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Introduction to Group Theory Solutions 5 Unit website: http://people.maths.bris.ac.uk/~majcr/IGT.html Questions 1,2,3 to be handed in for tutorials. 1. Let U11 be the group of non-zero elements of Z/11Z under multiplication (mod 11). (a) For each element of U11 find its inverse. (b) Prove that U11 is cyclic by finding a generator. Solution: (a) U11 = {[1], [2], [3], [4], [5], [6], [7], [8], [9], [10]}, since 11 is prime and so hcf(a, 11) = 1 for all 0 < a < 11. [1] is its own inverse. [2][6] = [12] = [1], so [2] and [6] are inverses of one another. [3][4] = [12] = [1], so [3] and [4] are inverses of one another. [5][9] = [45] = [1], so [5] and [9] are inverses of one another. [7][8] = [56] = [1], so [7] and [8] are inverses of one another. [10][10] = [100] = [1] (or, simpler, [10] = [−1] and [−1][−1] = [1]), so [10] is its own inverse. (b) Calculating the powers of [2]: [2]2 = [4], [2]3 = [8], [2]4 = [16] = [5], [2]5 = [2][5] = [10], [2]6 = [2][10] = [20] = [9], [2]7 = [2][9] = [18] = [7], [2]8 = [2][7] = [14] = [3], [2]9 = [2][3] = [6], and [2]10 = [2][6] = [12] = [1]. So ord([2]) = 10, and so U11 is cyclic, with [2] as a generator. (We’ll soon see a short cut for answering questions like this). [Other generators are [6], [7] and [8].] 2. Let Un be the multiplicative group of elements of Z/nZ with multiplicative inverses. List the elements of Un and find the order of each element in the following cases: (1) (2) (3) (4) n = 9. n = 12. n = 14. n = 15 In which of these cases is Un cyclic? Solution: (1) U9 = {[1], [2], [4], [5], [7], [8]}, since hcf(a, 9) = 1 unless 3 divides a. [1] is the identity and has order 1. [2]2 = [4], [2]3 = [8], [2]4 = [16] = [7], [2]5 = [2][7] = [14] = [5], [2]6 = [2][5] = [10] = [1], so ord([2]) = 6. [4]2 = [16] = [7], [4]3 = [4][7] = [28] = [1], so ord([4]) = 3. [5]2 = [25] = [7], [5]3 = [5][7] = [35] = [8], [5]4 = [5][8] = [40] = [4], [5]5 = [5][4] = [20] = [2], [5]6 = [5][2] = [10] = [1], so ord([5]) = 6. [7]2 = [49] = [4], [7]3 = [7][4] = [28] = [1], so ord([7]) = 3. [8]2 = [64] = [1], so ord([8]) = 2. Since |U9 | = 6 and there are elements of order 6, U9 is cyclic. (2) U12 = {[1], [5], [7], [11]} since hcf(a, 12) = 1 unless a is even or divisible by 3. [1] is the identity and has order 1. [5]2 = [25] = [1], so ord([5]) = 2. [7]2 = [49] = [1], so ord([7]) = 2. [11]2 = [−1]2 = [1], so ord([11]) = 2. Since |U12 | = 4 but there are no elements of order 4, U12 is not cyclic. (3) U14 = {[1], [3], [5], [9], [11], [13]}. [1] is the identity and has order 1. [3]2 = [9], [3]3 = [27] = [13], [3]4 = [3][13] = [39] = [11], [3]5 = [3][11] = [33] = [5], [3]6 = [3][5] = [15] = [1], so ord([3]) = 6. [5]2 = [25] = [11], [5]3 = [5][11] = [55] = [13], [5]4 = [5][13] = [65] = [9], [5]5 = [5][9] = [45] = [3], [5]6 = [5][3] = [15] = [1], so ord([5]) = 6. [9]2 = [81] = [11], [9]3 = [9][11] = [99] = [1], so ord([9]) = 3. [11]2 = [121] = [9], [11]3 = [11][9] = [99] = [1], so ord([11]) = 3. [13]2 = [−1]2 = [1], so ord([13]) = 2. Since |U14 | = 6 and there are elements of order 6, U14 is cyclic. (4) U15 = {[1], [2], [4], [7], [8], [11], [13], [14]}. [1] is the identity and has order 1. [2]2 = [4], [2]3 = [8], [2]4 = [16] = [1], so ord([2]) = 4. [4]2 = [16] = [1], so ord([4]) = 2. [7]2 = [49] = [4], [7]3 = [7][4] = [28] = [13], [7]4 = [7][13] = [91] = [1], so ord([7]) = 4. [8]2 = [64] = [4], [8]3 = [8][4] = [32] = [2], [8]4 = [8][2] = [16] = [1], so ord([8]) = 4. [11]2 = [121] = [1], so ord([11]) = 2. [13]2 = [−2]2 = [4], [13]3 = [−2][4] = [−8] = [7], [13]4 = [−2][7] = [−14] = [1], so ord([13]) = 4. [14]2 = [−1]2 = [1], so ord([14]) = 2. Since |U15 | = 8 but there are no elements of order 8, U15 is not cyclic. 3. Prove that [323] ∈ Z/1451Z has a multiplicative inverse and find it using Euclid’s algorithm. Solution: Applying Euclid’s Algorithm: 1451 323 159 5 = = = = (323 × 4) + 159 (159 × 2) + 5 (5 × 31) + 4 (4 × 1) + 1, so hcf(323, 1451) = 1, and so [323] does have a multiplicative inverse. Working backwards through the calculation above: 1 = = = = = = = 5−4 5 − (159 − (5 × 31)) −159 + (5 × 32) −159 + (323 − (159 × 2)) × 32 (323 × 32) − (159 × 65) (323 × 32) − (1451 − (323 × 4)) × 65 −(1451 × 65) + (323 × 292), and so [292] is the multiplicative inverse of [323], since 323 × 292 ≡ 1 (mod 1451). 4. Let (S, ∗) be a set with a binary operation satisfying all the group axioms except for the existence of inverses. I.e, ∗ is associative, and S has an identity element e such that e ∗ x = x = x ∗ e for all x ∈ S. (a) Prove that if G is the set of elements x ∈ S that have an inverse in S (i.e., such that there is x−1 ∈ S such that xx−1 = e = x−1 x), then (G, ∗) is a group. (b) Give some examples of groups you have seen that arise naturally in this way. Solution: (a) If x, y are invertible, then (xy)(y −1 x−1 ) = e = (y −1 x−1 )(xy), so xy is invertible, with inverse y −1 x−1 . So G is closed. Associativity holds in G, since it does in S. The identity e is invertible, being equal to its own inverse, so e ∈ G, and G has an identity. If x is invertible, then x−1 is invertible with inverse x, so x ∈ G ⇒ x−1 ∈ G, and so every element of G has an inverse in G. So G is a group. (b) Some examples: • (R \ {0}, ×) in (R, ×). • ({1, −1}, ×) in (Z, ×). • (Un , ×) in (Z/nZ, ×). • GL(n, R), the group of invertible n×n matrices, in (Mn (R), ×). 5. Let p > 3 be a prime. (a) What is the order of U3p ? (b) If [a] ∈ U3p prove that a2 ≡ 1 (mod 3) and ai ≡ 1 (mod p) for some i with 0 < i < p. [Hint: Consider possible orders of [a] considered as an element of U3 and Up .] (c) Deduce that U3p is not cyclic. Solution: (a) Of the numbers 0, 1, . . . , 3p − 1, p are divisible by 3 and another two (p and 2p) are diisible by p. So 2p − 2 are coprime to 3p, and so |U3p | = 2p − 2. (b) If hcf(a, 3p) = 1 then hcf(a, 3) = 1, and so the congruence class of a (mod 3) is in U3 , every element of which has order 1 or 2, and so a2 ≡ 1 (mod 3). Also hcf(a, p) = 1, and since |Up | = p − 1, every element of Up has order less than p, and so ai ≡ 1 (mod p) for some 0 < i < p. (c) Taking i as in (b), aj ≡ 1 (mod 3p) if j is even (so that we are guaranteed aj ≡ 1 (mod 3)) and divisible by i (so that aj ≡ 1 (mod p)). So if i < p − 1 then aj ≡ 1 (mod 3p) for j = 2i < 2p − 2, and if i = p − 1 then aj ≡ 1 (mod 3p) for j = p − 1 < 2p − 2. In either case, ord([a]) < 2p − 2 = |U3p |, and so U3p is not cyclic, having no element whose order is |U3p |. 6. Prove that if p and q are two odd primes with p 6= q, then Upq is not cyclic (generalizing question 5). Solution: Follow the same strategy as in question 5. Of the integers 0, 1, . . . , pq−1, p are divisible by q, q are divisible by p, and one (zero) is diisible by both, so |Upq | = pq−p−q+1 = (p−1)(q−1). Let [a] ∈ Upq . The order np of a as an element of Up is at most p − 1. The order nq of a as an element of Uq is at most q − 1. So n = lcm(np , nq ) < (p − 1)(q − 1) since either np < p − 1 or nq < q − 1 or np = p − 1 and nq = q − 1 in which case lcm(np , nq ) divides (p − 1)(q − 1)/2. But an ≡ 1 (mod p) and an ≡ 1 (mod q), and so an ≡ 1 (mod pq) and so ord([a]) ≤ n < |Upq |, and so Upq is not cyclic. c University of Bristol 2017. This material is copyright of the University unless explicitly stated otherwise. It is provided exclusively for educational purposes at the University and is to be downloaded or copied for your private study only.