survey of all eight-element groups

survey of all eight-element groups



           Let G be any group of order 8. If g has an element of order 8, then G ~= Z . Let us assume now that G has no element of order 8; hence all the elements ไม่เท่ากับ e in G have order 2 or 4. 


1.) If every x ไม่เท่ากับ e in has order 2, let a, b, c be three such elements. Prove that 

G = {e, a, b, c, ab, bc, ac, abc } . Conclude that G ~= Z2* Z2 * Z2  .



Proof 
ให้ f :g -----> Z2*Z2*Z2
กำหนดโดย  f(e) = (0,0,0)
f(a) = (1,0,0)			f(ab) = (1,1,0)
f(b) = (0,1,0)			f( bc) = (0,1,1)
f(c) = (0,0,1)			f(ac) =(1,0,0)
f (abc) =(1,1,1)





(G,·)	
e
a
b
c	
ab
bc
ac
abc


E	
e
a
b
c
ab
bc
ac
abc


a
a
e
ab
ac
b
abc
c
bc


b	
b	
ab
e
bc
a
c
abc
ac


c	
c
ac
bc
e
abc
b
a
ab


ab
ab
b	
a
abc
e
ac
bc
c


bc
bc
abc
c
b
ac
e
ab
a


ac
ac
c
abc
a	
bc
ab
e
b


abc
abc
bc
ac
ab
c
a
b
e







(Z2*Z2*Z2 , +)	
(0,0,0)
(1,0,0)
(0,1,0)
(0,0,1)
(1,1,0)
(0,1,1)
(1,0,1)
(1,1,1)


(0,0,0)
(0,0,0)
(1,0,0)
(0,1,0)	
(0,0,1)	
(1,1,0)
(0,1,1)	
(1,0,1)	
(1,1,1)


(1,0,0)
(1,0,0)
(0,0,0)
(1,1,0)
(1,0,1)	
(0,1,0)	
(1,1,1)
(0,0,1)
(0,1,1)


(0,1,0)
(0,1,0)
(1,1,0)
(0,0,0)
(0,1,1)	
(1,0,0)	
(0,0,1)
(1,1,1)	
(0,0,1)


(0,0,1)
(0,0,1)	
(1,0,1)	
(0,1,1)	
(0,0,0)
(1,1,1)	
(0,1,0)	
(1,0,0)	
(1,1,0)


(1,1,0)
(1,1,0)
(0,1,0)	
(1,0,0)	
(1,1,1)	
(0,0,0)
(1,0,1)	
(0,1,1)
(0,0,1)


(0,1,1)
(0,1,1)	
(1,1,1)	
(0,0,1)	
(0,1,0)
(1,0,1)	
(0,0,0)	
(1,1,0)	
(1,0,0)


(1,0,1)
(1,0,1)
(0,0,1)
(1,1,1)	
(1,0,0)	
(0,0,1)	
(1,1,0)	
(0,0,0)	
(0,1,0)


(1,1,1)
(1,1,1)	
(0,1,1)	
(1,0,1)	
(1,1,0)	
(0,0,1)	
(1,0,0)
(0,1,0)	
(0,0,0)


จากตารางดังกล่าวจะเห็นได้ว่า  G ~= Z2*Z2*Z2 



         In the remainder of this exercise set, assume G has an element  a of order 4. 
         Let H = [a ]= {e, a, a2, a3}. 
         If b E G is not in H, then the coset  Hb = {b, ab, a2b, a3b}.   
         By Largrange's theorem, G is the union of He = H and Hb; hence
                        G = {e, a, a2, a3,b, ab, a2b, a3b}
2.)Assume there is in Hb an element of order 2. (Let b be this element.) If ba = a2b, prove that
b2a = a4b2, hence a = a4, which is impossible. (why?) Conclude that either ba = ab or ba = a3b.
Proof 

ให้ 	ba = a2b
	จะต้องพิสูจน์ว่าb2a = a4b2
	จาก	b2a = b(a2b)
                                                  = (a2b)(ab)
                                                  = a2(ba)b
                                                  = a2a2(bb)
                                                  = a4b2
                       เนื่องจาก order b = 2 
                                   และ  b2a = a4b2
                                             ea = a4e
                                                a = a4= e ( ซึ่งเป็นไปไม่ได้ เนื่องจาก order a  =  4   แต่  a4=a ซึ่งขัดแย้ง)
3.) Let b be as in part2. Prove that if ba = ab , then G ~= Z4*Z2.
Proof 
                               ให้   f : G--->Z4*Z2
          กำหนดโดย 	f(e) = ( [0],[0] )          f(b) = ( [2],[1] )
                                          f(a) = ( [1],[1] )         f(ab) = ( [3],[0] )
                                          f(a2)= ( [2],[0] )      f(a2b) = ( [0],[1] )
                                          f(a3) = ( [3],[1] )      (a3b) = ( [1],[0] )



(G,·)
e	
a	
a2
a3	
b
ab	
a2b
a3b


e	
e	
a
a2
a3	
b
ab
a2b
a3b


a
a	
a2
a3
e
ab
a2b
a3b
b


a2
a2
a3
e
a
a2b
a3b
b
ab


a3
a3
e
a	
a2
a3b
b
ab
a2b


b
b
ab
a2b
a3b
e	
a
a2
a3


ab
ab
a2b
a3b
b	
a
a2
a3
e


a2b	
a2b
a3b	
b	
ab
a2
a3
e
a
 

a3b
a3b
a	
ab
a2b
a3
e	
a
a2





	(Z4*Z2 , +)
		(0,0)
		(1,1)
		(2,0)	
		(3,1)	
		(2,1)
		(3,0)
		(0,1)
		(1,0)


		(0,0)	
		(0,0)	
		(1,1)
		(2,0)
		(3,1)
		(2,1)
		(3,0)	
		(1,0)
		(1,0)


		(1,1)
		(1,1)
		(2,0)
		(3,1)
		(0,0)
			(1,1)
		(0,1)	
		(1,0)
		(2,1)


		(2,0)
		(2,0)
		(3,1)
		(0,0)	
		(1,1)
		(0,1)
		(1,0)
		(2,1)	
		(3,0)


		(3,1)
		(3,1)
		(0,0)
		(1,1)
		(2,0)	
		(1,0)
		(2,1)
		(3,0)
		(0,1)


		(2,1)
		(2,1)
		(3,0)
		(0,1)
		(1,0)
		(0,0)	
		(1,1)
		(2,0)
		(3,1)


		(3,0)	
		(3,0,)
		(0,1)
		(1,0)
		(2,1)	
		(1,1)	
		(2,0)	
		(3,1)	
		(0,0)


		(0,1)	
		(0,1)
		(1,0)
		(2,1)	
		(3,0)
		(2,0)
		(3,1)	
		(0,0)
		(1,1)


		(1,0)	
		(1,0)
		(2,1)	
		(3,0)
		(0,1)	
		(3,1)
		(0,0)	
		(1,1)
		(2,0)




จากตารางจะเห็นได้ว่า G ~= Z4*Z2
4. )Let b be as in part 2. Prove that if ba = a3b, then G ~= D4.
Proof 

ให้  f : G ------> D4
กำหนดโดย	f(e) = 
                      f(a) =
                    f(a2) = 
                    f(a3) = 
                       f(b) =
                     f(ab) =
                  f(a2b) = 
                  f(a3b) = 
                  f(ab4) = 




(D5,·)
















































































































G, *)
e	
a
a2
a3
b
ab
a2b
a3b


e	
e	
a	
a2
a3
b	
ab
a2b
a3b


a
a
a2
a3
e	
ab
a2b
a3b
b


a2
a2
a3
e	
a	
a2b
	a3b	
b
ab


a3
a3
e	
a
a2
a3b	
b	
ab
a2b


b
b
a3b
a2b
ab
e	
a3
	a2
a


ab
ab
b	
a3b
a2b
a	
e	
a3	
a2


a2b
a2b
ab	
b	
a3b
a2
a	
e	
a3


a3b
a3b
a2b
ab
b	
a3
a2
a	
e
5.) Now assume the hypothesis in part 2 is false. Then b, ab, a2b, and a3b all have order 4.
Prove that b2 = a2. (Hint : What is the order of b2 ? What element in G has the same order?)
Proof 
เนื่องจาก	b มี order 4 	จะได้ว่า
((b2)) = e
ดังนั้น b2 จึงมี order 2
จาก		b2  มี order 2  จะได้ว่า
		b2  <> a2b
		b2 <> a3b
		b2 <> ab
		b2 <>  b

         เนื่องจาก b, ab, a2b, a2b มี order 4 ดังนั้นจึงไม่เท่ากับ b2 และ b2  น a3
         เนื่องจาก a มี order 4  นั่นคือ  a4 = e
                                                          e = (a4 )3 = a4a4a4
                                                              = a12
                                                              = a3a3a3a3
                                                              = (a3)4
                   และจาก (a3)2 = a3a3 = a3a a2 = a2
                                          จึงสามารถสรุปได้ว่า b2 = a2
6.) Prove: If ba = ab, then (a3b)2 = e, contrary to the assumption that ord(a3b) = 4.
If ba = a2b, then a = b4= e, which is impossible. Thus, ba = a3b.
Proof 
                ถ้า ba = ab 
                                   จะได้ว่า  (a3b)2 = a3ba3b
                                                            = a3a3bb
                                                            = a4a2b2
                                                            = ea2b2
                                                            = a2b2
                                                            = a2a2 (เนื่องจาก b2 = a2)
                                                            =  a4= e    ซึ่งขัดแย้งกัน
                                                      ba <>ab


               ถ้า  ba = a2b
                                      จะได้ว่า  a = ea = b4a ( เพราะ b มี order 4)
                    b3ab = b3a2b = b2baab = b2a2bab
                                                            = b2a2 a2b b
                                                            = b2a2 a2b2
                                                            = b2 b2 b2 b2
                                                            = b4 b4
                                                            = e e
                                                            = e
                                                       ba = a3b.
7.) The equation a4 = b4= e, a2 = b2, and ba = a3b completely determine the table of G .
Write this table. (G is known as the quarternion group Q.)
Proof
(G,·) e a a2 a3 b ab a2b a3b
e e a a2 a3 b ab a2b a3b
a a a2 a3 a4 ab a2b a3b a4b
a2 a2 a3 a4 a5 a2b a3b a4b a5b
a3 a3 a4 a5 a6 a3b a4b a5b a6b
b b ba ba2 ba3 b2 bab ba2b ba3b
ab ab aba aba2 aba3 ab2 abab aba2b aba3b
a2b a2b a2ba a2ba2 a2ba3 a2b2 a2bab a2ba2b a2ba3b
a3b a3b a3ba a3ba2 a3ba3 a3b2 a3bab a3ba2b a3ba3b

                                               จาก a4 = b4 = e , a2 = b2 และ ba = a3 b

สามารถเขียนเป็นตารางใหม่ได้ดังนี้

(G,·) e a b2 ab2 b ab b3 ab3
e e a b2 ab2 b ab b3 ab3
a a b2 ab2 e b b3 ab3 b
b2 a2 ab2 e a ab a3b b ab
ab2 a3 e a b2 b3 b ab b3
b b ab3 b3 ab ab3 a e ab2
ab ab b ab3 b3 b2 b2 a e
b3 a2b ab b ab3 e ab2 b2 a
ab3 a3b b3 ab b a e ab2 b2

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