Subgroups of Symmetric Groups Determined Using Signed Brauer Diagrams Ram Parkash Sharma and Rajni Parmar Department of Mathematics, Himachal Pradesh University Summerhill, Shimla (India) Email: [email protected] Abstract The group 푆 푛 of signed Brauer diagrams with no horizontal edges becomes a subgroup of 푆2푛 . Therefore, it is natural to find the elements of 푆 푛 which are 푛−1 even permutations in 푆2푛 . There are 2 푛! such elements that form a subgroup 2−1 3−1 of 푆2푛 . Consequently, 푆푛 has a class of subgroups of order 1, 2 2!, 2 3!, . ., 푛 푛 −1 푛 푛 2 2 ! of even permutations and hence 2 2 ! | 푛!. Finally, using these 2 2 푛 results it is shown that every group of order 푛 푛 − 1 … ( + 1) has a 2 푛 푆푦푙표푤 2 − 푠푢푏푔푟표푢푝 of order 2 2 . Keywords: Signed Brauer algebras; symmetric groups; even permutations. AMSC: 20B30, 20B35. 1. Introduction Parvathi and Kamaraj [2] defined signed Brauer algebras 퐵 푛 푥 over ℂ, (푥 ∈ ℂ) having basis consisting of signed diagrams 푑 , which consist of two rows of 푛 points labelled {1,2, … , 푛} in first row and {(푛 + 1), (푛 + 2), … ,2푛} in second row, with each dot joined to precisely one other dot (distinct from itself). Each edge of the diagram 푑 has a direction: either positive or negative. A positive vertical (horizontal) edge is denoted by ↓ (→) and a negative vertical (horizontal) edge is denoted by ↑ (←). An element in 퐵 5 푥 is given as: To define multiplication in 퐵 푛 (푥), it is enough to define the product of two signed diagrams 푑1 and 푑2, to do so place 푑1 above 푑2 and join corresponding points (interior signed loops are deleted). Let 푑3 be the signed diagram obtained in this way. An edge in 푑3 is positive (negative) according as the number of negative edges involved to form this edge is even (odd). Similarly, a loop β in 푑1휊푑2 is so called positive (negative). The multiplication is then defined 2푛1+푛2 푑1휊푑2 = 푥 푑3 where 푛1 and 푛2 are the numbers of positive and negative loops respectively formed placing 푑1 above 푑2. The set of signed Brauer diagrams 퐵 푛 contains the set of Brauer diagrams 퐵푛 (each Brauer diagram corresponds to same underlying diagram with all positive edges). For more detail on Brauer and signed Brauer algebras, one can refer to [3,4,5,6]. The Brauer diagrams with no horizontal edges form a permutation group, by identifying a Brauer diagram with a permutation 휋 connecting the ith upper vertex to the (i)th lower vertex. In a similar way, the signed Brauer diagrams with no horizontal edges form a group 푆 푛 which corresponds to the wreath product of Z2 by 푆푛 [see 1]. The group 푆푛 becomes a subgroup of 푆2푛 . In order to identify 푆 푛 as a subgroup of 푆2푛 , first divide the set 1,2,. n,. ,2n into two equal parts that is the numbers n and n. To distinguish the numbers less than or equal to 푛 and greater than 푛 , we fix the notation r1,r2,. ,rn for the numbers less than or equal to n and 푠1, 푠2, … , 푠푛 for the numbers greater than n. We fix another notation as: For each i 1,2,. ,2n, we denote by i the number given by: i n, 1 i n, i . i n, n i 2n ∗ ∗ Thus 푟푖 > 푛; 푖 = 1,2, … , 푛 and 푠푗 ≤ 푛; 푗 = 1,2, … , 푛. The elements of 푆 푛 are identified in 푆2푛 as follows: Let 휎 ∈ 푆 푛 have a negative edge ∗ ∗ between 푟푖 and 푠푘. Then the number 푟푖(≤ 푛) ⟼ 푠푘 (> 푛) and 푟푖 ⟼ 푠푘 under 휎 in 푆2푛 and vice-versa. If we have a positive edge ∗ ∗ between 푟푖 and 푠푘 . Then 푟푖 ⟼ 푠푘 and 푟푖 ⟼ 푠푘 . Example 1.1. Consider the eight elements of 푆 2 given below: According to the above definition, the elements of 푆 2 are identified in 푆4 as follows: 푒, 2 4 , 1 3 , 1 3 2 4 , 1 2 3 4 , 1 2 3 4 , 1 4 3 2 , (1 4 )(2 3) . As 푆 푛 is a subgroup of 푆2푛 , we started this paper with the aim of finding the elements of 푆 푛 which correspond to even permutations in 푆2푛 and prove that the elements of 푆 푛 having even number of negative edges (see Theorem 3.1) are even permutations in 푆2푛 . Since the product of two negative edges in 퐵 푛 gives a positive edge and product of negative and positive edges is a negative edge, it is easy to see that these elements form a subgroup of 푆 푛 and hence of 푆2푛 . The total number of elements with even number of negative directions in 푆 푛 is given by: n n n n n ( c0 c2 c4 c6 ... cn )n!,if n is even n n n n ( c0 c2 c4 ... cn1)n!,if n is odd n1 2 n!. 푛−1 Thus, 푆2푛 has a subgroup of order 2 푛!. In particular, 푆푛 has a subgroup of 푛 −1 푛 order 2 2 ! for 푛 ≥ 2 and hence in general it has a class of subgroups of 2 푛 −1 푛 order 1, 22−12!, 23−13!, … , 2 2 ! of even permutations. In consequence of 2 푛 푛 it, 2 2 ! | 푛! ; hence 2 푛 푛 2 2 | 푛 푛 − 1 … + 1 . 2 Further, we observe that (Corollary 3.4) 푛 +1 푛 2 2 ∤ 푛 푛 − 1 … + 1 . 2 푛 Therefore, every group of order 푛 푛 − 1 … ( + 1) has a 푆푦푙표푤 2 − 푠푢푏푔푟표푢푝 2 푛 of order 2 2 . 2. Even Permutations of 푆 푛 in 푆2푛 First, we prove that the elements of 푆 푛 having positive direction of all the edges are even permutations in 푆2푛 . We, very often, use ri s j for (ri ) s j and ri s j for (ri ) s j when 휎 is understood. Lemma 2.1. If 휎 ∈ 푆 푛 has positive direction of all the edges, then 휎 is an even permutation in 푆2푛 . Proof. Let 휎 ∈ 푆푛 has all edges of positive direction. Then ri n, goes to some rj (s j ) n under 휎. Consequently, ri goes to some rj (s j ) s j and both ri , s j n. In this way, we get r1 r2 r3... ri, and r r r. 1 2 r3 ... i Note that ri r1 and ri r1 because this is only possible if we have a negative edge between r i and r1. Since direction of all the edges of 휎 is * positive, there is only one possibility of having ri r1 and hence ri r1 using a positive edge between 푟푖 and Therefore, corresponding to every cycle r1 r 2 r3. ri written using the numbers ≤ 푛, there is another cycle r1 r 2 r3 ...ri ) of the numbers greater than n having same length. So 휎 is an even permutation in 푆2푛 . Example 2.2. Consider an element of 푆 5 having all edges of positive direction. This element becomes (1 3 5 2 4)(6 8 10 7 9 ) in 푆10. For 휎 ∈ 푆 푛 with negative direction of all the edges, any cycle 푐푖 , in the product of cycles of 휎, no two numbers 푟푖 , 푟푗 occur together and same is true for 푠푖 , 푠푗 also. With this observation, we have Lemma 2.3. Let 휎 ∈ 푆 푛 has negative direction of all the edges. Then, length of each cycle in the product of 휎 in 푆2푛 is always even. Proof. Since all the edges are of negative direction, any number ri n goes to a number s j n. Since s j n, s j can be ri . In this case, we get a transposition (ri ri ) . Let (r1 s1...) be a cycle in the product of . In this cycle no ri ,rj or si , s j occur together as all the edges are of negative direction. Moreover the last number will be some sk n. For, if we have a cycle of the type (r1 s1 r2 s2 ... rk ) . Then * rk r1 gives a positive edge between rk and r1 . Hence any cycle in the product of 휎 will be of the type (r1 s1 r2 s2 ...rk1 sk ) having length 2k. The above result does not depend on whether 푛 is even or odd. However, the number of cycles in the product of 휎 ∈ 푆 푛 having negative direction of all the edges depends upon n as observed in the following examples. Example 2.4. For n to be odd, consider an element of 푆 5 having all edges of negative direction. This element becomes (1 7 3 9)(6 2 8 4)(5 10) in 푆10. Clearly, length of each cycle in the product of 휎 is even, but number of cycles is 3. Example 2.5. For n to be even, consider an element of 푆 6 having all edges of negative direction which becomes (1 8 4 9)(7 2 10 3)(5 12)(6 11) in 푆12. Here, again the length of each cycle is even, but the number of cycles is also even. With these observations, we prove a general result regarding the number of cycles of 휎 ∈ 푆 푛 having negative direction of all the edges. Lemma 2.6. Let 휎 ∈ 푆 푛 has negative direction of all the edges. Then the number of disjoint cycles in the product of 휎 is even if n is even and odd if n is odd.