Math 120: Homework 6 Solutions


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1 Math 120: Homewor 6 Solutions November 18, 2018 Problem 4.4 # 2. Prove that if G is an abelian group of order pq, where p and q are distinct primes then G is cyclic. Solution. By Cauchy s theorem, G has elements x and y of order p and q respectively. Let z = xy. We will show that z generates G. First note that z q = x q y q = x q. Since x has order p and p q, x q has order p. Similarly z p has order q. The order of z must therefore be a multiple of both p and q, in other words, a multiple of pq. By Lagrange s theorem, the order of z divides G = pq, so pq is exacctly the order of z. Thus z is a generator of G and G is cyclic. Problem 4.4 # 13. Let G be a group of order 203. Prove that if G has a normal subgroup H of order 7 then H Z(G). Solution. We have 203 = We are assuming that H has order 7 and is normal. We then have a homomorphism φ : G Aut(H) which is the action by conjugation. In other words, φ(g) is the automorphism c g Aut(H) defined by c g (x) = gxg 1. Now Aut(H) has order 6 by Proposition 16 on page 135 of Dummit and Foote. Therefore the image of φ is a subgroup of an order 6 that is isomorphic to G/ er(φ); so its order divides both 6 and 203. Since 6 and 203 are coprime, this means that φ is the trivial map, so c g is the identity automorphism of H for all g. That is, gxg 1 = c g (x) = x for all x H and g G. Therefore H is contained in the center of G. Problem 4.5 # 13. Prove that a group of order 56 has a normal psylow subgroup for some prime p dividing 56. Solution. Suppose that G = 56. The 7Sylow has either 1 or 8 conjugates, since the number of 7Sylows is 1 mod 7 and divides 56. Thus either 1
2 the 7Sylow is normal or it has 8 conjugates P 1,, P 8. Each P i contains 6 elements of order 7, and these are all distinct. So G has 8 6 = 48 elements of order 7. Now let Q be a 2Sylow, so Q = 8. There are precisely 8 elements that are not of order 7, so Q = {g G g does not have order 7}. From this we see that the elements of Q are permuted by conjugation, so hqh 1 = Q for all h, and Q is normal. Problem 4.5 # 25. Prove that if G is a group of order 385 then Z(G) contains a 7Sylow subgroup and an 11Sylow subgroup is normal in G. Solution. Since 385 = , the number of 7Sylows divides 55 and is 1 mod 7; therefore the 7Sylow P is normal. Also the number of 11 Sylows divides 35 and is 1 mod 11, so the 11Sylow is also normal. But we have to show that the 7Sylow is central. This is somewhat similar to We have a homomorphism θ : G Aut(P ) in which θ(g) is conjugation by P. The image is a subgroup of Aut(P ), which has order 6, which is isomorphic to G/ er(θ); hence it has order dividing both 6 and 386. Since these are coprime, θ is trivial, meaning that θ(g) = 1 P for all P. Thus if x P we have gxg 1 = θ(g)x = x, and so P is central. Problem 7.3 # 25. Assume R is a commutative ring with 1. Prove that the binomial theorem n ( ) n (a + b) n = a b n (1) =0 holds in R, where the binomial coefficient ( n ) is interpreted as of the identity of R taen ( n ) times. Solution. The identity ( ) ( ) ( ) n n 1 n 1 = + (2) 1 is true in Z, so applying the characteristic homomorphism (see next problem) it is true in R. We interpret ( ) n 1 a = 0 if a < 0 or a > n 1, and then the identity is true even in the exceptional cases = 0 or = n. (In both thse case, both sides of the identity are 1.) 2
3 Now (1) is clear if n = 1; if n > 1 then by induction n 1 ( ) n 1 (a + b) n 1 = a b n 1 =0 n 1 ( ) n 1 n 1 ( ) n 1 (a+b) n = a(a+b) n 1 +b(a+b) n 1 = a +1 b n 1 + a b n. =0 The coefficient of a m b n m has two contributions. From the first sum with = m 1, there is a coefficient of ( n 1 m 1), which we can omit if m = 0; and from the second term with = m, there is a contribution of ( ) n 1 m. Collecting the terms involving a m b n m together and using (2), the coefficient of a m b n m is ( n m) and we obtain (1). Problem 7.3 # 26. The characteristic of a ring is the smallest positive integer such that = 0 (n times) in R. If no such positive integer exists then characteristic of R is said to be zero. For example Z/nZ is a ring of characteristic n and Z is a ring of characteristic zero. (a) Prove that the map Z R defined by = ( times) if > 0, 0 if = 0, ( ) ( times) if < 0 is a ring homomorphism whose ernel is nz where n is the characteristic of R. (b) Determine the characteristics of the rings Q, Z[x] and Z/nZ[x]. (c) Prove that if p is a prime of R and if R is a commutative ring of characteristic p then (a + b) p = a p + b p for all a, b R. Solution. First we consider (a). Lemma 1 Let G be any group and let x G. Then there is a unique group homomorphism ϕ : Z G such that ϕ(1) = x. Proof The subgroup x of G generated by x is cyclic so by Theorem(1) 4 on page 56 of Dummit and Foote it is either isomorphic to Z/nZ or to Z, and furthermore in this isomorphism the chosen generator 1 of Z (written additively) can be made to go to the chosen generator x of x. We can compose this isomorphism with the natural map Z Z/nZ, or with the 3 (3)
4 identity map Z Z, to obtain a homomorphism Z x such that 1 x; and then we may compose this with the inclusion map x G to obtain the homomorphism ϕ. We apply this Lemma with G being the additive group of R and we obtain a homomorphism c : Z R of additive groups such that c(1) = 1. We claim that c is given by (3). Indeed, if > 0 then c() = c( ) = c(1) + + c(1) = ; if < 0 then c() = c( ) and since > 0 what we have already shown gives c( ) = ( times) and so the case < 0 is proved, while c(0) = 0 is trival. To show that c is a ring homomorphism, we first prove that c(ab) = c(a)c(b) if a, b > 0. Indeed, c(ab) = (ab times) while c(a)c(b) is (a times) times (b times). Multiplying these out using the distributive law and 1 1 = 1 gives c(ab) = c(a)c(b). We have to prove other cases of c(ab) = c(a)c(b). If either a or b = 0, both sides are 0; if a < 0 and b > 0 then c(ab) = c(( a)b) = c( a)c(b) = c(a)c(b) where we have used the fact that a > 0; and the remaining cases are similar. This proves (a). (b) It is clear that the characteristics of Q, Z[x] and Z/nZ[x] are 0, 0 and n, repectively. (c) We apply the characteristic map c to the binomial theorem in the form p 1 ( ) p (a + b) p = a p + b p + a b p. Each integer ( p ) is a multiple of p since it equals p(p 1) (p + 1), so it is in the ernel of c and may be discarded. (a) If G is any group (written additively) and x is an element, there is a homomorphism (c) Prove that if p is a prime of R and if R is a commutative ring of characteristic p then (a + b) p = a p + b p for all a, b R. Let c denote this characteristic homomorphism. First note that c is a homomorphism from the additive group of Z to the additive group of R. Indeed, Problem 7.3 # 34. Let I and J be ideals of R. (a) Prove that I + J is the smallest ideal of R containing both I and J. (b) Prove that IJ is an ideal contained in I J. (c) Give an example where IJ I J. 4 =1
5 (d) Prove that if R is commutative and I + J = R then IJ = I J. Solution. (a) First note that I + J is an ideal that obviously contains both I and J. If K is an ideal that contains both I and J, and if x I, y J then x, y K and so x + y K. This proves that I + J K. Therefore I + J is the smallest ideal containing both I and J. (b) Recall that IJ is defined to be the set of finite sums N i=1 x iy i where x i I and y i J. By construction it is closed under addition, and it is also true that RIJ IJ since RI I and JR J. Thus IJ is an ideal. Moreover IJ IR I and IJ RJ J since I, J are ideals. Therefore IJ I J. (c) Let R = Z and let I = J = (p) where p is a prime. Then IJ = (p 2 ) while I J = (p). They are not the same. (d) By (b) we have IJ I J, so we have to prove that I J IJ. Thus let a I J; we will prove that a IJ. We are assuming that I + J = R, so we may write 1 = x + y with x I and y J. Then a = a 1 = ax + by. Now ax = xa IJ since x I and a J. Similarly ay IJ and so a = ax + by IJ. Problem 7.4 # 4. Assume that R is commutative. Prove that R is a field if and only if 0 is a maximal ideal. Solution. Suppose that R is a field. To show that 0 is maximal, we must show that if I is an ideal of R then either I = 0 or I = R. If I 0 let x be an nonzero element of R. Then x is a unit since R is a field so R = Rx RI = I proving that I = R. Thus 0 is maximal. Conversely, suppose that 0 is a maximal ideal. We must show that R is a field. Let 0 x R. We will show that x is a unit. The ideal Rx is nonzero, and since 0 is maximal, we must have Rx = R. Therefore 1 Rx proving that 1 = yx for some y R and therefore x is a unit. This proves that R is a field. Problem 7.4 # 5. Prove that if M is an ideal such that R/M is a field then M is a maximal ideal. Solution. This follows from the previous problem applying the Lattice Isomorphism Theorem (Problem 3.3 #2 from HW4). Since the ideals of R/M are in bijection with the ideals of R that contain M, and the 0 ideal of R/M 5
6 corresponds to M, the Lattice Isomorphism Theorem implies that M is maximal in R if and only if 0 is maximal in R/M, which is true if and only if R/M is a field. Problem 7.5 # 3. Let F be a field. Prove that F contains a unique smallest subfield F 0 and that F 0 is isomorphic to either Q or Z/pZ for some prime p. Solution. In Exercise 7.3 #26 we constructed a homomorphism ϕ : Z F such that ϕ(1) = 1. Let p be the ernel of ϕ. Since ϕ(z) is a subring of a field, it is an integral domain. By the first isomorphism theorem, ϕ(z) = Z/p, and therefore p is a prime ideal. The prime ideals of Z are (0), and (p) where p is a prime integer. There are thus 2 cases. First, suppose that p = 0. Then ϕ is injective, by Corollary 16 on page 263 of Dummit and Foote, the smallest field F 0 of F that contains ϕ(z) = Z is isomorphic to the field of fractions Q of Z. Any subfield of F contains 1, hence the image of ϕ, and so F 0 is the smallest subfield of F. If p = (p), then ϕ(z) = Z/(p) is already a field, and it is a subfield of F. This is the field F 0 in this case. Since any subfield of F contains 1, it contains ϕ(z), and so F 0 is the smallest subfield of F. 6
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