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Homework 3 SolutionChapter 3. LetQbe the group of rational numbers under addition and letQ∗be the group of nonzero rational numbers under multiplication. InQ, list the elements in〈 12 〉. InQ∗, list the elements in〈 12 〉. InQ,〈 1 2〉={n· 1 2|n∈Z}={···,− 2 ,− 3 2,− 1 ,− 12, 0 , 12, 1 , 32, 2 ,···}={n 2...

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MATH 3005 Homework Solution

Han-Bom Moon

Homework 3 Solution Chapter 3.

2. Let Q be the group of rational numbers under addition and let Q∗ be the group of nonzero rational numbers under multiplication. In Q, list the elements in h 12 i. In Q∗ , list the elements in h 12 i. In Q, 1 3 1 1 3 n 1 h i = {n · | n ∈ Z} = {· · · , −2, − , −1, − , 0, , 1, , 2, · · · } = { | n ∈ Z}. 2 2 2 2 2 2 2 In Q∗ ,  n 1 1 1 1 1 i = { | n ∈ Z} = { n | n ∈ Z} = {· · · , 4, 2, 1, , , · · · } = {2n | n ∈ Z}. h 2 2 4 2 2 4. Prove that in any group, an element and its inverse have the same order. If |a| = n, an = e. So (a−1 )n = (an )−1 = e−1 = e. Therefore |a−1 | ≤ n = |a| by the definition of order. By the same reason, |a| = |(a−1 )−1 | ≤ |a−1 |. So we obtain |a| = |a−1 |. Now suppose that |a| = ∞. We need to show that |a−1 | = ∞ as well. If not, |a−1 | = n for some n > 0. Then e = (a−1 )n = (an )−1 , or equivalently, an = e−1 = e. Therefore |a| ≤ n and arise a contradiction. Hence |a−1 | = ∞. 6. In the group Z12 find |a|, |b|, and |a + b| for each case. (a) a = 6, b = 2 a = 6, a + a = 12 = 0 ⇒ |a| = 2 b = 2, b + b = 4, 3 · b = 6, 4 · b = 8, 5 · b = 10, 6 · b = 12 = 0 ⇒ |b| = 6 a + b = 8, 2 · (a + b) = 16 = 4, 3 · (a + b) = 24 = 0 ⇒ |a + b| = 3 (b) a = 3, b = 8 a = 3, 2 · a = 6, 3 · a = 9, 4 · a = 12 = 0 ⇒ |a| = 4 b = 8 ⇒ |b| = 3 (see (a).) a + b = 11 = −1 ⇒ |a + b| = | − 1| = |1| = 12

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MATH 3005 Homework Solution

Han-Bom Moon

(c) a = 5, b = 4 a = 5, 2 · a = 10, 3 · a = 15 = 3, 4 · a = 8, 5 · a = 13 = 1, 6 · a = 6, 7 · a = 11, 8 · a = 16 = 4, 9 · a = 9, 10 · a = 14 = 2, 11 · a = 7, 12 · a = 12 = 0 ⇒ |a| = 12 b = 4, 2 · b = 8, 3 · b = 12 = 0 ⇒ |b| = 3 a + b = 9 = −3 ⇒ |a + b| = | − 3| = |3| = 4 15. If a is an element of a group G and |a| = 7, show that a is the cube of some element of G. If |a| = 7, a7 = e. Then (a5 )3 = a15 = a2·7+1 = (a7 )2 a = e2 a = a. So a is the cube of a5 . 18. Suppose that a is a group element and a6 = e. What are the possibilities for |a|? Provide reasons for your answer. If a6 = e, then |a| must be at most 6. So only 1, 2, 3, 4, 5, 6 are all possibilities. Furthermore, if a4 = e, then a2 = a2 e = a2 a6 = a8 = (a4 )2 = e2 = e. So |a| = 4 is impossible. Similarly, if a5 = e, then a3 = a3 e2 = a3 (a6 )2 = a15 = (a5 )3 = e3 = e. So |a| = 5 is impossible as well. In summary, only 1, 2, 3, 6 are possible. (They are all divisors of 6. Of course, it is not an accident.) 19. If a is a group element and a has infinite order, prove that am 6= an when m 6= n. Suppose that am = an . Without lose of generality, we may assume that m ≥ n. Then am−n = am (an )−1 = am (am )−1 = e. If m > n, then from am−n = e, we know that |a| < ∞. Therefore m = n. 24. Suppose n is an even positive integer and H is a subgroup of Zn . Prove that either every member of H is even or exactly half of the members of H are even. Suppose that H is a subgroup of Zn . If H consists of even integers, then there is nothing to prove. Now assume that there is a ∈ H, which is an odd integer. Let E = {x ∈ H | x is even} and O = {x ∈ H | x is odd}. We need to show that the number of elements of E is equal to that of O. Take a map f : H → H, given by f (x) = x + a mod n in Zn . We claim that if x is even, then f (x) is odd and if x is odd, then f (x) is even. This is trivial if x +a < n. If x + a ≥ n, then x + a mod n is odd if and only if x + a is odd because n is an even integer. So the map f has a property that f (E) ⊂ O and f (O) ⊂ E .

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MATH 3005 Homework Solution

Han-Bom Moon

Moreover, f is injective. Indeed, if f (x) = f (y), then x + a = y + a so by cancellation, x = y. So from f (E) ⊂ O, we know that the number of elements of E is less than or equal to that of O. Similarly, from f (O) ⊂ E, the number of elements of O is less than or equal to that of E. Thus they have the same number of elements. 31. For each divisor k > 1 of n, let Uk (n) = {x ∈ U (n)|x mod k = 1}. List the elements of U4 (20), U5 (20), U5 (30), and U10 (30). Prove that Uk (n) is a subgroup of U (n). Let H = {x ∈ U (10)|x mod 3 = 1}. Is H a subgroup of U (10)?

U4 (20) = {1, 9, 13, 17}, U5 (20) = {1, 11}. U5 (30) = {1, 11}, U10 (30) = {1, 11}. Suppose that k is a divisor of n. Step 1. Uk (n) = 6 ∅. Because 1 ∈ Uk (n), Uk (n) 6= ∅. Step 2. a, b ∈ Uk (n) ⇒ ab ∈ Uk (n). If a, b ∈ Uk (n), then a mod k = 1 and b mod k = 1, or equivalently, a = xk + 1 and b = yk + 1 for two integers x, y ∈ Z. By using division algorithm, if we write ab = qn + r with 0 ≤ r < n, then ab = r in U (n). Also n = uk for some u ∈ Z. Now r = ab − qn = (xk + 1)(yk + 1) − qn = xyk 2 + xk + yk + 1 − quk = k(xyk + x + y − qu) + 1 so ab = r = 1 mod k. Thus ab ∈ Uk (n) as well. Step 3. a ∈ Uk (n) ⇒ a−1 ∈ Uk (n). In U (n), a−1 is the solution b ∈ Zn of ax = 1 mod n. For such b, ab = pn + 1 for some p ∈ Z. Since a mod k = 1, a = qk + 1. Then pn + 1 = (qk + 1)b = qk + b. Also n = uk for some u ∈ Z. Now b = qk + b = pn + 1 = puk + 1. Thus b = 1 mod k and b ∈ Uk (n). If n = 10 and k = 3, then H = {1, 7}. But 7 · 7 = 9 ∈ / H. So H is not closed under the multiplication and H is not a subgroup of U (10). 32. If H and K are subgroups of G, show that H ∩ K is a subgroup of G. (Can you see that the same proof shows that the intersection of any number of subgroups of G, finite or infinite, is again a subgroup of G?) Step 1. H ∩ K = 6 ∅. Because e ∈ H and e ∈ K, e ∈ H ∩ K. Thus H ∩ K 6= ∅. Step 2. a, b ∈ H ∩ K ⇒ ab−1 ∈ H ∩ K. 3

MATH 3005 Homework Solution

Han-Bom Moon

If a, b ∈ H ∩ K, then a, b ∈ H and a, b ∈ K. Since H and K are both subgroups, ab−1 ∈ H and ab−1 ∈ K. Thus ab−1 ∈ H ∩ K. Therefore by subgroup test 2, H ∩ K ≤ G. Note that the same proof holds for arbitrary number of subgroups (even for infinitely many!). 34. Let G be a group, and let a ∈ G. Prove that C(a) = C(a−1 ). If x ∈ C(a), ax = xa. Then x = a−1 ax = a−1 xa and xa−1 = a−1 xaa−1 = a−1 x. Thus x ∈ C(a−1 ). Therefore C (a) ⊂ C (a−1 ). By applying the same idea for a−1 , we have C (a−1 ) ⊂ C ((a−1 )−1 ) = C(a). Thus C(a) = C(a−1 ). 40. In the group Z, find (a) h8, 14i; (b) h8, 13i; (c) h6, 15i; (d) hm, ni; (e) h12, 18, 45i. We would like to show a general statement (for (d)): hm, ni = hgcd(m, n)i. Let d = gcd(m, n). Then m = ad and n = bd for some a, b ∈ Z. So m = a · d ∈ hd i and n = b · d ∈ hdi. Therefore hm, ni ⊂ hdi. On the other hand, if d = gcd(m, n), then there are two integers x, y ∈ Z such that d = xm + yn. So d = x · m + y · n ∈ hm, ni. Hence hdi ⊂ hm, ni. Thus we have hdi = hm, ni. (a) h8, 14i gcd(8, 14) = 2 ⇒ h8, 14i = h2i (b) h8, 13i gcd(8, 13) = 1 ⇒ h8, 13i = h1i = Z (c) h6, 15i gcd(6, 15) = 3 ⇒ h6, 15i = h3i (d) hm, ni hm, ni = hgcd(m, n)i (e) h12, 18, 45i gcd(12, 45) = 3 ⇒ h12, 45i = h3i 18 = 6 · 3 ⇒ 18 ∈ h3i 12, 18, 45 ∈ h3i ⇒ h12, 18, 45i ⊂ h3i 12, 45 ∈ h12, 18, 45i ⇒ h3i = h12, 45i ⊂ h12, 18, 45i So h12, 18, 45i = h3i. 4

MATH 3005 Homework Solution

Han-Bom Moon

42. If H is a subgroup of G, then by the centralizer C(H) of H we mean the set {x ∈ G | xh = hx for all h ∈ H}. Prove that C(H) is a subgroup of G. Step 1. C(H) 6= ∅. Because eh = he for all h ∈ H, e ∈ C(H). Step 2. a, b ∈ C(H) ⇒ ab ∈ C(H). For all h ∈ H, abh = ahb = hab because a, b ∈ C(H). So ab ∈ C(H). Step 3. a ∈ C(H) ⇒ a−1 ∈ C(H). For all h ∈ H, ah = ha. Then aha−1 = haa−1 = h and a−1 h = a−1 aha−1 = ha−1 for all h ∈ H. So a−1 ∈ C(H). By subgroup test 1, C(H) ≤ G. 51. Let a be a group element of order n, and suppose that d is a positive divisor of n. Prove that |ad | = n/d . Suppose that n = kd. Then (ad )k = adk = an = e. So |ad | ≤ k = n/d. If |ad | = m, then (ad )m = adm = e so dm ≥ n and |ad | = m ≥ n/d. Therefore |ad | = n/d . 75. Let H be a subgroup of a group G. Prove that the set H Z (G) = {hz|h ∈ H, z ∈ Z(G)} is a subgroup of G. Step 1. H Z(G) 6= ∅. Since both H and Z(G) are subgroups of G, e ∈ H and e ∈ Z(G). So e = ee ∈ H Z (G) and H Z (G) 6= ∅. Step 2. a, b ∈ H Z (G) ⇒ ab−1 ∈ H Z (G). From a, b ∈ H Z(G), a = h1 z1 and b = h2 z2 where h1 , h2 ∈ H and z1 , z2 ∈ Z(G). Then −1 −1 −1 −1 −1 ab−1 = h1 z1 (h2 z2 )−1 = h1 z1 z −1 2 h2 = h1 z1 h 2 z2 = h1 h 2 z1 z2 . −1 −1 = h h−1 z z −1 ∈ H Z(G). Because h1 h−1 1 2 1 2 2 ∈ H and z1 z2 ∈ Z(G), ab

By subgroup test 2, HZ(G) ≤ G. 79. Let G = GL(2, R). #! " 1 1 . (a) Find C 1 0 " #! # " a b 1 1 If , then ∈C 1 0 c d "

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MATH 3005 Homework Solution

Han-Bom Moon

So a +b = a +c, a = b+d , c+d = a, and"b = c. These conditions are equivalent # a b to b = c and a = b+d. Conversely, for ∈ G with b = c and a = b+d, c d " # " #! a b 1 1 then by the same equation, ∈C . Therefore c d 1 0 " # ) #! ("  1 1 a b  C ∈ G  a = b + d, b = c = c d 1 0 "

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#" # " # 0 1 a b c d = . 1 0 c d a b # b ∈ G with b = c and a = d, then d #! 0 1 . Therefore 1 0 # )   ∈ G  a = d, b = c

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#" # " k 0 a b ka kb = 0 k c d kc kd (" ) #  k 0  So  k 6= 0 ⊂ Z(G). 0 k " # a b Conversely, if ∈ Z(G), then c d " #" # " a b 1 1 = c d 1 0 " #! # " a b 1 1 so . Similarly, ∈C 1 0 c d " # " #" a b 0 1 = 1 0 c d 6

with k 6= 0 is in Z(G), because #

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MATH 3005 Homework Solution so

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Han-Bom Moon

0 1 1 0

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Therefore # a = d. From a = b + d and a = d, b = c = 0. " a = b#+ d," b = c and a 0 a b = . Because this matrix is in G = GL(2, R), a 6= 0. Hence c d 0 a So we have (" ) #  k 0  Z(G) ⊂  k 6= 0 0 k and hence,

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