4107_midterm2_2009_solutions

4107_midterm2_2009_solutions - Selected solutions to Math...

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Selected solutions to Math 4107 midterm 2, Fall 2009 December 11, 2009 1. You can look these up. 2. This is a routine computation, so I won’t bother to do it. 3. The center of S 2 is all of S 2 , since S 2 is cyclic of order 2, which is abelian. We claim that the center of S n for n 3 is just { e } , the identity: To see this, suppose ϕ is an element of the center. If ϕ is not the identity, then there exists a, b ∈ { 1 , 2 , ..., n } , a n = b , such that ϕ ( a ) = b. So, if we let c ∈ { 1 , 2 , ..., n } be distinct from a and b (possible, since n 3), we have that ϕ ( a c ) ϕ - 1 = ( b ϕ ( c )) n = ( a c ) . This contradicts the fact that ϕ was in the center; and so, the center contains only the identity. 4. If | G | = pq , q > p , then G has Sylow- p and Sylow- q subgroups of orders p and q , respectively. But how many Sylow subgroups? The number of Sylow- q ’s is 1 , q + 1 , 2 q + 1 , ... . But it can’t be 2 q + 1 or higher, since if any
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This note was uploaded on 10/23/2011 for the course MATH 4107 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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4107_midterm2_2009_solutions - Selected solutions to Math...

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