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Unformatted text preview: Modern Algebra Practice Exam  Solutions Disclaimer: This practice exam is not intended to reflect the content of Wednesdays midterm. It is simply a list of problems left over from the preparation of the actual exam, and should serve to indicate the general format and difficulty level thereof. Solutions will be posted Monday evening. Problem 1. Let G be a finite group. Show that the number of elements in G of order greater than 2 must be even. Conclude that any group of even order must contain an element of order 2. Solution. This is really just a simple counting argument. Let S be the set of all elements of G with order greater than two. If S = then  S  = 0 and were finished. So assume S 6 = . Since  x  =  x 1  for any x G , we see that x S if and only if x 1 S . Moreover, if x S then x 2 6 = e so that x 6 = x 1 . It follows that S can be written as the disjoint union of two element sets of the form { x,x 1 } , and hence that  S  is even. Suppose that  G  is even. Since e has order 1, e 6 S . It follows that G \ S 6 = . So <  G \ S  =  G    S  . Since  G  and  S  are both even, it follows that  G \ S  is a nonzero even integer, i.e. is at least 2. Thus, there is aneven integer, i....
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 Fall '10
 RyanDaileda
 Algebra

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