Test1. Due Friday, March 13, 2015.


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1 1 Abstract Algebra Professor M. Zuker Test1. Due Friday, March 13, Euclidean algorithm and related. (a) Suppose that a and b are two positive integers and that gcd(a, b) = d. Find all solutions m and n to am bn = 0. Hint: Note that for each solution, am and bn are common multiples of a and b. am bn = 0 am = bn. Thus, am and bn are common multiples of a and b for every solution. Let l = lcm(a, b) = ab d. Then all common multiples of a and b are of the form kl, where k is any integer. If am = bn, then am = bn = kl for some k Z. This is equivalent to m = kb/d and n = ka/d. On the other hand, for any k Z, m = kb/d and n = ka/d satisfy am = bn = kl. Thus all solutions are m = kb/d and n = ka/d for k Z. (b) If am bn = d, find all solutions m and n to am bn = d. Hint: Note that a(m m) b(n n) = 0. If am bn = d and am bn = d, then by subtraction, a(m m) b(n n) = d d = 0, so M m and N n must be of the form kb/d and ka/d, respectively, where k Z. That is, m = M kb/d and n = N ka/d. In words, if am bn = d, then a(m + kb/d) b(n + ka/d) = d, so (M + kb/d, N + kb/d) is a solution for every k Z and all solutions are of this form. (c) Show that 1966 and 2017 are relatively prime and compute integers m and n such that 1966m n = = = = = = = = = so gcd(2017, 1966) = 1. A solution for m and n may be computed by backtracking in the above computation or by multiplying 2 2 matrices. Backtracking:
2 2 Abstract Algebra Professor M. Zuker 1 = = 3 1 (5 1 3) = = 2 (23 4 5) 1 5 = = ( ) = = 11 ( ) 9 28 = = ( ) = = 771 ( ) = = ( 791) Thus m = 791 and n = 771 is a solution. (d) Referring to (c) above, compute a solution where m > 0 and m is as small as possible and a solution where n > 0 and n is as small as possible. Hint: If you use the extended Euclidean algorithm, the solution you find will satisfy one of these conditions, so you need only compute a second solution. Any solution is of the form m = k and n = k. If k = 0, n is positive. The next smaller value of n is < 0, so ( 791, 771) is the solution with the smallest positive value for n. The next larger value of m is = 1226, for which n = = 1195, so (1226, 1195) is the solution with the smallest positive value for m. 2. True or false. (a) In the group Z m, if 0 a b < m, then a = b a = b. False. In fact, a = b gcd(a, m) = gcd(b, m). (b) If the least common multiple of two positive integers a and b equals a or b, then either a b or b a. True. ab/ gcd(a, b) = a gcd(a, b) = b, so b a. ab/ gcd(a, b) = b gcd(a, b) = a, so a b. (c) Suppose that gcd(a, b) = 1 for positive integers a and b. Then for any positive integers m and n, a m and b n are relatively prime. True. a = K i=i pk i i and b = L i=i ql i i, where each p i and each q i are distinct primes. We know that no p i equals some q i. Thus a m is a product of K distinct primes p i to higher powers and b n is a product of L distinct primes q i to higher powers. The primes remain distinct so gcd(a m, b n ) = 1.
3 3 Abstract Algebra Professor M. Zuker (d) For positive integers a, b and c, suppose that gcd(a, b) = 1 and that c ab. Then c = d 1 d 2 where d 1 a, d 2 b and gcd(d 1, d 2 ) = 1. True. a = K i=1 pk i i and b = L i=1 ql i i where the p i s and q i s are distinct primes. If d a, then d = d 1 d 2, where d 1 = K i=1 pk i i, 0 k i k i and d 2 = L i=1 ql i i, 0 l i l i. d 1 a, d 2 b and gcd(d 1, d 2 ) = 1. (e) If σ S n, then σ and σ 1 always have the same number of orbits. True. The inverse of a cycle (i 1, i 2,... i k ) is (i k, i k 1,... i 2, i 1 ), which is a cycle. σ(i) = i σ 1 (i) = i. Thus, σ and σ 1 have the same orbits. This is stronger than the same number of orbits. (f) If σ S n, then σ and σ 2 always have the same number of orbits. False. Counterexample. In S 4, σ = (1, 2, 3, 4) has one orbit, but σ 2 = (1, 3)(2, 4) has two orbits. (g) In D 5, let H = ρ 2, τρ 3. Then H = D 5. True Since gcd(2, 5) = gcd(1, 5) = 1, ρ = ρ 2, so H contains all powers of ρ. ρ 2 H and τρ 3 H implies that τρ 3 ρ 2 = τ H. Thus H contains ρ i and τρ i for 0 i < 5, so H = D 5. (h) If H and K are subgroups of G, then HK is a subgroup of G. False. If h 1, h 2 H and k 1, k 2 K, then there is no reason why (h 2 k 2 )(h 1 k 1 ) should equal h 3 k 3 for some h 3 H and k 3 K. Counterexample. In S 3, let H = {ι, (1, 2)} and let K = {ι, (1, 3)}. Then HK = {ι, (1, 2), (1, 3), (1, 3, 2)} However, (1, 2)(1, 3)(1, 2)(1, 3) = (1, 2, 3) / HK. (i) If H and K are normal subgroups of G, then HK is a subgroup of G. True. In fact, it s true if just one of H and K is normal. H normal implies that gh = Hg for all g G. In particular, kh = Hk for any k K. If h 1, h 2 H and k 1, k 2 K, then If h H and k K, (h 1 k 1 )(h 2 k 2 ) = h 1 (k 1 h 2 )k 2 = h 1 (h 3 k 1 )k 2 because H is normal. h 3 H = (h 1 h 3 )(k 1 k 2 ) HK. (hk) 1 = k 1 h 1 = h 1 k 1 because H is normal. h 1 H
4 4 Abstract Algebra Professor M. Zuker (j) If H and K are normal subgroups of G, then HK is a normal subgroup of G. True. HK is Normal if ghk = HKg for all g G. This is equivalent to ghkg 1 = HK for all g G. If ghkg 1 ghkg 1, then ghkg 1 = (gh)kg 1 = (h 1 g)kg 1 since H is normal = h 1 (gk)g 1 = h 1 (k 1 g)g 1 since K is normal = (h 1 k 1 )gg 1 = h 1 k 1 HK. In the above, h 1 H and k 1 K. To prove that HK is normal, we need H and K to be normal. 3. For each set G and binary operation, decide the following. Is G a group. If not, what properties fail? If so, is G Abelian? If it is Abelian, is it cyclic? (a) G = Q + and is ordinary multiplication. G is a noncyclic Abelian group. {[ ] } 1 n (b) G = n Z and is matrix multiplication. 0 1 by G is a cyclic group. In fact (G, ) (Z, +). If f : Z G is defined f(n) = [ 1 n 0 1 then f is an isomorphism. {[ ] } a b (c) G = a, b, c, d Z and ad bc = ±1 and is matrix multiplication. c d G is a nonabelian group (and cannot be cyclic, of course). (d) G = {a + b 3 R a, b Q} and is ordinary multiplication. Note that G excludes 0. G is a noncyclic Abelian group. The binary operation is welldefined, since a + b 3)(c + d 3) = (ac + 3bd) + (ad + bd) 3 G. (a + b 3)(a b 3) = a 2 3b 2. If a 2 3b 2 = 0, then a = 0 b = 0 and b = 0 a = 0. If ab 0, then 3 = a Q, which is false. Thus, a + b 3 = 0 b a = b = 0. Then (a + b 3) 1 = a b 3, where a = a and b = b. a 2 3b 2 a 2 3b 2 (e) G = {n Z 60 gcd(n, 60) = 1}. is addition. G is not a group. In fact, G is not even closed under addition, since gcd(7, 60) = gcd(13, 60) = 1, but gcd(7 + 13, 60) = gcd(20, 60) = 20. ],
5 5 Abstract Algebra Professor M. Zuker (f) G = {n Z 60 gcd(n, 60) = 1}. is multiplication. G is in fact an Abelian group. If m and n are both relatively prime to 60, then neither m nor n contain prime factors in common with 60. Thus, mn contains no prime factor in common with 60. is clearly commutative. If gcd(a, 60) = 1, then gcd(a n, 60) = 1 for all n 0 (see the appropriate true/false question above). If m is the smallest integer such that a m a h mod 60 for some h < m, then a m h 1 mod 60 = a 0, so h = 0. Thus a m 1 is the multiplicative inverse of a. G = ϕ(60) = ϕ(5)ϕ(3)ϕ(4) = = 16. G is not cyclic. In fact, the maximum order of any a G is Let σ = ( i σ(i) ) (a) Compute the decomposition of σ into disjoint cycles. σ = µ 1 µ 2 µ 3 µ 4, where µ 1 = (1, 16, 3, 4, 18, 20), µ 2 = (2, 13, 9, 8), µ 3 = (5, 6, 14, 17, 10) and µ 4 = (7, 19, 15, 11) (b) How many orbits does σ have. σ has 5 orbits. Each of the four disjoint cycles above comprise 19 of the 20 numbers between 1 and 12. σ(12) = 12, so {12} is the fifth orbit. (c) Is σ an even or an odd permutation? µ 1 is a 6cycle (odd), mu 2 is a 4cycle (odd), mu 3 is a 5cycle (even) and mu 4 is a 4cycle (odd). Thus, the sign of σ is ( 1)( 1)(1)( 1) = 1, so σ is odd. (d) Compute σ. µ 1, µ 2, µ 3 and µ 4 have orders 6, 4, 5 and 4, respectively. The least common multiple of these integers is the least common multiple of 4 and the least common multiple of 6 and 5, which is the least common multiple of 4 and 30, which is 60. Thus, σ = 60. (e) Compute the cycle decomposition of σ 2. Since the cycles permute disjoint sets of numbers, they commute. Thus, σ 2 = (µ 1 ) 2 (µ 2 ) 2 (µ 3 ) 2 (µ 4 ) 2. µ 2 1 = (1, 3, 18)(4, 20, 16) µ 2 2 = (2, 9)(8, 13) µ 2 3 = (5, 14, 10, 6, 17) µ 2 4 = (7, 15)(11, 19) Thus σ 2 = (1, 3, 18)(4, 20, 16)(2, 9)(8, 13)(5, 14, 10, 6, 17)(7, 15)(11, 19) (f) Let µ = (1, 13, 7, 18, 3, 20, 5, 11)(2, 15, 5, 9, 7, 1, 10, 19) Is µ even or odd. You must give a reason. µ is the product of two kcycles, so it is even. The fact that k = 8, making both cycles odd permutations, is not relevant.
6 6 Abstract Algebra Professor M. Zuker 5. In S n for n > 2, let H = {ι, (1, 2)}, where ι is the identity in S n and (1, 2) is a transposition. H is clearly a subgroup of S n. Prove that H is not a normal subgroup of S n. Hint: It suffices to find a single π S n such that πh Hπ. (1, 3)H = {(1, 3), (1, 3)(1, 2)} = {(1, 3), (1, 2, 3)} and H(1, 3) = {(1, 3), (1, 2)(1, 3)} = {(1, 3), (1, 3, 2)}. Since (1, 3)H H(1, 3), H is not normal. 6. A n and 3cycles. We know that 3cycles are even permutations. The object is to show that A n is generated by 3cycles. This problem is broken into parts to assist you. (a) In S 4, write the double transition (1, 2)(3, 4) as the product of two 3cycles. Hint: A 3cycle will leave one of the four numbers fixed. The hint is useful. Select the first 3cycle to place 1 into position 2. Select the second 3cycle to leave position 2 fixed and to place 2 into position 1. The first 3cycle can be σ 1 = (1, 2, 3). The second 3cycle σ 2 must satisfy (a) σ 2 (2) = 2 to keep position 2 fixed, and (b) σ 2 (3) = 1 to move 2 (which is in position 3 after σ 1 is applied) into position 1. That is, σ 2 = (3, 1, 4). There are choices for σ 1, but given σ 1, σ 2 is uniquely determined. Check: (3, 1, 4)(1, 2, 3) = (1, 2)(3, 4). It works! Of course, other choices for σ 1 are possible. They are: (1, 2, 4), (2, 1, 3), (2, 1, 4), (3, 4, 1), (3, 4, 2), (4, 3, 1), (4, 3, 2). (b) In S n for n > 3, suppose that i, j, k and l are distinct integers between 1 and n. Write µ = (i, j)(k, l) as the product of two 3cycles. Replace 1, 2, 3 and 4 above by i, j, k and l. Then µ = (k, i, l)(i, j, k) = (i, j)(k, l). (c) In S n for n > 2, suppose that i, j and k are distinct integers between 1 and n. Let µ = (i, k)(i, j). Compute the cycle decomposition of µ. I did this one in class. µ = (i, j, k). (d) Show that any even permutation is a product of 3cycles. Hint: If µ A n, then µ is a product of 2k transpositions for some k > 0 (unless µ = ι). Show that µ is a product of at most 2k 3cycles. An even permutation µ can be written as a product of an even number of swaps (transpositions), say 2k swaps. Then µ is a product of k pairs of swaps. Consider each pair: Case 1: The two swaps involve just two distinct numbers, say i and j. This gives (i, j)(i, j) = ι, the identity permutation. That is, this pair may be deleted (or written as (i, j, k)(i, k, j) for some k not equal to i or j). Case 2. The two swaps involve three distinct numbers, say i, j and k. As shown above (i, k)(i, j) is a 3cycle. Note that the first swap may be written as (i, j) without loss of generality where i is the number that is repeated in the second swap. Case 3 The two swaps involve four distinct numbers, say i, j, k and l. As shown above, this is the product of 2 3cycles.
7 7 Abstract Algebra Professor M. Zuker Thus, µ may be written as the product of at most 2k 3cycles. It may also be written as the product of exactly 2k 3cycles. Why? Case 1, the trivial case, can be written as the product of 2 3cycles. Case 2: A single 3cycle (i, j, k) may also be written as (i, k, j)(i, k, j), which is the square of the inverse of (i, j, k). Case 3: This is already the product of 2 3cyles.
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