# sol6 - Algebra Assignment 6 Solutions 1 Let σ G → G be...

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Unformatted text preview: Algebra , Assignment 6 Solutions 1. Let σ : G → G be an automorphism of G . Let x,y ∈ G with y = σ ( x ). (a) Assume x s = e . Prove y s = e . Solution: y s = σ ( x ) s = σ ( x s ) = σ ( e ) = e (b) Assume x s negationslash = e . Prove y s negationslash = e . Solution: As above, y s = σ ( 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 σ ∈ S n be given by σ ( i ) = n +1 − i (so 1 ··· n is “reversed.”) When n = 403 is σ even or odd? For which n is σ even? Solution: We flip 1 , 403, 2 , 402 all the way to 201 , 202 so there are 201 flips so σ 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 σ = (12)(34) and γ = (56)(13) find τ ∈ S 6 with τ- 1 στ = γ Solution: We send 1234 into 5613. The other 56 go into the other 24 in either order: τ = 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 σ = (123) and γ = (13)(578) there is no τ with τ- 1 στ = γ Solution: Such a τ exists if and only if σ,γ have the same cycle sizes with the same multiplicity: these are 3 for σ and 2 , 3 for γ . (c) Set σ = (12)(34). In S 8 precisely how many γ have the property that τ- 1 στ = γ for some τ ? Solution: We must have γ = ( 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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sol6 - Algebra Assignment 6 Solutions 1 Let σ G → G be...

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