Survey of All Six - Element Groups


           Let G be any group of order 6 . By Cauchy's thorem, G has an element " a " of order 2 and an element " b " of order 3. The element (e, a, b, b2, ab, ab2) Are all distinct; and since G has only six element, these are all the element in G, thus " ba " is one of the elements e , a , b , b2, ab or ab2 


1.)Prove that " ba " cannot be equal to either e , a , b or b2   Thus, ba = ab or ba = ab2


Proof 
	Case 1  ถ้า      ba = e
		        ba·a = ea
		          ba2 = a2
		           be = a
 			 b = a	(ซึ่งเกิดการขัดแย้ง)
	Case 2  ถ้า      ba = a
		        ba·a = a·a
		          ba2 = a2
		           be = e
 			 b = e	(ซึ่งเกิดการขัดแย้ง)
	Case 3  ถ้า      ba = b
		        ba·a = ba
		          ba2 = ba
		           be = ba
 			 b = ba	(ซึ่งเกิดการขัดแย้ง)
	Case 4  ถ้า      ba = b2
		        ba·a = b·b
			a = b (จาก การตัดออกทางด้านซ้าย จะเกิดการขัดแย้ง)

ดังนั้น  	ba = ab or ba = ab2

Either of these two equations completely determines the table of G

2.) If " ba " = " ab " Prove that G ~= Z6


Proof.....
                             Z6 ={ [0] , [1] , [2] , [3] , [4] , [5] }
                             G  ={ e , a , b , b2, ab , ab,2 }
                             F: G -----> Z6  
                             กำหนดโดย 	f(e) = [0]
                                                f(a) = [3]
                                                f(b) = [2]
                                                f(b2) = [4]
                                                f(ab) = [5]
                                                f(ab2) = [1]
                               และจาก " ba " = " ab " 
                                     
    สามารถแสดงเป็นตารางดังนี้









(G,·)    	e ab2 b a	b2	ab



e     e	ab2	b	a	b2	ab


ab2
ab2	
b	
a	
b2
ab
e


b	
b	
a	
b2
ab
e
ab2


a
a	
b2	
ab
e	
ab2
b


b2
b2
	ab
e	
ab2
b
a


ab
ab	
e	
ab2
b	
a
b2
  






 (Z,+)
[0]
[1]
[2]	
[3]
[4]	
[5]


[0]	
[0]
[1]
[2]
[3]
[4]
[5]


[1]
[1]
[2]
[3]
[4]
[5]
[0]


[2]
[2]
[3]
[4]
[5]	
[0]
[1]


[3]
[3]
[4]
[5]
[0]
[1]
[2]


[4]
[4]
[5]	
[0]
[1]
[2]
[3]


[5]
[5]
[0]
[1]
[2]
[3]	
[4]




จากตารางจะได้ว่า G ~=Z6




3.) If " ba " = " ab2", Prove that G ~= S3


Proof.....
                  S3  = {(1) , (1 2) , (1 3) , (1 2 3) , (1 3 2) , (2 3) }
                  G  = { e , a , b , b2, ab , ab2 }
                  F: G --------> S3
                  กำหนดโดย f(e) = (1)
                                    f(a) = (1 2)
                                    f(b) = (1 2 3)
                                    f(b2) = (1 3 2)
                                    f(ab) = (1 3)
                                    f(ab2) = (2 3)
                       และจาก " ba " = " ab2"
                                   สามารถแสดงเป็นตารางได้ดังนี้









(G,·)
e
a	
ab
b
b2	
ab2


e
e
a	
ab
b	
b2
ab2


a
a
e
b
ab
ab2
b2


ab
ab
b2
e
ab2
a
b


b	
b
ab2
a
b2	
e	
ab


b2	
b2
ab
ab2
e	
b	
a


ab2
ab2
b
b2
a
ab	
e






(S3,+)
(1)
(1 2)
(1 3 )
(1 2 3)
(1 3 2)
(2 3)


(1)
(1)
(1 2)
(1 3 )
(1 2 3)
[4]
(2 3)


(1 2)
	(1 2)	
(1)
(1 2 3)
(1 3 )
(2 3)
(1 3 2)


(1 3 )
(1 3 )
(1 3 2)
(1)
(2 3)
(1 2)
(1 2 3)


(1 2 3)	
(1 2 3)
(2 3)
(1 2)
(1 3 2)
(1 3 2)
(1 3)


(1 3 2)
(1 3 2)
(1 3 )	
(2 3)
(1)
(1 2 3)
(1 2)


(2 3)
(2 3)	
(1 2 3)
(1 3 2)
(1 2)
(1 3 )
(1)




จากตารางจะได้ว่า G~= S3







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survey of all eight-element groups