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2 Solution. Since it is possible to have at most two disjoint 3-cycles in S 7, we conclude from part (a) that if σ S 7 has σ 3 then σ (a b c) or σ (a b c)(d e f) where a, b, c, d, e, f are distinct elements of {1, 2, 3, 4, 5, 6, 7}. The number of distinct 3-cycles in S 7 is ; 3 and the number of distinct products of two disjoint 3-cycles in S 7 is 1 2 ( )( ) Thus the number of elements of order 3 in S 7 is [12 Points] Let G, N, and K be the subgroups of GL(2, Q) given by G {[ ] a b : a, b, d Q; ad 0 }, N {[ ] 1 x : x Q }, and K {[ ] 1 x : x Z }. 0 d (The operation is matrix multiplication.) (a) Prove that N is a normal subgroup of G. (You do not need to prove that N is a subgroup of G. Just prove that it satisfies the condition to be normal.) Solution. The map ϕ : G Q Q defined by ([ ]) a b ϕ (a, d) 0 d is a group homomorphism and N Ker(ϕ). Hence N is a normal subgroup of G. (b) Prove that K is a normal subgroup of N. (Again you do not need to prove that K is a subgroup of N.) Solution. Since [ ][ ] 1 x 1 y [ ] 1 x + y [ 1 y ][ ] 1 x, it follows that N is an abelian group, and hence every subgroup, including K, is normal. (c) Prove that normality of subgroups is not transitive by showing that K is not a normal subgroup of G. [ ] [ ] Solution. If A K and B G, then 0 2 [ ][ ][ ] BAB /2 0 2 /2 [ ] 1 1/2 / K. Hence K is not normal in G. Math 4201 May 9,

3 4. [16 Points] Make a list of abelian groups of order Your list should contain exactly one group from each isomorphism class of abelian groups of order 72. That is, if G is an abelian group of order 72, then G is isomorphic to exactly one group one your list. Be sure to explain how you arrived at your list. Here are two groups of order 72. Identify them on your list. (a) G 1 Z 6 Z 12 (b) G 2 Z 36 Z 2 Solution. We will use elementary divisors to write our list of groups of order 72. The order 72 can be written as a product of prime powers in exactly 6 ways: Consequently every abelian group of order 72 must be isomorphic to one of the following groups 1. Z 3 2 Z Z 3 2 Z 2 Z Z 3 2 Z 2 Z 2 Z 2 4. Z 3 Z 3 Z Z 3 Z 3 Z 2 Z Z 3 Z 3 Z 2 Z 2 Z 2 Since G 1 Z 6 Z 12 Z3 Z 3 Z 3 Z 4 Z3 Z 3 Z 2 Z 2 2, we see that G 1 is isomorphic to number 5 on the above list, while G 2 Z 36 Z 2 Z9 Z 4 Z 2 Z3 2 Z 2 Z 2 2, so that G 2 is isomorphic to group number 2 on the list. 5. [15 Points] Let K be a Galois extension of a field F such that Gal F (K) Z 2 S 3. How many intermediate fields L are there such that (a) [L : F] 4, (b) [L : F] 6, (c) Gal L (K) Z 4? Solution. By the Fundamental Theorem of Galois Theory, there is a one-to-one correspondence between fields L intermediate between F and K and subgroups of the Galois group Gal F K. The correspondence is given by L Gal L K and the indexes are related by [L : F] [Gal F K : Gal L K]. We use this correspondence and knowledge of the subgroup structure of Gal F (K) Z 2 S 3 to answer questions (a), (b), and (c). (a) To answer this question we need to count the number of subgroups of Z 2 S 3 of index 4. That is, since Z 2 S 3 12, we want subgroups of order 3. Since 3 is prime, a subgroup of order 3 is just the cyclic subgroup generated by an element of order 3. In the direct product Z 2 S 3, the order of an element (a, b) is given by (a, b) lcm { a, b }. This is equal to 3 if and only if a 1 (since a Z 2 ) and b 3. The only elements of S 3 of order 3 are the 3-cycles. Since there are exactly two 3-cycles in S 3 and they both generate the same subgroup (1 2 3), it follows that there is exactly one subgroup H of Z 2 S 3 of order 3, namely H {0} (1 2 3). Hence there is exactly one field L between F and K with [L : F] 4. Math 4201 May 9,

4 (b) [L : F] 6 [Gal F K : Gal L K] 6 Gal L K 2. Thus we need to look for subgroups H of Z 2 S 3 of order 2, so that H (a, b) where a 1 or 2, and b 1 or 2 and we must have (a, b) not equal to the identity element. Since a Z 2 there are two choices for a (namely 0 and 1), and since b S 3 there are 4 choices for b (namely (1), (1 2), (1 3), and (2 3)). Hence there are subgroups of Z 2 S 3 of order 2, so there are 7 intermediate fields with [L : F] 6. (c) There are no such fields L since there are no elements of the Galois group Z 2 S 3 of order 4. Reason: The elements of S 3 have orders 1, 2, or 3 and the elements of Z 2 have orders 1 and 2. It is not possible to get 4 as the least common multiple of any of these numbers. 6. [12 Points] Let F Z 3 be the finite field with 3 elements and let f(x) x 2 +2x+2 F[x]. Let u be a root of f(x) in some extension field of F and let L F(u). (a) Show that f(x) is irreducible in F[x]. Solution. Since deg f(x) 3 it is only necessary to check that f(x) has no roots in F Z 3. But f(0) 2, f(1) 2, and f(2) f( 1) 1, so f(x) has no roots in Z 3, and hence is irreducible in F[x]. (b) Find all the roots of f(x) in L. Solution. By hypothesis u is one root of f(x) in L. Find the other by the remainder theorem: Dividing f(x) by x u gives f(x) (x u)(x + u + 2) so the second root is u 2 u + 1 2u + 1. (c) How many elements does L have? Solution. [L : Z 3 ] [F(u) : F] 2 so L is a 2-dimensional vector space over the base field F Z 3. Hence L (d) Find the order of u as an element of the multiplicative group L of non-zero elements of L. Solution. Since f(u) u 2 + 2u it follows that u 2 2u 2 u + 1 L. Then u 4 (u + 1) 2 u 2 + 2u so that u 8 1. Hence the order of u divides 8 and u 4 1 so we conclude that u [25 Points] Let f(x) x 4 7x Q[x]. (a) Find the factorization of f(x) into irreducible polynomials in Q[x]. Solution. f(x) x 4 49 (x 2 2)(x 2 5). (b) Find all the roots of f(x) in C. Solution. The roots are ± 2 and ± 5. (c) Find the splitting field E of f(x) over Q. Math 4201 May 9,

5 Solution. E Q( 2, 5). (d) Find [E : Q], and give a linear basis of E as a vector space over Q. Solution. By the tower theorem [E : Q] [E : Q( 2)][Q( 2) : Q] and the proof of the tower theorem gives the following as a basis of E over Q: { B 1, 2, 5, } 10. (e) Find the Galois group Gal Q E, and for each σ Gal Q E, give the action of σ on the basis found in Part (d). Solution. If σ Gal Q E then σ( 5) ± 5 and σ( 2) ± 2 since σ must take a root of an irreducible polynomial over Q to another root of the same polynomial, and since x 2 5 and x 2 2 are irreducible over Q. Since these data give 4 possibilities for the action of σ on the two generators of E over Q and since Gal Q E [E : Q] 4 this accounts for all the elements of Gal Q E, which we will denote by Gal Q E {id, α, β, γ}. The action of the elements on all the basis elements of E over Q are given in the following table: id α β γ and the multiplication table of Gal Q E is: id α β γ id id α β γ α α id γ β β β γ id α γ γ β α id from which we conclude that Gal F E Z 2 Z 2. (f) Find all subgroups H of Gal Q E and draw the lattice of subgroups. Solution. The only proper subgroups are the cyclic subgroups generated by a nonidentity element. Thus the subgroups of Gal Q E are: id, α, β, γ, and Gal Q E. The lattice of inclusion relations is: Gal Q E α β γ id Math 4201 May 9,

6 (g) For each subgroup H, find the fixed field E H, and draw the lattice of subfields of E. Solution. The fixed fields for each subgroup can be read off from the table giving the action of each automorphism of E on the basis elements. Thus, E id E, E α Q( 2), E β Q( 5), E γ Q( 10), and E GalQ E Q, and the lattice of subfields is: E Q( 2) Q( 5) Q( 10) Q Math 4201 May 9,

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