View of the Symmetric Groups S 3, S 4, Quaternion Group Q 8

Transcription

1 View of the Symmetric Groups S 3, S 4, Quaternion Group Q 8 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 S n 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 S 3, S 4, Q 8. to be a way to show similar problems. Key words: Representation. Symmetric Groups, subgroup, Cosets, Quaternion Group Q 8 representation of S 3 and S 4, 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 S 3 and S 4 by helping composition of permutation functions. Then we show Tow representations equivalence of Q 8, with helping TieTze Transformations. Then we show by helping algorithm Todd and Coxter, that representation is representation Q 8, 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 S x.[13],[18],[19] 1-2- Point: can be proved, S x 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 S x with the S n is shown and the symmetric Group of degree n is called and every element S n is a objective Function to form f=, f i X, that it is called a permutation 1-2 according what was said. S n is a symmetric Groups of degree n, all permutation of n distinct objects.[10] 1-4- Point: S n in the symmetric Group with n! members. There fore S 3 and S 4, 6 and 24 respectively are members.[10],[18] 1-5- Agreement: if X={1,2,,n} and f S n is a permutation, and f x 1 to x 2 and x 2 to x 3 and and x n-1 to x n and. to x 1 to win, Then it f=(x 1 x 2 x k ) show, x i X and this permutation is call a k-sycle and if k=2 Then it is call 2-sycle or transposition. If f=(x 1 x 2 x k ) Then {x 1, x 2,, x k } is call orbit of f. and f function act on the every element self, k times, Then it receives to primary element or (x i ) = x i ; k n 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 S n. 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 S n has transpositions.[15],[12] 1-8- Theorem: The S n Generate by (n-1) transpositions (12), (13),, (1n) or S n = <(12), (13),, (1n)>.[15],[12] 1-9- Lema: every cycle in the S n can be written to form finitely number transposition product. For example (123579)= (12)(13)(15)(17)(19). [15] Result: every permutation can be written to form finitely number transposition product. [ 13] Definition: every permutation as the product of an even number transposition is called a even permutation.[13] 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] Definition: G=<X R> is called finite representation if X and R are finitely sets (of course: This representation is not unique). [1] Theorem: every Group has a representation and every finitely Group is with finitely representation.[3] Theorem: the set of even permutations in the S n is normal subgroup of S n and that slow with An.[1],[3] COPY RIGHT 2013 Institute of Interdisciplinary Business Research 622

2 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). Q 8 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 a 4 = 1, a 2 =b 2, b -1 ab=a -1 >.[7],[8] 2-2- Lema: symmetric Groups S n, n 2 Generally show with n-1 Generators is as follows: S n =<x 1, x 2,, x n-1 x 2 i= (x j, x j+1 ) 3 = (x k, x e ) 2 =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) S 3 = <X,y x 2 =y 2 =(xy) 3 =1> S 4 = <X,y x 4 =y 2 =(xy) 3 =1> Representation of symmetric Group S 3 : we show with choice sui table Generators when S 3 = <x,y x 2 =y 2 =(xy) 3 =1> Dissolving: we know the symmetric Group S 3 written to form under set : S 3 ={,,,,, } And or to form set of cycles: S 3 = {(1), (12), (13), (23), (123), (132)} We suppose x=(12), y=(13), we show that X={x,y} is Generators set of S 3 and x,y Result relations x 2 =y 2 =(xy) 3 =1 and another every reaction is identical relations before and so, <x,y x 2 =y 2 =(xy) 3 =1> is representation of S 3. First' x,y S 3 and as we have I) x 2 = (12)(12)=(1) II) xy= (12)(13)=(123) III) yx= (13)(12)=(132) IV) yxy= (13)(12)(13)= (132)(13)=(23) Finally, x and y Generate S 3 and X={x,y} is Generator set of S 3. Secondly, as y 2 =(13)(13)=(1) and (xy) 3 =(123) 3 =(1), so x and y result, x 2 =y 2 =(xy) 3 =1. Thirdly if we make every relation with x and y that is equal before relations or equal one of elements of S 3. For example xyx= (123)(12)=(23) xyx=yxy S 3 (xy) 2 =(xy)(xy)=xyx.y= yxy.y=yx.y 2 =yx(1)=yx S 3 (yxy) 2 = yxy.yxy=yx.y 2.xy=yx.x.y=yx 2 y=y.y=y 2 =(1) S 3 Accordingly, every another relation of x and y equql before relations or that is one of elements of S 3 then: S 3 = <x,y x 2 =y 2 =(xy) 3 =1>.[5] Exersis: with choise x=(12) and y=(123) you show <x,y x 2 =y 3 =(xy) 2 =1> is a representation of S 3 or S 3 =<x,y x 2 =y 3 =(xy) 2 =1> Representation S 4 :[5] II) We show that S 4 is equal to representation S 4 =<x,y x 4 =y 2 =(xy) 3 =1> Solution: According to (1-4) the symmetric Group S 4 has 24 members: S 4 ={(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 S 4 equal X={x,y} and then we show view S 4 is equal to S 4 =<x,y x 4 =y 2 =(xy) 3 =1> The first part of the proof: y 2 =(12)(12)=(1) with regard to the assumption, (1), x, y <x,y> xy=(1234)(12)=(234) (xy) 3 =(1) x 2 = (1234)(1234)=(13)(24) x 4 =(1234) 4 =(1) accordingly: x 4 =y 2 =(xy) 2 =1 or we make relations x 3 =(1234)(1234)(1234)=(1432) yx=(12)(1234)=(134) x 2 y=x.xy=(1234)(234)=(1324) x 3 y=(1432)(12)=(143) COPY RIGHT 2013 Institute of Interdisciplinary Business Research 623

3 = (1324)(1324)=(12)(34) yx 2 =yx.x=(134)(1234)=(1423) yx 3 =(12)(1432)=(243) (xy)(x 3 y)=(234)(143)=(142) (x 3 y)(xy)=(143)(234)=(123) [(x 3 y)(xy)] 2 =(123) 2 =(132) (xy)(x 2 y)=(13) (x 2 y)(xy)=(14) (yx 2 )(yx)=(24) and (x 2 )(yx 2 )=(34) and ( )(yx 2 )=(124) (x 3 )(yx 3 )=(1342) and [(x 3 )(yx 3 )] 3 =(1342) 3 =(1243) [(x 2 y)(xy)][(x 3 )(yx 3 )] 2 =(23) and [(x 3 )(yx 3 )] 2 =(1342) 2 =(14)(23) Accordingly, we find complete element of S 4 then X={x,y} is Generator set of S 4,but every combination of x and y is one of elements of S 4 for example: (yx) 2 =(43) and (x 3 y) 2 =(134) and (yx 3 ) 2 =(234) Then with relations: x 4 =y 2 =(xy) 3 =1 and S 4 was make with x and y,accordingly: S 4 =< x,y x 4 =y 2 =(xy) 3 =1> III) we show that tow views are.[2] 1).=<a,b a 4 =1, a 2 =b 2, b -1 ab=a -1 > 2) =<a,b aba=b, bab=a> We know that ={±1, ±i, ±j, ±k} and we have the following relations. A) i 2 =j 2 =k 2 =-1 B) ij=k, ji=-k, jk=i, kj=-i, ik=-j, ki=j and is a group of order 8. Assuming i=a and k=b we show that the views above of is satisfied with the actions defined and X={a,b} is a generating set group. Since a 4 =i 4 =1 then 1,a,b and because a 2 =i 2 =-1, ab=ik=-j, ba=j, a 3 =-i, a 2 b=a.ab=-k then X={a,b} is a generating set 8 group and because a 4 =1, a 2 =b 2 =-1 and b -1 =-k then, b -1 ab=(-k)(-j)=ky=-i and or b -1 ab=a -1 accordingly, a 4 =1, a 2 =b 2, b -1 ab=a -1 and because each combination of a,b is a member of for example: a 3 b=j Then <a,b a 4 =1, a 2 =b 2, b -1 ab=a -1 > Now we show =<a,b aba=b, bab=a> by assumption a=i, b=k,and we saw =<a,b a 4 =1, a 2 =b 2, b -1 ab=a -1 > But with this relation, we show Now: =<a,b aba=b, bab=a> Since: a 4 =1 and a 2 =b 2 b 4 =1 and since: b -1 ab=a -1 b -1 aba=a -1.a b -1 aba=1 aba=b since: aba=b aba 2 =ba abb 2 =ba ab 3 =ba ab 4 =bab a(1)=bab bab=a accordingly: =<a,b aba=b, bab=a> exercise: you show that view =<a,b aba=b, bab=a> will result in the following display: 8=<a,b a 4 =1, a 2 =b 2, b - 1 ab=a -1 > COPY RIGHT 2013 Institute of Interdisciplinary Business Research 624

4 Problem 4: With method Todd and Coxter we show that symmetric Group <a,b aba=b, bab=a> is 8 order Group and aqual.[7],[8], [18],[19] Solution: Todd and Coxter method, define tables and the result. Directed: aba= b abab -1 =1 ************************************************** b a b a -1 a b a b -1 Definition Result Raw Table * a b b=2 1a=4 1 A a=3 5a=2 1 B b=4 4a=6 2 B b=5 8b=3 2 A b=6 7b=1 3 B a=7 6b=7 3 A a=7 7a=5 4 B b=8 5 B 8 A B 8a=1 5 A 3a=7 6 B ******************************************************* Counting the table is filled will the above definitions the above representation, introduces a G8 member. This tables is filled from Tow related aba=b and bab=a Following this table all members have then =<a,b aba=b, bab=a> COPY RIGHT 2013 Institute of Interdisciplinary Business Research 625

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