Test1. Due Friday, March 13, 2015.

1 Abstract Algebra Professor M. Zuker Test1. Due Friday, March 13, 2015. 1. 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 + 2017n = 1. 2017 = 1 1966 + 51 1966 = 38 51 + 28 51 = 1 28 + 23 28 = 1 23 + 5 23 = 4 5 + 3 5 = 1 3 + 2 3 = 1 2 + 1 2 = 2 1 + 0 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 Abstract Algebra Professor M. Zuker 1 = 3 1 2 = 3 1 (5 1 3) = 2 3 1 5 = 2 (23 4 5) 1 5 = 2 23 9 5 = 2 23 9(28 1 23) = 11 23 9 28 = 11 (51 1 28) 9 28 = 11 51 20 28 = 11 51 20(1966 38 51) = 771 51 20 1966 = 771 (2017 1 1966) 20 1966 = 771 2017 21 1966 = ( 791) 1966 + 771 2017 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 = 791 + 2017k and n = 771 1966k. If k = 0, n is positive. The next smaller value of n is 771 1966 < 0, so ( 791, 771) is the solution with the smallest positive value for n. The next larger value of m is 791 + 2017 = 1226, for which n = 771 1966 = 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 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. Counter-example. 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. Counter-example. 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 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 non-cyclic 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 non-abelian 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 non-cyclic Abelian group. The binary operation is well-defined, 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 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) = 4 2 2 = 16. G is not cyclic. In fact, the maximum order of any a G is 4. 4. Let σ = ( i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 σ(i) 16 13 4 18 6 14 19 2 8 5 7 12 9 17 11 3 10 20 15 1 ) (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 6-cycle (odd), mu 2 is a 4-cycle (odd), mu 3 is a 5-cycle (even) and mu 4 is a 4-cycle (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 k-cycles, so it is even. The fact that k = 8, making both cycles odd permutations, is not relevant.

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 3-cycles. We know that 3-cycles are even permutations. The object is to show that A n is generated by 3-cycles. 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 3-cycles. Hint: A 3-cycle will leave one of the four numbers fixed. The hint is useful. Select the first 3-cycle to place 1 into position 2. Select the second 3-cycle to leave position 2 fixed and to place 2 into position 1. The first 3-cycle can be σ 1 = (1, 2, 3). The second 3-cycle 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 3-cycles. 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 3-cycles. 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 3-cycles. 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 3-cycle. 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 3-cycles.

7 Abstract Algebra Professor M. Zuker Thus, µ may be written as the product of at most 2k 3-cycles. It may also be written as the product of exactly 2k 3-cycles. Why? Case 1, the trivial case, can be written as the product of 2 3-cycles. Case 2: A single 3-cycle (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 3-cyles.