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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 negationslash = 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 )) negationslash = ( 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 pair of Sylow- q ’s shared a non-idenitty element they would have to be the same subgroup; so, if there were 2 q + 1 or more of them, then they would, collectively, contribute at least (2 q + 1)( q 1) > pq = | G | elements of order q , impossible. So, there is only one Sylow-

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