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! just read out that equation in English … … so what, for example, does ata^-1 do to a(2) ?
Ok if a is defined by 1--->a(1), 2-----a(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)
ah! no, you're misunderstanding the notation for a cycle … (1,2....k) means that it sends 1 to 2, 2 to 3, … and k to 1. so t(2) = … ?
no!! 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!)
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).
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! (just got up :zzz: …) do you mean we have (a(1), a(2), a(3),...a(k))? anyway, before we go any further, i need to know: what exactly is S?