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aa_hw_3_Math 113

# aa_hw_3_Math 113 - Homework 3 Introduction to Abstract...

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Introduction to Abstract Algebra Homework 3 September 27, 2010 Problem 1 (a) Inverses exist, since matrices A with det ( A ) = 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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Introduction to Abstract Algebra Homework 3 September 27, 2010 We quickly see that we require a = 0 and c = 0 , and the third equality reduces to nb = 0 , which is only possible if also b = 0 . Thus, all elements, besides the identity matrix, have
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