103a_05f_1s

103a_05f_1s - and hence not a group. 4. (a) Clearly G is a...

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Math 103A Midterm Exam Solutions 26 October 2005 1. (a) True (b) False (c) False (d) True (e) False. 2. (a) α = (152)(34) (b) | α | is the least common of its cycle lengths when in disjoint form; that is, lcm(3 , 2) = 6. 3. (a) Let a, b G n . By the de±nition of G n , there are g, h G such that a = g n and b = h n . Thus ab = g n ( h n ) - 1 = g n ( h - 1 ) n . Since G is abelian, this equals ( gh - 1 ) n , which is in G n . (b) If a is a rotation a 3 = e . If b is a re²ection b 2 = e and so b 3 = b . Thus ( D 3 ) 3 contains the identity and all re²ections but no rotations. Since the product of two di³erent re²ections is a rotation, ( D 3 ) 3 is not closed under multiplication
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Unformatted text preview: and hence not a group. 4. (a) Clearly G is a subset of the complex numbers under multiplication, so all we need to do is show that it is a subgroup. One proof: Suppose z = e ri and w = e si are in G . Then zw-1 = e ( r-s ) i is also in G . An alternate proof: Suppose z, w ∈ G . Then | zw-1 | = | z | / | w | = 1 and so zw ∈ G . (b) There is exactly one such subgroup, namely a e 2 πi/n A . In other words, the n th roots of unity....
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This note was uploaded on 06/18/2009 for the course MATH 103A taught by Professor Rogalski,daniel during the Winter '08 term at UCSD.

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