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 n 1 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.
Proof. For given 휎 ∈ 푆 푛 with negative direction of all the edges, note that 휎 fixes no ri or s j , as (ri ) ri if and only if we have a positive edge between 푟푖 and ri . Similarly (s j ) s j if and only if we have a positive edge between s j and s j . Therefore all the numbers 1,2,3,...,n,n 1,...2n will be involved in the cycles of 휎. We discuss different cases for involvement of even or odd number of numbers in a cycle of 휎. Here we also use the fact observed in Lemma 2.3 that in any cycle of 휎 ∈ 푆푛 , no ri , rj or si , s j occur together as all the edges are of negative direction. Case (i) Suppose we want to involve two numbers from each side of n in the product of cycles of 휎, then the only possibility is ri s j (s j ri ) and ri s j . In this case, we get two cycles (ri s j ) and (ri s j ). The possibility of having these four numbers in a single cycle is ruled out because these four numbers will be in one cycle if and only if we have a positive edge between s j and rj . Further, if we want to use four numbers from each side of n, then
r1 s1 r2 s2, and r1 s1 r2 s2.
(Note that without loss of generality, we can start with r1, because if we start with s 1 then we are writing the second row in place of first). Here we take si ri , otherwise we get two cycles of length 2, which we have already discussed). These sequences will terminate if s2 r1 and s2 r1 , because s2 r1 , s2 r1 being on the same side of n as 휎 has all negative edges. So we get two cycles r1 s 1 r 2 s2 and r1 s1 r 2 s2. In general, the involvement of an even number of numbers from each side is possible if and only if we take
r1 s1 r2 s2 ... ri si , and
r1 s1 r2 s2 ... ri si . Again, in this case, we get two cycles r1 s 1 r 2 s 2 . . . . ri si and r1 s1 r 2 s2. . . . . ri si . Case (ii) Now suppose that only one number from either side of n is involved in a cycle of 휎 , then ri ri and we get only one cycle (ri ri ) , a negative transposition. Three numbers can be involved from either side of n in a cycle of 휎 if and only if r1 s1 r2,s1 r1 and r2 s1 and r1 s1 r2. These sequences will terminate if r 2 r1 and r2 r1, because r2 r1 and r 2 r1 being on the same side of n. So, we get a single cycle r1 s 1 r 2 r1 s1 r2. In general, if an odd number of 푟푖 and 푠푖 are to be involved from either side of n in a cycle of 휎, then r1 s1 r 2 s2 r3 s3 r4. . . . . si1 ri and r 1 s1 r2 s2 r3 s3 r4. . . si1 ri . These sequences will terminate if r i r1 and ri r1, because ri r1 and r i r1 being on the same side of n. So we get a single cycle r1 s1r2 s2. . . si1 r i r1 s1 r 2 s2. . . . si1 ri of even length. Therefore we conclude that there are two type of cycles in the product of : (i) cycles having even number of 푟푖 ≤ 푛 and (ii) cycles having odd number of 푟푖 ≤ 푛. The cycles of first kind exist in pair. A cycle of second kind, having length 2t, corresponds to some odd number t which exists in 휆 ⊢ 푛. Therefore if n is even, then corresponding to an odd number t in the partition 휆 ⊢ 푛 , there exists 푡′ an odd number in the same partition 휆 ⊢ 푛. Hence we get a pair of cycles of type 2t and 2t’ (see case (ii)). Hence, the number of cycles in the product of 휎 will be even if n is even and odd if 푛 is odd.
As an easy consequence of the above result, we have
Corollary 2.7. Any cycle in the product of 휎 ∈ 푆 푛 having all the edges of negative direction, ends with some 푠푗 if it starts with 푟푖.
Corollary 2.8. Let 휎 ∈ 푆 푛 with n even, has negative direction of all the edges. Then, 휎 is not a 2푛 –cycle in 푆2푛. Proof. This falls in the case (i) of the above theorem. In this case, for each cycle, there corresponds another cycle of the same length, in the product of 휎. Hence any such 휎 cannot be a single cycle of length 2n.
Using the above results proved for 휎 ∈ 푆 푛 having negative direction of all the edges, we have
Proposition 2.9. Let 휎 ∈ 푆 푛 with n even, has negative direction of all the edges. Then 휎 is an even permutation in 푆2푛. Proof. By lemma 2.6, number of disjoint cycles of 휎 in 푆2푛 is even if and only if n is even. Also by lemma 2.3, length of each cycle of 휎 is even. Even number of disjoint cycles with even length will give even number of transpositions, hence 휎 is an even permutation in 푆2푛.
3. Main Results
Now we prove the main results of this paper. n Theorem 3.1. For 푛 ≥ 2, let n 2m r,0 m 2 and 0 r n. Then 휎 ∈ 푆푛 is an even permutation in 푆2푛 if and only if 휎 has 2m edges of negative direction. Proof. First, we analyze the role of positive and negative edges to get a cycle in the product of 휎 in 푆2푛. One negative edge between ri and ri gives a transposition (ri ri ) in 푆2푛. Thus, if we have such 2m negative edges, then we get an even number of transpositions (r1 r1 ), (r2 r2 ), (r3 r3 ),...,(r2m r2m ) in 푆2푛 corresponding to these negative edges. The remaining positive edges give pair of cycles of equal length as observed in Lemma 2.1 and hence 휎 is an even permutation in 푆2푛 . Suppose some mixed positive and negative edges give a cycle in the product of .
Note that two numbers ri , rj or si , s j in a cycle occur if and only if there is a positive edge and two numbers ri , s j or si , rj occur if and only if there is a negative edge. Case (i) Suppose a sequence of the type
r1 .... si ... rj occurs inside a cycle of . In this sequence, there are even number of changes from ri to si and s j to ri (possibly no change). As observed above this sequence occurs if and only if there exists a sequence of even number of negative edges (possibly no negative edge). But then we also have r1 ... si ... rj . In order to complete the cycles either rj r1 and hence rj r1 or rj r1 and hence rj r1. Here we are not taking the case rj rj because that will remove rj and rj from this cycle and single cycles have been discussed earlier. Subcase (i) rj r1 and rj r1 . This is only possible with an additional positive edge between rj and r1 . Therefore the number of negative edges used remain even. Hence in this case, we have used even number of negative edges with some positive edges and we get two cycles (r1....si ... rj ) and (r1 ....si ....rj ) contributing an even number of transpositions in the product of . Subcase (ii) rj r1 and rj r1. This is only possible with an additional negative edge between rj and r1 . Hence in this case, number of negative edges used changes to odd and we get a single cycle (r1....si ... rj r1 ....si ....rj ). Clearly the length of this cycle is even as number of ri = number of ri and number of si = number of si . This contributes an odd number of transpositions in the product of 휎. Case (ii) Now consider a sequence of the type
r1 ... ri ... si ... s j .
Here, there are odd number of changes from ri to si and s j to rj (both included). Hence this sequence occurs inside a cycle of 휎 if and only if there exists a sequence of odd number of negative edges. But then we also have r1 ... ri ... si ... s j . Again in order to complete the cycle either s j r1 and hence s j r1 or s j r1 and hence s j r1. Subcase (i) s j r1 and s j r1 . This is only possible with an additional negative edge between s j and r1 . Hence, in this case, number of negative edges used, changes to even and we get two cycles (r1....ri ....si ....s j ) and (r1 ....ri ... si ...s j ) contributing an even number of transpositions in the product of . Subcase (ii) s j r1 and s j r1. This is possible only with an additional positive edge between s j r1 . In this case number of negative edges used remains odd and we get a cycle (r1....ri ....si ....s j r1 ....ri ... si ....s j ) of even length. This contributes in product of an odd number of transpositions.
Conclusion: With the above observations, finally we conclude that an even number of negative edges (possibly zero) together with some positive edges contribute two cycles of same length in the product of 휎 and an odd number of negative edges (possibly one) contribute a product of odd number of transpositions for 휎 . If we take any odd number r in a partition of 2m , then there exists an another odd number r in this partition. Hence, by above discussion it follows that in the product of 휎, there are pairs of cycles of equal length contributing to even number of transposition and some more even number of transpositions coming from the pairs (r , r) of negative edges ( r and r are odd numbers).
Therefore, 휎 is an even permutation in 푆2푛.
Let 휎 ∈ 푆 푛 has 2푘 + 1 (푘 ≥ 1) edges with negative direction. Then in any partition of 2푘 + 1, the number of odd numbers will be odd. Therefore, from the above discussion, the total number of transposition in the product of 휎 in 푆2푛 will odd.
Theorem 3.2. Let 퐴 푛 = {휎 ∈ 푆 푛 | 휎 has even number of negative edges. Then 푛−1 퐴 푛 is a normal subgroup of 푆 푛 of order 2 푛!. Further 퐴 푛 = 푆 푛 ∩ 퐴2푛 and 푛−1 hence 푆2푛 has a subgroup of order 2 푛!. In particular, 푆푛 has a subgroup of 푛 −1 푛 order 2 2 of even permutations. 2 Proof. The multiplication defined in 퐵 푛 yields that 퐴 푛 ≤ 퐵 푛 . The result 퐴 푛 = 푛−1 푆 푛 ∩ 퐴2푛 follows from Theorem 3.1. The order of 퐴 푛 is 2 푛! as observed in the Introduction. Also, [푆 푛 : 퐴푛 ] = 2 implies that 퐴 푛 ⊲ 푆 푛 . Since 푆2푛 is a subgroup of 푆2푛+1, we have a more general result, 푆푛 has subgroups of order 1, 푛 −1 푛 22−12!, 23−13!, . . ., 2 2 ! for 푛 ≥ 2. 2
Example 3.3. For k = 2 and 3, we have 2=2.0+2, 2=2.1+0 and 3=2.0+3,
3=2.1+1. Therefore, 푆6 has two subgroups 퐴 2 and 퐴 3 of order 6 −1 6 2 2 2! + 2 2! = 22−12! = 4 and 3 3! + 3 3! = 2 2 ! 2 3! 24 퐶0 퐶2 퐶0 퐶2 2 respectively, which are obtained from the elements of 푆 2 and 푆 3 having even number of negative edges.
The subgroup 퐴 2 can be written from Example 1.1; that is,
{e,(13)(24),(12)(34),(14)(23)}. Here, we find the subgroup 퐴 3 as follows:
, ,
, ,
, ,
,
, ,
, ,
, , , ,
, ,
, ,
, ,
,
Finally, we obtain a result as an application of Theorem 3.2.
푛 Every group of order 푛 푛 − 1 … ( + 1) has a 푆푦푙표푤 2 − Corollary 3.4. 2 푛
푠푢푏푔푟표푢푝 subgroup of order 2 2 . 푛 푛 Proof. Using theorem 3.2, 2 2 ! | 푛!, hence 2 푛 −1 푛 2 2 | 푛 푛 − 1 … + 1 . We claim that 2 푛 +1 푛 2 2 ∤ 푛 푛 − 1 … + 1 , n≥ 2. 2 For, we show that
푛 푛 푛 + 1 + 2 …n = 2 2 푘, for some positive odd integer 푘. 2 2 The result is true for 푛 = 2. Suppose that the result is true for all integers 푡, 2 ≤ 푡 ≤ 푛.
Case(i) If 푛 is even, then 푛 n n 22 푘 = + 1 + 2 … 푛 2 2 implies
푛+1 푛 + 1 푛 + 1 2 2 푘 = + 1 + 2 … 푛, 2 2 which yields
푛+1 푛+1 푛+1 2 2 푘 푛 + 1 = + 1 + 2 … (푛 + 1), 2 2 where 푘 푛 + 1 is odd.
Case(ii) If 푛 is odd, then
푛 푛 푛 2 2 푘 = + 1 + 2 … 푛 2 2 implies
푛−1 n + 1 n + 3 2 2 푘 = … 푛, 2 2 which yields
푛+1 n + 1 n + 1 2 2 푘 = + 1 + 2 … 푛 + 1 . 2 2 푛 So every group of order 푛 푛 − 1 … ( + 1) has a 푆푦푙표푤 2 − 푠푢푏푔푟표푢푝 of 2 푛 order 2 2 .
References
[1] C. Musli, Representations of finitie groups, Hindustan book agency , 1993.
[2] M. Parvathi and M. Kamaraj, Signed Brauer's algebras, Comm.in Algebra, 26(3) (1998), 839-855.
[3] M. Parvathi and C. Selvaraj, Signed Brauer's algebras as Centralizer algebras, Comm. in Algebra, 27(12), 5985-5998 (1999).
[4] R.P. Sharma and V.S. Kapil, Irreducible S_{n}-Modules and A cellular structure of the Signed Brauer Algebras, Southeast Asian Bulletin of Mathematics, (2011), 35: 497-522.
[5] R. P. Sharma and V. S. Kapil, Generic irreducibles of the Brauer algebras, Contemporary Mathematics (AMS), 456(2008), 189-204.
[6] H. Wenzl, On the structure of Brauer's centralizer algebras, Ann. of Math., (1988):173-193.