Permutations, cycles
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. 
hi kathrynag! :smile:
just read out that equation in English … Quote:

Re: Permutations, cycles
Does it go to 2?

Re: 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) 
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a^{1} sends it to 2 … so what does ata^{1} send it to? 
Re: Permutations, cycles
a(t(2))
t(2)=2 a(2) 
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(1,2....k) means that it sends 1 to 2, 2 to 3, … and k to 1. :wink: so t(2) = … ? 
Re: Permutations, cycles
3
so we are left with a(3)=4 
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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 :zzz: … see you tomorrow!) 
Re: Permutations, cycles
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). 
Re: Permutations, cycles
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. 
hi kathrynag! :smile:
(just got up :zzz: …) Quote:
anyway, before we go any further, i need to know: what exactly is S? :confused: 
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