Homework 5 Solutions

Transcription

1 Homewor 5 Solutions Enrique Treviño October 10, Chater 5 Problem 1. Exercise 1 Suose that G is a finite grou with an element g of orer 5 an an element h of orer 7. Why must G 35? Solution 1. Let G = n. Since g is a subset of G, then g n. Therefore 5 n. Similarly h n, so 7 n. Since 5 n an 7 n, then 35 n. Since n is a ositive integer, then n 35. Problem 2. Exercise 3 Prove or isrove: Every subgrou of the integers has finite inex. Solution 2. To turn in. Problem 3. Exercise 5 List the left cosets of the subgrous in each of the following. a 8 in Z 24 b 3 in U8 c 3Z in Z A 4 in S 4 e A n in S n f D 4 in S 4 Solution 3. a The cosets are: {0, 8, 16}, {1, 9, 17}, {2, 10, 18}, {3, 11, 19}, {4, 12, 20}, {5, 13, 21}, {6, 14, 22}, {7, 15, 23}. b To turn in. c One coset is H = 3Z. Since 1 3Z, then 1H is a ifferent coset. 1H = {3n + 1 n Z}, i.e., 1H = {1, 4, 7, 10, 13,...} { 2, 5, 8,...}. Since 2 H 1H, then 2H is a ifferent coset. 2H = {3n + 2 n Z}, i.e., 2H = {2, 5, 8, 11,...} { 1, 4, 7,...}. Since these three cosets artition Z, there are no more cosets. To turn in. e If n 2, then A n is half the size of S n, so [S n : A n ] = 2. Therefore there are only two cosets. One coset is A n an the other coset is what is left, i.e., the other coset is S n \ A n = {σ S n σ A n } = {σ S n σ is an o ermutation }. So one coset is the even ermutations an the other is the o ermutations. If n = 1, then A 1 = S 1, so the only coset is A 1. f To turn in. 1

2 Problem 4. Exercise 8 Use Fermat s Little Theorem to show that if = 4n+3 is rime, there is no solution to the equation x 2 1 mo. Solution 4. Let G = Z, i.e., G is the multilicative grou moulo. Suose x 2 1 mo. Then x 0 mo, therefore x G. Since x 2 1 mo, then x 4 1 mo. If x 1 mo, then x 2 1 mo. Then 1 1 mo, so 2 0 mo, so = 2. But since = 4n + 3, 2. Therefore x 1 mo. Since x 1 mo an x 2 1 mo an x 4 1 mo, then the orer of x in G is 4. By Langrange s theorem 4 G. But the orer of G is 1 = 4n n + 2 is not a multile of 4, therefore we ve reache a contraiction. Therefore no x G satisfies x 2 1 mo. Solution using Fermat s last theorem: Above I inclue a solution using grou theory. Now, let s just use Fermat s little theorem. Suose that x 2 1 mo. Then x 0 mo, therefore by Fermat s little theorem x 1 1 mo. Therefore x 1 x 4n+2 x 2 2n+1 1 2n+1 1 mo. Therefore 1 1 mo, so 2 0 mo, so 2, so = 2. Since 2, then there is no solution to the equation x 2 1 mo. Problem 5. Exercise 11 Let H be a subgrou of a grou G an suose that g 1, g 2 G. Prove that the following conitions are equivalent. a g 1 H = g 2 H b Hg1 1 = Hg2 1 c g 1 H g 2 H g 2 g 1 H e g 1 1 g 2 H Solution 5. I will inclue the roof that b imlies. To turn in you have to rove a imlies c. Suose Hg1 1 = Hg2 1. We want to show g 2 g 1 H. Since e H an g1 1 = eg1 1, then g 1 1 Hg 1 Since Hg1 1 = Hg2 1, then g 1 1 Hg2 1. Therefore there exists h H such that g 1 1 = hg2 1. Therefore g 1 g 1 1 g 2 = g 1 hg 1 2 g 2 g 2 = g 1 h. Therefore g 2 g 1 H. This roves that Hg 1 1 = Hh 1 2 imlies g 2 g 1 H. Now let s rove the reverse irection. Suose g 2 g 1 H. Then there exists an h H such that g 2 = g 1 h. Therefore g1 1 g 2 = h, so g1 1 = hg2 1. We want to rove that Hg1 1 = Hg2 1. Let x Hg 1 1. Then there exists h H such that x = h g1 1. So x = h hg 1 2 = h hg 1 2. Since h h H because H is a subgrou of G, then h hg 1 Therefore Hg1 1 Hg2 1. Now, suose that x Hg 1 2. Therefore there exists h H such that x = h g 1 g 1 2 = h 1 g 1 1. Therefore x = h h 1 g 1 Therefore Hg1 1 = Hg2 1. This roves that b an are equivalent. Problem 6. Exercise 17 Suose that [G : H] = 2. If a an b are not in H, show that ab H. Solution 6. To turn in Hg2 1, so x Hg 1 2. Since g 2 = g 1 h, then 1. Since h h 1 H, then x Hg1 1. Therefore Hg 1 2 Hg

3 Problem 7. Exercise 18 If [G : H] = 2, rove that gh = Hg. Solution 7. To turn in. Problem 8. Exercise 22 Let n = e1 1 e2 2 e, where 1, 2,..., are istinct rimes. Prove that φn = n Solution 8. Let s rove that φmn = φmφn if gc m, n = 1. Let a m be relatively rime to m. Now consier {a, a + m, a + 2m,..., a + n 1m}. All of these numbers are relatively rime to m because a is relatively rime to m an m m. If we loo moulo n, then since m an n are relatively rime {a, a + m, a + 2m,..., a + n 1m} {0, 1, 2,..., n 1} mo n in a ifferent orer. Therefore there are φn numbers relatively rime to n in {a, a + m, a + 2m,..., a + n 1m}. Since there are φm ossibilities for a an for each a there are φn numbers mn relatively rime to m an n, then there are φmφn numbers relatively rime to mn, so φmn = φmφn. Now let s calculate φ. Among the numbers 1, 2, 3,..., the only numbers relatively rime to are, 2, 3,..., 1. Therefore φ = 1 = 1 1. Since φ satisfies that φab = φaφb whenever gc a, b = 1, then if n = e1 1 e2 2 e, we have: φn = φ e1 1 φe2 2 φe = e1 1 = n e e Alternative Solution: Let n = e1 1 e2 2 e. Consier the following sets: A 1 = {m n : 1 m}, A 2 = {m n : 2 m},, A = {m n : m}. If gcm, n 1, then m is ivisible by i for some i, so m A i. Therefore φn = n A 1 A 2 A. Now Therefore by inclusion-exclusion: A i1 A i2 A ir = n i1 i2 ir. φn = n A 1... A + A 1 A r A i1 A i2 A ir A 1 A = n n n... n + n r n n i1 i2 ir 1 2 = n r i1 i2 ir 1 2 = n

4 Problem 9. Exercise 23 Show that n = n φ for all ositive integers n. Solution 9. Let m {1, 2, 3,..., n}. Let gc m, n =. Then m = m an n = n where gc m, n = 1 an m n. Note that n = n/ an that n. Now consier all numbers m such that gc m, n =. The numbers satisfy that m/ is relatively rime with n/ an less than or equal to n/. Also, as long as those requirements are satisfie, then m, n =. Therefore there are n φ numbers m satisfying that gc m, n = whenever n. Since every number m {1, 2, 3..., n} has a gc with n that ivies n, then if for each n we count all numbers that have gc with n, we get n. This is because we are artitioning the set {1, 2,..., n} into the gc that each number has with n. Therefore n φ = n. But if n, then n n as well, so n n n φ = φ. n The conclusion follows. Alternative hrasing of the solution: Consier the relation on the set {1, 2, 3,..., n}, where a b if gc a, n = gc b, n. It is not har to show that is an equivalence relation. Therefore artitions the set {1, 2, 3,..., n}. Let C be the equivalence class of the number. Then C = {m n : gc m, n = }. C = whenever n, therefore C artitions {1, 2, 3,..., n} whenever ranges over the ivisors of n. Therefore n = n C. As roven above n C = φ. The roof conclues the same way as the original roof. One more solution: Let gn = φ. n Our goal is to show that gn = n. We will rove that g is multilicative, i.e., that if gc a, b = 1, then gab = gagb. Let ab. We re goig to nee to rove that there exist unique a an 2 b such that = 2. Let = gc a, an 2 = gc b,. Then an 2 are unique. Now let s show that 2 =. a Since = gc a,, then 1 = gc,. Now ab an is relatively rime so by Exercise 27 in Chater 2 one in HW 1, then b. Since a an b are relatively rime, then gc, b = gc, b = 2. But since b, then gc, b =. Therefore = 2. Therefore = 2. Oay, so we have roven that if ab an gc a, b = 1, then there exist unique a an 2 b. Therefore gab = ab φ = a 2 b φ 2 = a φ 2. 2 b 4

5 Now, since gc a, b = 1 an a an 2 b, then gc, 2 = 1. Since φab = φaφb whenever gc a, b = 1, then φ 2 = φ φ 2. Therefore gab = φ 2 = φ φ 2 = φ φ 2 = gagb. a 2 b a 2 b 1 a 2 b Now, g = φ = φ1 + φ + φ φ = 1 + = 1 + = = =. 1 Since g = an gab = gagb whenever gc a, b = 1, then gn = n. The roof is comlete. 5