# HOMEWORK 3 LOUIS-PHILIPPE THIBAULT

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1 HOMEWORK 3 LOUIS-PHILIPPE THIBAULT Problem 1 Let G be a group of order 56. We have that 56 = Then, using Sylow s theorem, we have that the only possibilities for the number of Sylow-p subgroups are: (1) n 2 (G) = 1 or 7; (2) n 7 (G) = 1 or 8. We will show that the case n 2 (G) = 7, n 7 (G) = 8 is impossible. Two different Sylow-7 subgroups intersect only in the identity, so none of the elements of order 7 in a given Sylow-7 subgroup is in another Sylow-7 subgroup. Also, all the Sylow-7 subgroups are conjugate, by Sylow s theorem, hence isomorphic. Then, if n 7 (G) = 8, we have that G has at least 8 6 = 48 elements of order 7. The remaining elements must form a Sylow-2 subgroup. So there are not enough elements of order 2 to form seven Sylow-2 subgroup, which is a contradiction. We have shown that n 2 (G) = 1 or n 7 (G) = 1. Suppose without lost of generality that n 2 (G) = 1. Then there is a unique Sylow-2 subgroup P 2. By the Sylow s theorem, every conjugate of P 2 is a Sylow-2 subgroups. So P 2 is equal to its conjugates. Hence P 2 is normal in G. Problem 2 Part 1. We have that G = (Z/5) 5 S 5. As a set G is the direct product of (Z/5) 5 and S 5, so G = (Z/5) 5 S 5 = = Part 2. A Sylow-5 subgroup of G has order 5 6. We claim that P = (Z/5) 5 (1, 2, 3, 4, 5) is a Sylow-5 subgroup of G. In fact, P = 5 6. Also, it is a subgroup of G, because it is closed under multiplication. Indeed, the multiplication is clearly closed in the first variable, since we have all of (Z/5) 5. It is also closed in the second variable, because (1, 2, 3, 4, 5) is a subgroup of S 5 and the multiplication in the second variable is the same as the multiplication in S 5. We claim that there are six Sylow-5 subgroups of G. Indeed, all Sylow-5 subgroups are conjugate of P. Conjugating in the first variable does not change the group (Z/5) 5. So, the number of Sylow-5 subgroups of G is equal to the number of groups conjugated to (1, 2, 3, 4, 5) = C 5. Every conjugate P of (1, 2, 3, 4, 5) is such that P = (1, 2, 3, 4, 5) = 5. Also, there are 4! = 24 elements of order 5 in S 5. Since each conjugate in S 5 preserves the cycle type, we have that every conjugate of P contains four 5-cycles and the identity. So there is 24/4 = 6 groups conjugated to (1, 2, 3, 4, 5). Hence n 5 (G) = 6. Problem 3 If Q was the semi-direct product of two of its proper subgroups, it would have to be of a group of order 4 with a group of order 2. The only group of order 2 is C 2 and the two only groups of order 4 are C 4 and C 2 C 2 = V 4. But V 4 is not a subgroup of Q, because V 4 has three elements of order 2 and Q has only one element of order 2. So if Q is a semi-direct product, then there is only two possibilities, namely 1

2 2 LOUIS-PHILIPPE THIBAULT (1) Q = C 4 C 2 ; (2) Q = C 2 C 4. We will show that all of these possibilities are impossible. First of all, the only subgroup of order 2 in Q is {+1, 1}. Moreover, Q/{+1, 1} = V 4. Indeed, Q/{+1, 1} = { 1, ī, j, k}. Since ī, j, k all have order 2, Q/{+1, 1} = V 4. So case 2 is impossible, because whenever a group G = N H, then G/N = H. We now analyze case 1. Aut(C 4 ) = C 2. We now imagine C 2 = {0, 1} as the additive cyclic group. So there is only one non-trivial homomorphism φ : C 2 Aut(C 4 ), namely the one sending 0 to the identity automorphism and 1 to φ 1, where φ 1 (0) = 0, φ 1 (1) = 3, φ 1 (2) = 2, φ 1 (3) = 1. Then, C 4 C 2 = {(0, 1), (1, 1), (2, 1), (3, 1), (0, 0), (1, 0), (2, 0), (3, 0)} as a set. Clearly, the identity has to be (0, 0). We have that (0, 1)(0, 1) = (0, 0), under the operation of the semidirect product. Also, (2, 1)(2, 1) = (0, 0). So there is two elements of order 2 in C 4 C 2, but Q has only one element of order 2. If φ is trivial, then C 4 C 2 = C 4 C 2. But (0, 1) and (2, 1) have order 2 in C 4 C 2, whereas Q has only one element of order 2. So case 1 is impossible. So, none of the possible semi-direct products of order 8 is isomorphic to Q. Problem 4 Suppose H = p α for a given α. Let H acts on G/H by left multiplication. Then, Orb(gH) = {hgh h H}. We have that Orb(gH) = 1 hgh = gh, h H g 1 Hg H g N G (H) gh N G (H)/H. Let g i H be representatives of the orbits that contain more than one element. Then, G/H = N G (H)/H + i Orb(g i H). Now we have that for all i, Orb(g i H) > 1 and Orb(g i H) H = p α. So Orb(g i H) 0 mod p for all i. Then G/H N G (H)/H mod p. Problem 5 We will start by part 2. We have that ( a)(a) = (a 2 ). Indeed, ( a)(a) + (a)(a) = ( a + a)(a) = 0, where we have used the distribution property. Then, ( a)( a) + ( (a 2 )) = ( a)( a) + ( a)(a) = ( a)( a + a) = 0. So ( a) 2 = a 2, where we have used the fact that ( (a 2 )) = a 2. For part 1, we just need to take a = 1.

3 HOMEWORK 3 3 Problem 6 Part 1. Let D be a finite integral domain. By definition, an integral domain is commutative, so we only need to check that every nonzero element of D has a multiplicative inverse. Let a 0 be an element of D. Then we have that {ax x D} = D. ( ) In fact, we have that whenever x y, ax ay, because D is a domain. Since D is finite, we have {ax x D} = D, and this implies the result ( ). In particular, there exists an x such that ax = 1. So a has an inverse. Since a was arbitrary, we have that every element of D has an inverse. Hence D is a field. Part 2. We have that an ideal P is prime in R if and only if R/P is an integral domain. Since R/P is a finite integral domain, it is a field (see part 1). We have proved in class that given a ring S and an ideal I, the quotient S/I is a field if and only if I is maximal. Then, using this theorem, P is maximal. Problem 7 Part 1. We first show that for every x R, 2x = 0. Indeed, 2x = x + x = (x + x) 2 = x 2 + 2x 2 + x 2 = x + 2x + x = 4x. Subtracting by 2x both side, we have 2x = 0. In particular, it means that x = x. Using this property, we prove the main result: x + y = (x + y) 2 = x 2 + xy + yx + y 2 = x + xy + yx + y. Subtracting by x and y both side, we have xy + yx = 0, so xy = yx = yx. Since x, y were arbitrary, we conclude that R is commutative. Part 2. Z/2 is clearly a Boolean ring. It is also a field, so it is an integral domain. Since Z/2 is the only ring up to isomorphism of order 2, suppose we have a Boolean ring R such that R > 2. Take a 0, 1 in R. Then, a(a 1) = a 2 a = 0, but a 0 and a 1 0, because a 1. So R is not an integral domain. Problem 8 Part 1. By the Bolzano-Weiestrass theorem, every bounded sequence has a converging subsequence. So, intuitively, we want to define a map φ : S R such that φ sends a sequence to the limit of one of its converging subsequence. We want to find a way to choose which subsequence to take. We will do this by using the fact that we want J to be in the kernel of φ. Define U ɛ,(an) = {i N a i < ɛ, a i (a n )}, and U J = {U ɛ,(an) ɛ > 0, (a n ) J}. We claim that the map φ : S R, (a n ) x, where x is chosen such that for all ɛ > 0, {i N a i x < ɛ, a i (a n )} U J (1) is well-defined, that is, x exists and is unique; (2) is a surjective homomorphism; (3) has kerφ = J. These three properties will complete the proof. Indeed, by the first isomorphism theorem, we will have S/J = R. We start by stating four properties of J and U J.

4 4 LOUIS-PHILIPPE THIBAULT a. We first notice that J cannot contain a sequence with a finite number of elements equal to 0 (or no element equal to 0), unless this sequence contains a subsequence that converges to 0. Otherwise, if (a n ) J has only a finite number of elements equal to 0 and no subsequences converging to 0, then J = S. Indeed, we can take (d n ) I such that (ã n ) = (a n ) + (d n ) J do not contain any zero, nor subsequences converging to 0. Then, for every (b n ) S, there exists a (c n ) S such that (b n ) = (ã n )(c n ). The sequence (c n ) can be chosen to be bounded because (ã n ) has no subsequence converging to 0. Thus, (b n ) J. Since (b n ) was arbitrary, J = S. In particular, U J and U J contains no finite sets. b. If U ɛ,(an) U J and U ɛ,(an) V, then V U J. Indeed, V is of the form V = U ɛ,(ã n), where ɛ > ɛ and (ã n ) J is such that U ɛ,(an) = U ɛ,(ãn). Note that every (ã n ) having the property U ɛ,(an) = U ɛ,(ãn) are in J, as it suffices to obtain it from a multiplication of (a n ) by an appropriate sequence in S. c. If (a n ), (b n ) J, U ɛ,(an), U ɛ,(b n) U j, then U ɛ,(an) U ɛ,(b n) U J. Indeed, U ɛ,(an) U ɛ,(b n) U ɛ+ɛ,(a n)+(b n). Then the result follows from b. d. If U U J, then U c U J. In fact, otherwise let (b n ) be the sequence such that b i = 0 for every i U c. Note that (b n ) have infinitely many 0, because otherwise U U I U J. Then, J[(b n )], the smallest ideal containing both J and (b n ) is not all of S. This contradicts the maximality of J. Indeed, if (s n ) S has sufficiently large entries at every index, it is clearly impossible to multiply (b n ) by a sequence of S to obtain (s n ), since (b n ) has infinitely many 0. Moreover, if (b n ) + (c n ) = (s n ), with (c n ) J, then there exists ɛ > 0 such that U ɛ,(cn) U. In fact, since (c n ) J, it has infinitely many small values. Those small values have to be at different indices that those of (b n ), since (s n ) has large values. So, by property b, U U J. Since this is impossible, it implies that (c n ) cannot be in J, so (s n ) J[(b n )] implies J[(b n )] S. We now prove the uniqueness. Suppose x 1 and x 2 are good candidate for φ(a n ). Then there exists ɛ > 0 such that {i N a i x 1 < ɛ} U J, {i N a i x 2 < ɛ} U J are disjoint. But, by properties a and c, this is impossible. We want to prove the existence. Suppose that there exists (a n ) such that for all convergent subsequences (a k ), there exists ɛ (ak ) > 0 such that Ũ ɛ(ak ),(a k ) = {i N a i x (ak ) < ɛ (ak )} U J, where x (ak ) is the limit of (a k ). Then, since the complement Ũ c ɛ (ak ),(a k ) of every Ũɛ (ak ),(a k ) is in U J (property d), then Ũ c ɛ (ak ),(a k ) U J (property c) is finite, where the intersection is taken over all the converging subsequences of (a n ). This contradicts property a. We now want to prove that φ is a surjective homomorphism. First, φ is clearly surjective, as (φ((r) i=0 ) = r for all r in R. If φ(a n) = x 1 and φ(b n ) = x 2, then (a n ) (x 1 ) J and (b n ) (x 2 ) J. So ((a n ) + (b n ) ((x 1 ) + (x 2 ))) J, where we view x 1, x 2 as sequences (x 1 ), (x 2 ), respectively. Thus, for all ɛ > 0, {i N a i + b i (x 1 + x 2 ) < ɛ} U J. Hence, φ(a n + b n ) = x 1 + x 2. Also, (x 2 )((a n ) (x 1 )) J, because J is an ideal. Then, ((a n )((b n ) (x 2 )) + (x 2 )((a n ) (x 1 ))) = ((a n )(b n ) (x 1 )(x 2 )) J, so φ(a n b n ) = x 1 x 2 as

5 HOMEWORK 3 5 before. So φ is a homomorphism. Finally, kerφ = J. This is clear, because J kerφ. Then, by maximality of J, J = kerφ. So S/J = R. Part 2. The last two parts are due to the fact that Lim J is a homomorphism. For the first part, let Lim J (a n ) = x. Then, ((a n ) (x)) J. So, (c)((a n ) (x)) J, because J is an ideal. Then, as in part 1, for all ɛ > 0, So, Lim j (ca n ) = cx. {i N ca i cx < ɛ} U J. Part 3. First of all, notice that all convergent sequences having limit 0 are in J. This is due to the fact that if (a n ) 0, then for all ɛ > 0, there exists N such that for all n > N, a n < ɛ. Then, take (b n ) I such that b n = 0 for all n such that a n < ɛ. We have that {i N a i < ɛ} = U ɛ,(bn) U J. Since ɛ > 0 was arbirary, Lim J (a n ) = 0. So, (a n ) J. Now take a convergent sequence (c n ) x. Then, ((c n ) (x)) is a sequence converging to 0. So ((c n ) (x)) J. Thus, Lim J ((c n ) (x)) = 0. Hence, Lim J (c n ) = Lim J (x) = x. Part 4. The answer is no. Indeed, take the sequences (( 1) n ) and (( 1) n+1 ). Then, if Lim J ((( 1) n )) = Lim J ((( 1) n+1 )), we have Lim J ((( 1) n )) (( 1) n+1 )) = 0. But, clearly Lim J ((( 1) n )) (( 1) n+1 )) = Lim J (2( 1) n ) = 2 or 2, depending on J.

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