Construction of Character Tables of Symmetric Group of Degree 3 and 4 (S 3, S 4,)

Full Length Research Paper

Construction of character tables of symmetric group of degree 3 and 4 (S 3, S 4,)

Isa Sani and Tijjani Bukar

1Federal University Gashua. Nigeria 2Department of Mathematics Yobe State University Damaturu Nigeria

Accepted 13 July, 2015

Since most problems in representation of finite groups are best solved by character tables, we have constructed character tables of symmetric group of degree 3 and 4 (S 3 and S 4). Lifting process and orthogonality relations were applied in deriving the solutions. The results revealed that the symmetric group of finite representation can be completely reducible using character tables.

Key words: Symmetric group, orthogonality, finite representation.

INTRODUCTION

In mathematics, especially the field of algebra which simple group cannot have a generalized quartenim group deals with group theory, the character of a representation However, one may see that one dwelt so much on the plays a very important role. It is a function on the group as its sylow – 2 – subgroups. Background of characters, which associates to each group element the trace of the this is because it will enable us acquire enough corresponding matrix. Characters were introduced by information that will guide us in the construction of George Frobenius (1849-1917). He initially developed various character tables. Thus, it is possible the one may representation of finite groups entirely based on study character and representatives of group with one characters. This is possible because a complex knowing much about character tables, but it is not representation of a group is determined up to possible to construct character table without enough (Isomorphism) by its characters. Group representation knowledge on character theory. Now for proper over a field of positive characteristics which is called understanding of this paper we need to explain some (Modular Representation) is very delicate, but Richard terminologies used. Moreover, there are some basic Brauer developed a very useful theory of characters theorem corollaries and lemma that we use to back our which helps a lot in this case as well. Hence, many deep points. theories on the structure of finite groups use the character of modular representations. Serre (1979), Definition 1 (Baker (2011)) stated that, characters are easy to tabulate and compute with, using character tables. A representation of a group G over a field F is a They expressed deep properties of irreducible homomorphism φ from G to GL (nF) [an nxn matrix with representations. Therefore they are essential tools in the entries in F) for some n, and the degree of φ is the integer classification of finite groups. Close to half of the proof of n. Therefore, if φ is a function from G → Gl (nF), then φ is Walter Felt Thompson’s theorem involve indicate a representation if and only if φ (gh) = φ (g) φ(h) Vg, h ∈G. calculations with character values. The Burnside theorems, Michio Suzuki, all use a lot of character theory Definition 2 (Momoh, 2010) in providing their theorems. For example, a theorem of Richard Brauer and Michio Suzuki stating that a finite Let G be a finite group and V a rector space over a field F. A homomorphism φ of G into the group GL (VF) of non singular transformation of V is called a representation of G. Therefore φ is a representation of G if for all gh ∈G *Corresponding author’s e-mail: [email protected]. Sani and Bukar 9

then φ(gh) = φ(g) (h), and the image of φ, im φ = GL (VF). over F. Furthermore, the G-module V over F is said to be completely reducible if V = U 1 ⊕ U2 ⊕… ⊕ Ui. Where U i ⇒ l=1, 2…1 are irreducible G-module over F. Equivalence Representation Definition 7 (Reducible representation Matrix Let φ: G → GL (mf) and θ:G φ GL (nF) be 2 Version) representation of G over F. we say that φ is equivalent to θ if n = m and there exists invertible nxn matrix T such The matrix representation A(x) is irreducible over F if ∃ a that for all g belongs to G, θ(g) = T -1 φ (g) T. (FG – non-singular matrix T over F such that Module): B(X) = T 1 A (X) T = C(x) 0  for all x ∈G   Definition 3  E(x) D(x)

Let V be a rector space over a field F and let G be a Where if A(X) ∈ GL (m), C(x) an r x r, D(X) an S x S group. Then V is an FG-module if multiplication Vg (for all Then E (x) is an S x r and m = str v∈V, g ∈G) is defined satisfying the following conditions. For all uv ∈ V, λ ∈ F and V gh ∈G, then We assume that the reader is familiar with Group theory and its properties, Group homomorphism, normal i. V ig∈G subgroups, conjugacy, cosets of a group and their likes, ii. V i (gh) = V ig) h because this will be relevant in our field of study of iii. V i I = V I character theory and construction of the character tables. iv. ( λ1V1 + λ2V2 +…. + λnVn) g = λ1(V 1g) + λ2 (V 2g) + …+

λn (Vng) RESULTS Definition 4 The irreducible characters of finite groups G are class (1). An FG-module V is said to be irreducible or simple if functions and the number of them is equal to the number it is non-zero and it has no FG-sub module apart from {0} of the conjugacy classes of G. it is convenient to record and V. all these values of all irreducible characters of G in a (2). An FG-module which is a direct sum of simple sub square matrix. The matrix is called the character table of modules is said to be completely reducible or semi- G. The entries in a character table are related to each simple other in a subtle way, many which are completed using orthogonality relations. Therefore, since our main concerned is the Now, let X 1, … XK be the irreducible character of G and representations of completely reducible finite groups, we let g 1,…g k be representation of the conjugacy classes of are now set to state our first major result that is G. Then, the K x K matrix where ij-entry is x i (gj) (for all ij Marscke’s theorem without proof. 1< I < = k, 1 < < k) is called the character table of G. Therefore it is usual to number the irreducible Theorem 1 (Marschke’s Theorem) characters and classes of G so that X i = 1 G, the trivial characters and g 1=1 the identity element of G. Also note Let G be a finite group and F a field whose characteristics that the character table, the rows are indexed by the does not divide the order of G (IGI). Then every FG-sub irreducible characters of G and the columns are indexed module U of an FG-module V, ∃ an FG sub module U of by the conjugacy classes or the conjugacy classes’ 1 2 representative. V such that V = U 1⊕ U2 Displayed below is an example of the character table of Definition 5 a finite group. Its column corresponds to the conjugacy classes of G while its row corresponds to the characters χ of the in equivalent irreducible representation of G. The An FG- module V is said to be completely reducible if V = i jth conjugacy class C is indicated by displaying a U ⊕ U ⊕…⊕ U where each U is an irreducible FG – j 1 2 n i representative C C . In the (ij)th entry we put χ (C ). sub module of V. j ∈ j i j

Definition 6 (Complete Reducibility) C1 C2 ………….. Cn χi χi(C 1) χi(C 2) ………….. χi(C n) A matrix representation A (X) over F is said to be χ2 χ2(C 1) χ2(C 2) ………….. χ2(C n) completely reducible if

A (X) ∼ diag A 1 (x), A 2(x), A i(x). χn χn(C 1) χn(C 2) ………….. χn(C n) Where A i(x) ⇒ i=1,2,… 1 are irreducible representation Scholarly J. Math. Comp. Sci. 10

Remark 1 if i = j δ ij =   0 if i ≠ j Webb (2012) stated that the orthogonality relations provide a check on the accuracy of our calculation and Where δ is the Kronecher delta function also enable us to complete the final row of the character ij table. The facts that the character degrees (dimension) divides order of G and that the sum of squares of the Example: dimension equals order of G also help in this case. Schur’s Lemma is another vital tool for the irreducible Suppose we are given the following part of character modules. table of a group order 12 which have exactly four conjugacy classes. Theorem 2 (Schur’s Lemma) (1) g 1 g 2 g1 g3 g4 Suppose that U and V are irreducible G-module over F. | CG ()g | 12 4 3 4 Then a G –homomorphism θ: V → U is; χ1 1 1 1 1 i. θ is the zero map or else 2 χ2 1 1 Ω Ω ii. θ is an isomorphism 1 1 2 Ω χ3 Ω PROOF e g χ4 f h

Let prove the following statements: That is If θ ≠ 0 then θ 2πi is a bijective mapping. That is ker θ = {0} and Im θ = U. Where Ω = e 3 . We shall use the column orthogonality But since Ker θ is a sub module of the irreducible G- relations to determine the last row of the character table. module V, we must have Ker θ = {0} or Ker θ = V. That is to find the value of a , b, c, d. Now, the entries However, the second alternative will amount to saying in the first column of the character tables are the degrees that θ = 0, hence we infer that Ker θ = {0} of the irreducible characters, so they are positive Next, since Im θ is a sub module of U 1 then Either Im θ = integers. Thus by column orthogonality relation {0} or Im θ = U. But we have already used the first r = s = ,1 the sum of squares of these numbers is. 12 . possibility. Therefore it follows that Im θ = U. Hence the last entry in the first column is 2 2 2 2 Theorem 3 (Matrix from of Schur’s Lemma) (2) 1 + 1 + 1 + e = 12 is e + 3 = 12 ⇒ e = ± .3 and since they are positive integers, it follows that e = .3 Let A (X) and B(X) be 2 irreducible representation over F Now for the second column, the orthogonalitry relation is. of a group G, and suppose ∃ a constant matrix T over F, such that TA(X) = B(X)T for all x ∈g then, 4 ∑ χi ()()g1 χ j gs = 0 (i) either T = 0 or i=1

(ii) T is non – singular so that Which gives . 1.1 + 1.1 + 1.1 + s. f = 0 ⇒ 3 f A(X) = T -1 B(X) T Therefore f = −1 That is A(X) and B(X) are equivalent. Now, before going for the proper construction of the Hence by considering the orthogonality relation between tables, let look at this definition and the theorem on the first column and the column 3 and 4 we obtain the orthogonality relations. complete character table as follows:

Definition 8 g g g g gi i 2 3 4

4 3 3 Let G be a finite group and let V 1, Vk be a complete | CG ()gi | 12 set of non-ismorphic irreducible CG-modules. If is the 1 1 1 1 χi χi 2 character of V i (1 ≤ i≤ k ) then inner product of χ2 1 1 Ω Ω χ and χ = δ ∀ij . 2 i j ij χ3 1 1 Ω Ω χ 3 −1 0 0 That is 4 Sani and Bukar 11

Definition 9 Tabulating the information we have:

Let A(X ) = aij (x) be a matrix representation of G of S3 1 (12 ) (123 ) degree M . The trace of A X , that is tr A X is called ( ) ( ) χ1 1 1 1 the character of A X . Therefore the character afforded ( ) χ2 1 −1 1 by the representation say φ:G → GL V ∀ g ∈ G is the a ( ) χ3 2 b function χr (g) = trace of (φ(g )) Thus χ : G → F, the trace of A(X ) denoted by φ(x) Here, we have chosen representative for the 3 conjugacy classes and labelled the column with them and is called the character of A(X ) . That is let the two entries as a and b that have not computed M yet. As before, these can be determined by column φ()()()()()x = a11 x +a22 x + ... + amm x = ∑aii x . orthogonality relation that is i=1 χ1(1)χ1(g)+ χ2 (1)χ2 (g)+ χ3(1)χ3 (g) = 0 Remark where g ≠ 1. (i) The above definition shows that the trace of the a square matrix is the sum of the elements in the main Hence the complete table is diagonal. (ii) if we consider the characteristics polynomial A(X ) S3 (1) (12 ) (123 ) denoted by d et (λI − A(x)), then the coefficient λm−1 is χ1 1 1 1 m χ 1 −1 1 equal to φ()()x =∑aii x . 2 i=1 2 0 1 χ3 − (iii) That if B(X ) is equivalent to A(X ), that is if B(X ) = T −1A(X )T for any non-singular matrix T , then Now suppose N is a normal subgroup of finite group G the matrix have the same characteristics polynomial. and N ≠ }1{ , Then G | N is called the factor group, which is smaller than the group G . Therefore we can use Now we start with construction of symmetric group of the character s of G | N to get to some of the characters degree three that is S . 3 of G by a process which is known as lifting and the The conjugacy classes of S3 are: normal subgroup enable us to do that.

Class Cycle -shape Definition 10 }1{ (1) {( 12 ,) (13 ), (23 )} (2) If N is normal in G ( that is N normal in G ) and χ {( 123 ,) (132 )} (3) is a character of G | N , then the character χ of G which is given by χ(g)= χ(Ng )(gCG ) is called the left Therefore, there are three conjugacy classes, so there are three irreducible representations. of χ to G . We have two linear characters that is We now construct the symmetric group of degree

4 (S4 ) lifting method. Let G = S4 (symmetric group ). 1. The trivial character χ1 and Let N = V4 ={( ),1 (12 )(34 ),(13 )(24 ),(14 )(23 )} so that 2. The sign or alternating character which we call χ2 . N is normal in G suppose we let a = N (123 ) and

There must be exactly one other character say D3 and b = N (12 ) . 2 Then G | N = 〈 a,b: a3 = b2 = N, b−1ab = a−1〉 since ∑Di = | G | = 6 , we must have D3 =2 because in the alternating character, we have Clearly, this means that G | N ≅ S3 or D6 therefore the 2 2 2 2 character table of G | N is ()()1 + 1 + D3 = 6 ⇒ D3 = 4 ∴ D = 4 = 2 Scholarly J. Math. Comp. Sci. 12

N N (1) N(12 ) N(123 ) So χ4 is irreducible. The result χ 4 χ2 is also irreducible by using the same computation. We have already shown χ1 1 1 1 what χ3 and χ1 ,χ2 are the linear and alternating χ2 1 −1 1 0 characters respectively. Hence, let χ5 = χ4 χ2 . Since χ3 2 −1

χ 4 have fine contumacy classes and we have produced five irreducible characters, we have now found the Now in order to calculate the left χ of a character χ of complete character table of S as shown below: G | N , we note that 12 34 N because 4 χ( )( )= χ( ) (12 )(34 )∈ N. and χ (1234 ) = χ(N(13 )) since gi 1 (12 ) 123 (12 )(34 ) (1234 )

1234 13 . Hence the lift of are N ( ) = N ( ) χ1 ,χ2 , χ3 | CG ()gi | 24 4 3 8 4

χ1 ,χ2 ,χ3 which is given by: χ1 1 1 1 1 1

χ2 1 −1 1 1 −1

1 (12 ) (123 ) (12 )(34 ) (1234 ) 0 0 χ3 2 −1 2 χ1 1 1 1 1 1 χ 4 3 1 0 −1 −1 1 −1 1 1 −1 χ2 3 0 χ5 −1 −1 1 2 0 −1 2 0 χ3

CONCLUSION Then χ1,χ2 ,χ3 are the irreducible characters of G .

Since χ1 ,χ2, χ3 is the irreducible character of G | N . The We have in this paper, constructed character table of symmetric group of degree 3 and 4. During the course of above example is part of the character table of S4 . construction, it is observed that, the use of orthogonality Remark: relation and lifting process gives the complete character table of a finite group which shows that equivalent representations have the same character table. The We have already produced three characters, χ1 ,χ2 ,χ3 following conclusions are drawn. of S4 by lifting character, of the factor group S 4 /V4 1. Characters are class function; that is they each take a above. We shall now use theorem 2 which deals with the constant value on a given conjugacy class product of character with a linear character to complete 2. Isomorphic representations have the same character the character table of S4 . Now let χ4 be the character. over a field of characteristics zero 3. If a character of finite G is restricted to a subgroup H, Then χ4 (g)=| fix (g)| −1 g ∈ S4 . By theorem 2, the then the result is also a character of H product χ4 χ2 is also a character of S4 . Therefore the value of χ2 , χ4 and χ2 χ4 are as follows: REFERENCES

gi 1 (12 ) (123 ) (12 )(34 ) (1234 ) Baker, A. (2011). Representation of finite groups, University of Glascow. E-mail address: a baker @ math.gla.ca.uk CG ()gi 24 4 3 8 4 URL.HTTP.//www.math.gla.ac.uk/ ajb. Lam, T.Y. (1998). An Article on Representation of finite groups. A χ2 1 −1 1 1 −1 Hundred Years Part I. University of California, Berkeley. 3 0 Momoh, S.U. (2010). A lecture Note on Group Representation Theory. χ4 1 −1 −1 University of Jos, Nigeria. − 3 −1 0 −1 1 Onwumere, O.O. (1998). A project on a Survey of Group χ4 χ2 Representations. Unpublished M.Sc. Thesis. University of Jos, Nigeria. Now let check for the irreducibility of χ that is Peter, W. (Printed Feb. 2012). Finite Group Representation for Pure Mathematician University of Minnesote, U.S.A. Serre, J.P. (1977). Linear Representations of Finite Groups. Spring V. (− 3)2 (− 1)2 (− 1)2 (1)2 Verlag ISBN 0-387-90190.6. 〈 χ 4 χ 2 〉 = + + 0 + + 24 12 8 4 9 1 1 1 = + + + = 1 24 12 8 4