oct13 - e G ∈ A we have θ e G ∈ θ A so θ A 6 =/0(ii...

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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 θ ( e G ) ∈ θ ( A ) , so θ ( A ) 6 = /0. (ii) Let x , y ∈ θ ( A ) . Then we may write x = θ a and y = θ b for some a , b ∈ A . Then xy = θ a θ b = θ ( ab ) . Since A is a subgroup, ab ∈ A and we deduce that xy ∈ θ A . (iii) Let x ∈ θ ( A ) . Then we may write x = θ a for some a ∈ A . Then x-1 = ( θ a )-1 = θ ( a-1 ) . Since A is a subgroup, a-1 ∈ A and we deduce that x-1 ∈ θ A ....
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.

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