Algebra homework 8 Homomorphisms, isomorphisms

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1 MATH-UA T.A. Louis Guigo Algebra homework 8 Homomorphisms, isomorphisms For every n 1 we denote by S n the n-th symmetric group. Exercise 1. Consider the following permutations: ( ) ( σ 1 =, σ = Compute σ 1 σ 3. For all k {1... 7}, σ 1 σ 3 (k) = σ 1 (σ 3 (k)). ( ) σ 1 σ 3 = ) ( , σ 3 = Recall that both σ 1 and σ 3 can be seen as elements of S 8 by putting σ 1 (8) = σ 3 (8) = 8. Compute the products σ 1 σ 2 and σ 2 σ 3 in S 8. ( ) σ 1 σ 2 = And: σ 2 σ 3 = ( ) 3. Decompose σ 1, σ 2 and σ 3 into products of disjoint cycles, and then write each of them as a product of transpositions. To answer this question, you can iterate σ until you complete a cycle. Inductively, iterate σ over an element that is not within the support of the previous cycles obtained. Do this algorithm until there is no element remaining (except fixed points). To get the decomposition into a product of transpositions, you can use the previous decomposition into a product of cycle, and decompose each cycle into a product of transpositions. σ 1 = (165)(234) = (16)(65)(23)(34). σ 2 = (1256)(478) = (12)(25)(56)(47)(78). σ 3 = ( ) = (14)(47)(72)(23)(36)(65). Exercise 2. Write the following permutations as products of disjoint cycles: ). 1

2 1. (1, 2, 7, 5)(2, 6, 1) This decomposition is not good since 1 and 2 are common elements in each factor of this product of cycle. As previously, every element that doesn t appear in this decomposition is a fixed point, so you just need to compute images of the element in the support of these cycles. You can use the previous algorithm. (175)(26) 2. (1, 5, 2)(2, 3)(5, 7)(1, 2, 3) Same technique, you should have: (13572) 3. (1, 3, 4) 100 There is a periodicity here, since (134) 3 = id. Because (mod 3), (134) 100 = (134) 4. (1, 3, 5, 6) 1 Observe that : (6531) = (1356) 1. Exercise 3. Let a 1,..., a k be distinct elements of {1,..., n}. Compute the inverse in S n of the cycle (a 1,..., a k ). The last question of the previous exercise can provide you with the intuition: (a 1... a k ) 1 = (a k... a 1 ) It is true since (a k... a 1 )(a 1... a k ) = (a 1... a k )(a k... a 1 ) = id (since elements in the support of the cycle is moved once and back to itself, and others are fixed points). Exercise 4. Compute the sets and E = {σ S 4, σ(1) = 3} F = {σ S 4, σ(2) = 2}. Are they subgroups of S 4? S 4 is composed of the following permutations (decomposed into products of disjoint cycles): The identity: id 2

3 2-cycles: (12), (13), (14), (23), (24), (34) Product of 2-cycles: (12)(34), (13)(24), (14)(23) 3-cycles: (123), (124), (132), (134), (142), (143), (234), (243) 4-cycles: (1234), (1243), (1324), (1342), (1423), (1432) Therefore: E = {(13), (13)(24), (132), (134), (1324), (1342)} F = {id, (13), (14), (34), (134), (143)} E can t be a subgroup of S 4 since it doesn t contain the identity. F is a subgroup of S 4, isomorphic to S 3. Indeed, it s a subset of S 4 containing the identity. It is stable by inversion since the inverse of the transpositions are themselves, and the two 3-cycles are inverse one of another. It is closed under composition. Exercise 5. Prove that if σ is a cycle of odd length, then σ 2 is also a cycle. Show that this is not true for cycles of even length by giving a counterexample. In fact it is a necessary and sufficient condition, for a length of cycle greater than or equal to 3. Let σ = (a 1 a 2... a 2k ). Then σ 2 = (a 1 a 3... a 2k 1 )(a 2 a 4... a 2k ). Conversely, assume that σ = (a 1 a 2... a 2k+1 ). It is not hard to see that σ 2 = (a 1 a 3... a 2k+1 a 2 a 4... a 2k ). These results can be proved by induction over the length of the cycle, for instance. But here, only the second implication was asked. A counterexample for a cycle of even length would be (1234), for which (1234) 2 = (13)(24), which is not a cycle. Exercise 6. Let σ S n. 1. Show that σ can be written as a product of at most n 1 transpositions. By theorem, there exist c 1,..., c k k disjoint cycles such that σ = c 1 c 2... c k. Since the cycles are disjoint, their lengths (n i for c i ) must add up to n (assuming that fixed points appear as trivial cycles in this decomposition). For any cycle in this decomposition, we have: (a 1 a 2... a ni ) = (a 1 a 2 )(a 2 a 3 )... (a ni 1a ni ). This is a decomposition in n i 1 transpositions. At the end of the day, if we perform this on each cycle of the product, σ can be decomposed into a product of k n=1 (n i 1) = n k n 1 transpositions. 3

4 2. Show that if σ is not a cycle, then σ can be written as a product of at most n 2 transpositions. If sigma is not a cycle k 2 in the previous proof, which shows that σ can be written as a product of at most n 2 transpositions. Exercise 7. Let σ = σ 1... σ k S n, where σ 1,..., σ k are disjoint cycles. Prove that the order of σ is the least common multiple of the lengths of the cycles σ 1,..., σ k. For this exercise, recall the crucial property that the order of a cycle is equal to its length. Let p be the order of σ. Note that, since disjoint cycles commute, we have σ p = σ p 1... σ p k. Let us show that for every i, the length of σ i divides p. Write σ i = (a 1,..., a m ) and write the Euclidean division of p by m: p = mq + r, where q, r are integers, 0 r < m. Then σ p i (a 1) = σ qm+r i (a 1 ) = (σi m ) q σi r (a }{{} 1 ) = a r+1. =id On the other hand, since the cycles σ 1,..., σ k are disjoint, none of the other cycles σ j, j i moves a 1 or a r+1, so have σ p (a 1 ) = a r+1. Since σ p is the identity, we have r = 0, so m indeed divides p. Thus, the order p of σ is indeed a common multiple of the lengths of the cycles σ 1,..., σ k. It remains to prove that it should be the smallest such multiple. For this, take any positive common multiple d of the lengths of the cycles σ 1,..., σ k. Them for any i, σ i is of order dividing d, so σi d = id. Then σ d = σ1 d... σk d = id. We therefore have proved equality between the sets and {positive integers d such that σ d = id} {positive common multiples of the lengths of σ 1,..., σ k }. This means that in particular the smallest elements of these sets, which are the order of σ and the least common multiple of the lengths of σ 1,..., σ k, respectively, are equal. Remark: note the similarity between this proof and the solution of exercise 7 of homework 5. Exercise Give a list of all possible orders of an element of S 4, then of S 5. According to the previous exercise, the order of a permutation only depends on the lengths of the cycles appearing in its decomposition into disjoint cycles. An element of S 4 is either the identity, or a transposition, or a product of two transpositions, or a cycle of length 3 or 4. Therefore, the possible orders are 1,2,3 or 4. In the same manner, elements of S 5 are of one of the following forms: cycle of length 5 (a, b, c, d, e), order 5 cycle of length 4 (a, b, c, d), order 4 cycle of length 3 (a, b, c), order 3 product of a cycle of length 3 and a transposition (a, b, c)(d, e), order lcm(3, 2) = 6 product of two transpositions (a, b)(c, d), order lcm(2, 2) = 2 transposition (a, b), order 2 identity id, order 1. 4

5 2. Show that an element of S 7 can t have order 8, 9 or 11. Assume σ S 7 has order 8. Then σ cannot be a cycle, because if it were a cycle, it would be of length 8, but S 7 only contains cycles of length at most 7. Therefore, we may write σ = σ 1... σ r as a product of disjoint cycles with r 2, each σ i being different from the identity. By exercise 7, we now that the orders of σ 1,..., σ r divide 8. Moreover, none of them can be equal to 8 because there are no cycles of length 8, nor to 1, because we assumed these cycles to be different from the identity. Therefore, these orders are equal to 2 or 4. But the least common multiple of 2 and 4 is 4, so σ should then be of order 4, a contradiction. Exactly the same argument applies to show that σ S 7 cannot be of order 9: It cannot be a cycle because S 7 does not contain any cycles of length 9. Therefore it is a product of at least two disjoint cycles, which have to be of length dividing 9 but different from 1 and 9. Thus, it is a product of cycles of length 3, but then it would be itself of order 3, a contradiction. The argument is even simpler to show that we don t have elements of order 11, because 11 is a prime number. 3. Give a list of all possible orders of an element of S 7. Possible decompositions into a product of disjoint cycles in S 7 are the following. (abcdef g), order 7 (abcdef), order 6 (abcde), order 5 (abcde)(f g), order 10 (abcd), order 4 (abcd)(ef g), order 12 (abcd)(ef), order 4 (abc), order 3 (abc)(def), order 3 (abc)(de), order 6 (abc)(de)(fg), order 6 (ab), order 2 (ab)(cd), order 2 (ab)(cd)(ef), order 2 id, order 1 Therefore, the possible orders are 1, 2, 3, 4, 5, 6, 7, 10 and 12. We are certain we didn t miss any possible orders because we proved 8,9 and 11 to be impossible in the previous question. 5