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Adv Alegbra HW Solutions 47

Adv Alegbra HW Solutions 47 - (i Prove that if α and β...

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47 (ii) Compute the parity of f . Solution. f = ( 26107 )( 3945 ) . Since 4-cycles are odd, f is even. (iii) Compute the inverse of f . Solution. f 1 = ( 71062 )( 5493 ) . 2.29 (i) Prove that α is regular if and only if α is a power of an n -cycle. Solution. If α = ( a 1 a 2 ··· a k )( b 1 b 2 ··· b k ) ··· ( c 1 c 2 ··· c k ) is a product of disjoint k -cycles involving all the numbers between 1 and n , show that α = β k , where β = ( a 1 b 1 ··· z 1 a 2 b 2 ··· z 2 ... a k b k ··· z k ). (ii) Prove that if α is an r -cycle, then α k is a product of ( r , k ) disjoint cycles, each of length r /( r , k ) . Solution. Absent. (iii) If p is a prime, prove that every power of a p -cycle is either a p -cycle or ( 1 ) . Solution. Absent. (iv) How many regular permutations are there in S 5 ? How many regu- lar permutations are there in S 8 ? Solution. Absent. 2.30
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Unformatted text preview: (i) Prove that if α and β are (not necessarily disjoint) permutations that commute, then (αβ) k = α k β k for all k ≥ 1. Solution. We prove f rst, by induction on k ≥ 1, that βα k = α k β . The base step is true because α and β commute. For the inductive step, βα k + 1 = βα k α = α k βα ( inductive hypothesis ) = α k αβ = α k + 1 β. We now prove the result by induction on k ≥ 1. The base step is obviously true. For the inductive step, (αβ) k + 1 = αβ(αβ) k = αβα k β k ( inductive hypothesis ) = αα k ββ k ( proof above ) = α k + 1 β k + 1 ....
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