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Unformatted text preview: Algebra , Assignment 6 Solutions 1. Let : G G be an automorphism of G . Let x,y G with y = ( x ). (a) Assume x s = e . Prove y s = e . Solution: y s = ( x ) s = ( x s ) = ( e ) = e (b) Assume x s negationslash = e . Prove y s negationslash = e . Solution: As above, y s = ( x s ). An automorphism is injective (and surjective, for that matter) so if y s = e then x s = e . The required statement is the contrapositive. (c) Show o ( x ) = o ( y ) (You can assume they are both finite.) Solution: : o ( x ) is the least s for which x s = e and o ( y ) is the least s for which y s = e . Since the values s are the same for both x,y the least value s is the same for both. 2. Let S n be given by ( i ) = n +1 i (so 1 n is reversed.) When n = 403 is even or odd? For which n is even? Solution: We flip 1 , 403, 2 , 402 all the way to 201 , 202 so there are 201 flips so is odd. More generally it depends on whether n is even or odd. When n = 2 k + 1 there are k flips so it is even when k is even otherwise odd. When n = 2 k there are k flips so it is even when k is even otherwise odd. 3. Here we write perutations in terms of their disjoint cycles. (a) Given = (12)(34) and = (56)(13) find S 6 with  1 = Solution: We send 1234 into 5613. The other 56 go into the other 24 in either order: = parenleftbigg 1 2 3 4 5 6 5 6 1 3 2 4 parenrightbigg or parenleftbigg 1 2 3 4 5 6 5 6 1 3 4 2 parenrightbigg (b) Prove that with = (123) and = (13)(578) there is no with  1 = Solution: Such a exists if and only if , have the same cycle sizes with the same multiplicity: these are 3 for and 2 , 3 for . (c) Set = (12)(34). In S 8 precisely how many have the property that  1 = for some ? Solution: We must have = ( ab )( cd ). So there are 8 7 6 5 choices for a,b,c,d except we are overcounting by an eight factor as we can flip ab , flip cd , flip ab with cd . So the number is 7 6 5 = 210. 4. Consider the symmetries of the square. Use the table and notation on4....
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 Fall '08
 JOSEPHS
 Algebra

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