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Kuperberg (10/17/07) Math 150a: Modern Algebra First Midterm Solutions 1. Show that every finite group G has an even number of elements of order 3. Solution: In general, for any element g of any group, the orders of g and g - 1 are the same. This comes from the fact that g n = e g - n = e . Now, if g G has order 3, then g - 1 also has order 3, and we can begin to count off such elements (of order 3) in pairs. There won’t be any double counting, since g = g - 1 would imply that g has order 2. Since G is a finite group, we are also guaranteed to be finished counting at some point. Thus, we will have an even number of elements of order 3. ± 2. Consider a product a = xyzw of four transpositions in some symmetric group S n . Can this product be a 4-cycle? No. A k -cycle must have signature ( - 1 ) k - 1 . So an even permu- tation, such as a , cannot be a cycle of even length. Can it be a 5-cycle? Yes. Here’s an example: ( 1 5 )( 1 4 )( 1 3 )( 1 2 ) = ( 1 2 3 4 5 ) . Can it be a 9-cycle?
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This homework help was uploaded on 02/01/2008 for the course MATH 150A taught by Professor Kuperberg during the Spring '03 term at UC Davis.

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MIDTERM 1 SOLUTION - Kuperberg (10/17/07) Math 150a: Modern...

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