aa_hw_3_Math 113 - Homework 3 September 27, 2010...

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Homework 3 September 27, 2010 Problem 1 (a) Inverses exist, since matrices A with det( A ) Ó = 0 are invertible. If A is invertible, then det( A ) = det( A - 1 ) - 1 , so it is closed under inverses. The identity is the identity matrix. Generally, for matrices A,B in the group, det( AB ) = det( A ) det( B ) = ± 1, so the group is closed under multiplication, satisfying the last requirement. (b) Again, inverses exist because the determinant is non-zero. Again, since det( A ) = 1 for some matrix A in the group, we have that det( A - 1 ) - 1 = det( A ) = 1, so 1 / det( A - 1 ) = 1 so det( A - 1 ) = 1, showing that the inverses are indeed in the group. The identity is again the identity matrix, and the argument for closure is the same as before. Problem 2 The Klein 4 group Z 2 × Z 2 is not isomorhic to Z 4 , since all non-identity elements in Z 2 × Z 2 have order 2, while e.g. 3 Z 4 has order 4. Problem 3 We wish to show that every element of the Heisenberg group has infinite order. Consider some matrix in the group A = 1 a b 0 1 c 0 0 1 ,a,b,c R , then we can write out powers of A : A 2 = 1 a b 0 1 c 0 0 1 1 a b 0 1 c 0 0 1 = 1 2 a 2 b + ac 0 1 2 c 0 0 1 A 3 = 1 2 a 2 b + ac 0 1 2 c 0 0 1 1 a b 0 1 c 0 0 1 = 1 3 a 3 b + 3 ac 0 1 3 c 0 0 1 A 4 = 1 3 a 3 b + 3 ac 0 1 3 c 0 0 1 1 a b 0 1 c 0 0 1 = 1 4 a 4 b + 6 ac 0 1 4 c 0 0 1 . In general, we find that A n = 1 na nb + [2( n - 1)] ac 0 1 nc 0 0 1 . For this to be the identity matrix, we would require, for n N , that na = 0 nc = 0 nb + [2( n - 1)] ac = 0 . 1/5
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aa_hw_3_Math 113 - Homework 3 September 27, 2010...

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