ijcrb.webs.com OCTOBER 2013 INTERDISCIPLINARY JOURNAL OF CONTEMPORARY RESEARCH IN BUSINESS VOL 5, NO 6 View of the Symmetric Groups S3, S4, Quaternion Group Q8 Azim Niknahad Department of Mathematic, Islamic Azad University, Rasht branch, Rasht, Iran Abstract: In this paper, first we have expressed some of the definitions and theorems in the Group Theory of Algebra. Then we have shown symmetric Groups Sn with n-1 Generator, to from <X | R> where X is the set of Generators relations of R. and then we have to find a suitable choice of Generators for representation S3, S4, Q8. to be a way to show similar problems. Key words: Representation. Symmetric Groups, subgroup, Cosets, Quaternion Group Q8 representation of S3 and S4, counting Introduction: In 1936 AD, Tow scientists named todd and coxter algorithm for counting cosets of subgroup finite Group, and Too, Counting finite Groups found, That mechanical practices were used in problems of Groups. Then we find representation symmetric Groups S3 and S4 by helping composition of permutation functions. Then we show Tow representations equivalence of Q8, with helping TieTze Transformations. Then we show by helping algorithm Todd and Coxter, that representation is representation Q8, Too.[3],[4],[7],[8],[5] 1- Definitions and Theorems: 1-1- definition: if X is a finite set, then we show set of objective functions X to X with Sx.[13],[18],[19] 1-2- Point: can be proved, Sx is a symmetric Group, with operation composition functions. They say its symmetric Groups on X.[7] 1-3- Agreement: if X= {1,2,…,n}, Then Sx with the Sn is shown and the symmetric Group of degree n is called and every element Sn is a objective Function to form f= , fi X, that it is called a permutation 1-2 according what was said. Sn is a symmetric Groups of degree n, all permutation of n distinct objects.[10] 1-4- Point: Sn in the symmetric Group with n! members. There fore S3 and S4, 6 and 24 respectively are members.[10],[18] 1-5- Agreement: if X={1,2,…,n} and f Sn is a permutation, and f x1 to x2 and x2 to x3 and … and xn-1 to xn and . to x1 to win, Then it f=(x1 x2 … xk) show, xi X and this permutation is call a k-sycle and if k=2 Then it is call 2-sycle or transposition. If f=(x1 x2 …xk) Then {x1, x2,…, xk} is call orbit of f. and f function act on the every element self, k times, Then it receives to primary element or (xi) = xi; k≤n. 1-6- Theorem: every permutation can be writhen as product apart form syclics for example: The permutation = then =(14)(257)(368).[13],[17] 1-7- Point: if (ab) is a transposition in the Sn. Since a and b can be formed to n and n-1 as indicated. Then we have number n(n-1) transpositions and as (ab)=(ba) Then symmetric Group Sn has transpositions.[15],[12] 1-8- Theorem: The Sn Generate by (n-1) transpositions (12), (13), …, (1n) or Sn= <(12), (13), …, (1n)> .[15],[12] 1-9- Lema: every cycle in the Sn can be written to form finitely number transposition product. For example (123579)= (12)(13)(15)(17)(19) . [15] 1-10- Result: every permutation can be written to form finitely number transposition product. [ 13] 1-11- Definition: every permutation as the product of an even number transposition is called a even permutation.[13] 1-12- Definition: if G is a Group and X and X is Generated set of G and G is Free Group on the X and R≤G is a relations set and X Generate G, by relations of R, Then we show G with G=<X | R>. [13] 1-13- Definition: G=<X | R> is called finite representation if X and R are finitely sets (of course: This representation is not unique). [1] 1-14- Theorem: every Group has a representation and every finitely Group is with finitely representation.[3] 1-15- Theorem: the set of even permutations in the Sn is normal subgroup of Sn and that slow with An.[1],[3] COPY RIGHT © 2013 Institute of Interdisciplinary Business Research 622 ijcrb.webs.com OCTOBER 2013 INTERDISCIPLINARY JOURNAL OF CONTEMPORARY RESEARCH IN BUSINESS VOL 5, NO 6 2- Meet some of the groups and their views and solving problems _________________________________________________ 2-1- Fibonacci Groups: Generally show a fibunacci is F (r,n) that n is number Generator and r is number relations between elements of Generator (They r and n number of relations and Generators are productive for example: members). Q8 is a fibonacci Group with representation F(3,3) to form under: 1) F(3,3)=<a,b,c│ab=c, bc=a, ca=b> 2) F(3,2)=<a,b│a4= 1, a2=b2, b-1ab=a-1> .[7],[8] 2-2- Lema: symmetric Groups Sn , n≥2 Generally show with n-1 Generators is as follows: 2 3 2 Sn=<x1, x2, …, xn-1│x i= (xj, xj+1) = (xk, xe) =1, 1≤ I ≤n-1, 1 ≤ j <n-2, 1≤ j < n-2, 1≤ L< k-1<n-1>. [7],[8] 2-3- Point: 2-2- can be writhen according to (and we will show in this paper)(we wrote the 2-2) 2 2 3 S3= <X,y│x =y =(xy) =1> 4 2 3 S4= <X,y │x =y =(xy) =1> ________________________________________________________ 2 2 3 Representation of symmetric Group S3: we show with choice sui table Generators when S3= <x,y│x =y =(xy) =1> Dissolving: we know the symmetric Group S3 written to form under set : S3={ , , , , , } And or to form set of cycles: S3= {(1), (12), (13), (23), (123), (132)} 2 2 3 We suppose x=(12), y=(13), we show that X={x,y} is Generators set of S3 and x,y Result relations x =y =(xy) =1 2 2 3 and another every reaction is identical relations before and so, <x,y│x =y =(xy) =1> is representation of S3. First' x,y S3 and as we have I) x2= (12)(12)=(1) II) xy= (12)(13)=(123) III) yx= (13)(12)=(132) IV) yxy= (13)(12)(13)= (132)(13)=(23) 2 Finally, x and y Generate S3 and X={x,y} is Generator set of S3. Secondly, as y =(13)(13)=(1) and (xy)3=(123)3=(1) , so x and y result , x2=y2=(xy)3=1. Thirdly if we make every relation with x and y that is equal before relations or equal one of elements of S3. For example xyx= (123)(12)=(23) xyx=yxy S3 2 2 (xy) =(xy)(xy)=xyx.y= yxy.y=yx.y =yx(1)=yx S3 2 2 2 2 (yxy) = yxy.yxy=yx.y .xy=yx.x.y=yx y=y.y=y =(1) S3 Accordingly, every another relation of x and y equql before relations or that is one of elements of S3 then: S3= <x,y│x2=y2=(xy)3=1> .[5] _______________________________________ 2 3 2 Exersis: with choise x=(12) and y=(123) you show <x,y│x =y =(xy) =1> is a representation of S3 or 2 3 2 S3=<x,y│x =y =(xy) =1> ____________________________________________________ Representation S4:[5] II) We show that S4 is equal to representation 4 2 3 S4=<x,y│x =y =(xy) =1> Solution: According to (1-4) the symmetric Group S4 has 24 members: S4={(1), (34), (23), (234), (243), (24), (12), (12)(34), (123), (1234), (1243), (124), (132), (1342), (13), (134), (13)(24), (1324),(1432) (142), (143), (14), (1423), (14)(23)} Now assume, x=(1234), y=(12), we show that the generator set S4 equal X={x,y} and then we show view S4 is equal to 4 2 3 S4=<x,y│x =y =(xy) =1> The first part of the proof: y2=(12)(12)=(1) with regard to the assumption , (1), x, y <x,y> xy=(1234)(12)=(234) (xy)3=(1) x2= (1234)(1234)=(13)(24) x4=(1234)4=(1) accordingly: x4=y2=(xy)2=1 or we make relations x3=(1234)(1234)(1234)=(1432) yx=(12)(1234)=(134) x2y=x.xy=(1234)(234)=(1324) x3y=(1432)(12)=(143) COPY RIGHT © 2013 Institute of Interdisciplinary Business Research 623 ijcrb.webs.com OCTOBER 2013 INTERDISCIPLINARY JOURNAL OF CONTEMPORARY RESEARCH IN BUSINESS VOL 5, NO 6 = (1324)(1324)=(12)(34) yx2=yx.x=(134)(1234)=(1423) yx3=(12)(1432)=(243) (xy)(x3y)=(234)(143)=(142) (x3y)(xy)=(143)(234)=(123) [(x3y)(xy)]2=(123)2=(132) (xy)(x2y)=(13) (x2y)(xy)=(14) (yx2)(yx)=(24) and (x2)(yx2)=(34) and ( )(yx2)=(124) (x3)(yx3)=(1342) and [(x3)(yx3)]3=(1342)3=(1243) [(x2y)(xy)][(x3)(yx3)]2=(23) and [(x3)(yx3)]2=(1342)2=(14)(23) Accordingly, we find complete element of S4 then X={x,y} is Generator set of S4 ,but every combination of x and y is one of elements of S4 for example: (yx)2=(43) and (x3y)2=(134) and (yx3)2=(234) Then with relations: 4 2 3 x =y =(xy) =1 and S4 was make with x and y ,accordingly: 4 2 3 S4=< x,y│x =y =(xy) =1> ____________________________________________________ III) we show that tow views are .[2] 4 2 2 -1 -1 1) .=<a,b│a =1, a =b , b ab=a > 2) =<a,b │aba=b, bab=a> We know that ={±1, ±i, ±j, ±k} and we have the following relations. A) i2=j2=k2=-1 B) ij=k, ji=-k, jk=i, kj=-i, ik=-j, ki=j and is a group of order 8 .