
#1
Oct1210, 05:13 PM

P: 607

1. The problem statement, all variables and given/known data
Let t be an element of S be the cycle (1,2....k) of length k with k<=n. a) prove that if a is an element of S then ata^1=(a(1),a(2),...,a(k)). Thus ata^1 is a cycle of length k. b)let b be any cycle of length k. Prove there exists a permutation a an element of S such that ata^1=b. 2. Relevant equations 3. The attempt at a solution We assume t is an element of S and a is an element S. By definition of elements of S if t is in S, we have a determined by t(1), t(2),...,t(n) Furthermore if a is in S, we have a(1), a(2)....a(n). That's as far as I get. 



#2
Oct1210, 05:21 PM

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P: 26,167

hi kathrynag!
just read out that equation in English … 



#3
Oct1210, 05:33 PM

P: 607

Does it go to 2?




#4
Oct1210, 05:49 PM

P: 607

Permutations, cycles
Ok if a is defined by 1>a(1), 2a(2),k>a(k), then a^1 is defined as a(1)>1,a(2)>2, a(k)>k
Then ata^1 for a(2) is a(t(2))=a(2) 



#5
Oct1210, 05:49 PM

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a^{1} sends it to 2 … so what does ata^{1} send it to? 



#6
Oct1210, 05:51 PM

P: 607

a(t(2))
t(2)=2 a(2) 



#7
Oct1210, 05:58 PM

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(1,2....k) means that it sends 1 to 2, 2 to 3, … and k to 1. so t(2) = … ? 



#8
Oct1210, 05:59 PM

P: 607

3
so we are left with a(3)=4 



#9
Oct1210, 06:09 PM

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you're not told anything about a, are you? a(3) is just a(3) ! :bigginr: (and I'm off to bed … see you tomorrow!) 



#10
Oct1210, 06:21 PM

P: 607

I thought I could do this:
By definition of elements of S if t is in S, we have a determined by t(1), t(2),...,t(n) Furthermore if a is in S, we have a(1), a(2)....a(n). 



#11
Oct1210, 07:44 PM

P: 607

So I have ata^1=at(n)
because a^1 sends a(1)>1,a(2)>2,.....a(n)>n By defininition of t, we have a(1), a(2), a(3)...a(k). We go to k because our definition of t says k<=n. For b, Let b be any cycle of length k. We have (1,2.....k). I'm not sure how to show the rest. 


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