oct13 - e G A , we have ( e G ) ( A ) , so ( A ) 6 = /0....

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Math 3124 Thursday, October 13 October 13, Ungraded Homework 1. Problem 21.4 on page 109. Define a homomorphism ρ r : M ( 2 , Z ) M ( 2 , Z ) by ρ r ( x ) = xr for each x M ( 2 , Z ) , where r = ± 1 0 0 0 ² . Find ker ρ r . ker ρ r = { x M ( 2 , Z ) | xr = 0 } . Thus the matrix ± a b c d ² is in ker ρ r if and only if ± a b c d ²± 1 0 0 0 ² = ± 0 0 0 0 ² This is if and only if a = c = 0. Therefore ker ρ r consists of matrices of the form ± 0 m 0 n ² where m , n Z are arbitrary. 2. Problem 21.9 on page 109. Prove that if θ : G H is a homomorphism and A is a subgroup of G , then θ ( A ) is a subgroup of H . We check the three conditions for a subgroup. (i) Since
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Unformatted text preview: e G A , we have ( e G ) ( A ) , so ( A ) 6 = /0. (ii) Let x , y ( A ) . Then we may write x = a and y = b for some a , b A . Then xy = a b = ( ab ) . Since A is a subgroup, ab A and we deduce that xy A . (iii) Let x ( A ) . Then we may write x = a for some a A . Then x-1 = ( a )-1 = ( a-1 ) . Since A is a subgroup, a-1 A and we deduce that x-1 A ....
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