). Thus we need to show that each of these three is a product of the three listed above. Let the four letters being permuted by A, B, C and D. Then we have (A, B, C, D) (A, B, C, D) (A, B, C, D) (B, A, C, D) (B, C, A, D) (A, C, B, D) (A, C, D, B) (B, A, C, D) (B, C, A, D) (C, B, A, D) (A, D, C, B) (B, C, D, A) (B, D, C, A) (D, B, C, A) These three columns show that (13), (24) and (14) are each products of the three generators. 2. (5 points) Eisenstein s criterion says that the polynomial a(x) = x n + a 1 x n 1 + a 2 x n 2... + a n Z[x] is irreducible (over Z) if (i) each coefficient a i is divisible by a prime p and (ii) a n is not divisible by p 2. Give an example of a polynomial satisfying (i) but not (ii) that is reducible. Solution: One example is (x + p) 2 = x 2 + 2px + p 2. Page 1 of 5
3. (15 points) Let p be an odd prime, and let Prove that f(x) is an integer whenever x is. f(x) = x(x(p 1)/2 + 1)(x (p 1)/2 + p 1). 2p Solution: This amounts to showing that the numerator is always even and always divisible by p. Its two factors besides x differ by p 2, which is odd, so one of them must be even, so their product is divisible by 2. For divisbility by p, note that the numerator is x(x (p 1)/2 + 1)(x (p 1)/2 + p 1) = x p + px (p+1)/2) + (p 1)x x p x mod p, and x p x is divisible by p by Fermat s Little Theorem. 4. (10 points) Find a symmetric group S n with an element of order greater than 2n and identify the element. Solution: S 9 has en element of order 20, namely (12345)(6789). There are many other examples. 5. (10 points) Let G be a group of order pq where p and q are disinct primes. Show that every proper subgroup of G is cyclic. Solution: If H is a proper subgroup of G, its order must be less than pq and divide pq, so it must be 1, p or q. Hernce H is either trivial or of prime order, so it is cyclic. 6. (15 points) List the even permutations of order 2 in S 4 and say how many there are in S 5 and S 6. Solution: The only even permutations of order 2 in S 4 are the three double transpositions,, (13)(24) and (14). Double transpositions are also the only such permutations in S 5 and S 6. (There are triple transpositions in S 6 which have order 2, but they are odd.) In S 5 there are three for each of the five subsets with 4 elements, making 15 in all. In S 6 there are three for each of the fifteen subsets with 4 elements, making 45 in all. 7. (10 points) You are making necklaces with 7 beads each, and each bead can be any one of n colors. Use Burnside s formula to find the number r(n) of distinct necklaces. Find the smallest n such that this number exceeds 1000. Page 2 of 5
Solution: The relevant group here is the dihedral group D 14 and the set X of all possible necklaces has n 7 elements. The number of orbits r(n) is g D 14 X g /14. The find r, note that The identity element fixes all n 7 necklaces. Each of the six nontrivial rotations fixes the n necklaces in which all beads have the same color. Each of the seven reflections fixes one bead and interchanges three pairs of beads, so it fixes n 4 necklaces. Hence the number of orbits is r(n) = n7 + 7n 4 + 6n 14 = n(n3 + 1)(n 3 + 6) 14 so r(2) = 2 9 14/14 = 18, r(3) = 3 28 33/14 = 198 and r(4) = 4 65 70/14 = 1300. Note that r(n) = f(n), where f is the function of Problem 3 for p = 7. f(n) is the number of necklaces for an odd prime number p of beads and n colors. 8. (10 points) Describe the commutator subgroup C(G) of every simple a. abelian group b. nonabelian group Solution: a. In an abelian group, all commuators are trival, so C(G) is trivial b. The commutator subgroup is normal, so if G is simple, C(G) is either G or {e}. It cannot be the latter since the G is nonabelain, so C(G) = G 9. (10 points) Prove that the intersection of two normal subgroups of G is a normal subgroup. Solution: Let H 1 and H 2 be normal subgroups of G. Let g G and h H 1 H 2. Then ghg 1 H 1 since h H 1, and similarly ghg 1 H 2. Hence ghg 1 H 1 H 2, so H 1 H 2 is normal. 10. (10 points) Determine the number of elements of order 4 in a. C 8 C 8 b. C 4 C 4 C 4 Page 3 of 5
Solution: a. All elements of order 4 or less are in the subgroup C 4 C 4, which has order 16, and all elements of order 2 or less are in the subgroup C 2 C 2, which has order 4, so the number of lements of order 4 is 16 4 = 12. b. An element of C 4 C 4 C 4 has order 4 if it is not contained in C 2 C 2 C 2. The number of such elements is 4 3 2 3 = 56. 11. (10 points) Let n be an integer such that 6n 1 and 6n + 1 are both primes. (Such primes are called twin primes.) Show that every group G of order 36n 2 1 is cyclic. Solution: Let p = 6n 1 and q = 6n + 1, so G = pq. We will use the third Sylow theorem to show that there is exactly one subgroup of order p and one of order q. The number of p-sylow subgroups divides pq (so it is 1, p, q or pq) and is congruent to 1 modulo p. The only number that fits this description is 1. Hence the p-sylow subgroup is normal, so there is a homomorphism φ 1 from G onto the quotient group, which must be C q. Similarly, there is only one q-sylow subgroups and hence a homomorphism φ 2 to C p. Using these we get a homomorphism to C p C q = C pq which is onto and therefore an isomorphism. 12. (15 points) Prove that every finite integral domain is a field. Solution: See Theorem 19.11 on page 181 of Fraleigh. 13. (10 points) Describe a nonabelian group of order 55. Hint: Consider the group of 2 2 matrices (under multiplication) of integers modulo 11 of the form [ ] a b M = with a 0. Solution: The set of matrices of this form is a group under multiplication since [ ] [ ] [ ] a b a b aa ab = + b and [ a b ] 1 = [ a 1 a 1 b ] There are 10 possible values of b and 11 possible values of b, so there are 110 matrices of this form. We to find a 5-element subset of a-values that is closed under multiplication, so we need an element whose 5th power is 1. One such element is 3, so the subset is {3, 9, 5, 4, 1}. Hence our subgroup consists of matrices M in which a is a power of 3. Page 4 of 5
14. (10 points) Prove that if n is an odd integer, then n 2 1 modulo 8 and n 4 1 modulo 16. Solution: Let n = 2k + 1. Then n 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 4k(k + 1) + 1 and k(k + 1) is even since it is the product of two consecutive integers. This means that n 2 is congruent to 1 modulo 8. Since n 2 = 8s + 1, we have We also have n 4 = (8s + 1) 2 = 64s 2 + 16s + 1 1 mod 16. 15. (15 points) For which integers r is f(x) = x 4 +rx 3 +x+3 reducible over the integers? Prove your answer. Solution: If f(x) is reducible, it is either the product of two quardatic factors, or of a linear factor and a cubic one. In the former case the product of the constant terms must by 3, so we have f(x) = (x 2 + ax + 1)(x 2 + bx + 3) or f(x) = (x 2 + ax 1)(x 2 + bx 3). In the first of these, we have f(x) = (x 2 + ax + 1)(x 2 + bx + 3) = x 4 + (a + b)x 3 + (4 + ab)x 2 + (3a + b)x + 3 so ab = 4 and 3a + b = 1. The only possibility here is (a, b) = ( 1, 4), so r = a + b = 3 and x 4 + 3x 3 + x + 3 = (x 3 + 1)(x + 3) = (x + 1)(x 2 x + 1)(x + 3). In the second case of quadratic factors we have f(x) = (x 2 + ax 1)(x 2 + bx 3) = x 4 + (a + b)x 3 + ( 4 + ab)x 2 (3a + b)x + 3 so ab = 4 and 3a + b = 1. These lead to 3a 2 + a + 4, which has no integral solution. Now suppose f(x) has a root. It must be ±1 or ±3. We have f(1) = 1 + r + 1 + 3 = 5 + r f( 1) = 1 r 1 + 3 = 3 r f(3) = 81 + 27r + 3 + 3 = 87 + 27r f( 3) = 81 27r 3 + 3 = 81 27r This gives us one new value of r, namely r = 5, and x 4 5x 3 + x + 3 = (x 1)(x 3 4x 2 4x 3) Hence f(x) is reducible only for r = 3 and r = 5. Page 5 of 5