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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 zw1 = e ( rs ) i is also in G . An alternate proof: Suppose z, w ∈ G . Then  zw1  =  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.
 Winter '08
 Rogalski,Daniel
 Math, Algebra

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