oct03 - Math 5125 Monday, October 3 Monday, October 3...

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Unformatted text preview: Math 5125 Monday, October 3 Monday, October 3 Exercise 6.3.7 on page 220 Prove that the following is a presentation for the quaternion group of order 8: Q 8 = h a , b | a 2 = b 2 , a- 1 ba = b- 1 i . Q 8 is a group with 8 elements, namely { 1 ,- 1 , ± i , ± j , ± k } with identity 1, center { 1 ,- 1 } , i 2 = j 2 = k 2 =- 1, i j = k , jk = i , ki = j . Let F denote the free group on { a , b } . Let us map F to Q 8 by sending a to i and b to j . Since Q 8 is generated by { i , j } , we obtain an epimorphism θ : F → Q 8 such that θ a = i and θ b = j . Let K = ker θ . Now θ ( a 2 ) = i 2 =- 1 = j 2 = θ ( b 2 ) and θ ( a- 1 ba ) = i- 1 ji =- i ji =- ki =- j = j- 1 = θ ( b- 1 ) , so if R is the normal subgroup generated by a 2 b- 2 , a- 1 bab , we see that R ⊆ K . Since F / K ∼ = Q 8 , we see that | F / K | = 8, so we will be done if we can show that | F / R | < 16 (because by Lagrange | F / R | has to be a multiple of | F / K | = 8, and | F / R | = 8 if and only if R = K ). Since)....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.

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oct03 - Math 5125 Monday, October 3 Monday, October 3...

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