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

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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 = a , b | a 2 = b 2 , a - 1 ba = b - 1 . 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 a - 1 ba = b - 1 mod R , we see that a - 1 b 2 a = b - 2 mod R and hence a - 1 a 2 a = b - 2 = a - 2 mod R . We deduce that a 4 = b 4 = 1 mod R . Using ba = ab - 1 , we now see that every element of F can be written in the form a i b j mod
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