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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
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=
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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)= 2) F(3,2)= .[7],[8] 2-2- Lema: symmetric Groups Sn , n≥2 Generally show with n-1 Generators is as follows: 2 3 2 Sn=
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,
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= (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) .= 2) = 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 . 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 a4=i4=1 then 1,a,b and 2 2 3 2 4 because a =i =-1, ab=ik=-j, ba=j, a =-i, a b=a.ab=-k then X={a,b} is a generating set 8 group and because a =1, a2=b2=-1 and b-1=-k then, b-1ab=(-k)(-j)=ky=-i and or b-1ab=a-1 accordingly, a4=1, a2=b2, b-1ab=a-1 and because each combination of a,b is a member of for example: a3b=j Then Now we show = by assumption a=i , b=k,and we saw = But with this relation, we show Now: = Since: a4=1 and a2=b2 b4=1 and since: b-1ab=a-1 b-1aba=a-1.a b-1aba=1 aba=b since: aba=b aba2=ba abb2=ba ab3=ba ab4=bab a(1)=bab bab=a accordingly: = ______4 2 2 - exercise: you show that view = will result in the following display: 8= ______
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Problem 4: With method Todd and Coxter we show that symmetric Group 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 1 2 3 4 1 1 4 5 2 1 1b=2 1a=4 1 A 1 2 2 6 8 3 2 2 3 4 6 2 2a=3 5a=2 1 B 2 3 6 3 4 6 7 3 3 7 1 4 3 3b=4 4a=6 2 B 3 7 4 4 5 2 6 4 4 6 7 5 4 4b=5 8b=3 2 A 4 5 5 8 1 2 5 5 2 6 8 5 2b=6 7b=1 3 B 5 6 7 5 8 6 6 8 3 7 6 3a=7 6b=7 3 A 6 8 7 1 4 5 7 7 5 8 1 7 6a=7 7a=5 4 B 7 8 3 7 1 8 8 1 2 3 8 5b=8 5 B 8 8a=1 5 A A B 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 = ______
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