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 March 11th, 2012, 04:22 AM #1 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Proof on product of cycles (permutation groups) Hey everybody, I'm having a hard time with the following proof: s = (a1 a2 ... ak), a k-cycle. Prove that t * s * t^-1 = (t(a1) t(a2) ... t(ak)) I can't figure out where exactly to start... t * s * t^-1 = (b1 b2 ... bn) (a1 a2 ... ak) (bn ... b1), because the last cycle is just t^-1, t inverse. Then I noticed that for every element in t that is also in s, say bi, the following happens: bi-1 --> bi = aj ---> aj+1, for a certain i and j, so that bi = aj. For every element in t that is not in s, it gets projected onto itself, because then you have t * t^-1. (t(a1) t(a2) ... t(ak)) is just every element of s permuted according to tau, but with the sequence of s still intact. Is this correct? I feel I'm close, hints greatly appreciated! Greets, Kappie March 13th, 2012, 09:30 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Proof on product of cycles (permutation groups) let's look at 2 cases: j = t(ai) for some i, and j ? t(ai) for any i (here j is an element of {1,2,...,n} as the assumption is we are working in Sn). in the first case, we have: tst^-1(j) = t(s(t^-1(j))) = t(s(t^-1(t(ai)))) = t(s(ai)) = t(a(i+1))) (where we may have to take i+1 mod k) that is: tst^-1 takes t(ai) to t(a(i+1)) (more properly, i+1 (mod k)). in the second case, we want to show that tst^-1(j) = j. now if j is not the image of any ai under t, then t^-1(j) ? ai, for any ai. that is, s leaves t^-1(j) fixed. so tst^-1(j) = t(s(t^-1(j))) = t(t^-1(j)) = j, as desired. it's easier to see like this (t^-1 is left-most, then s, then t on the right): j= t(ai)--->ai--->ai+1--->t(a(i+1)) or: j--->t^-1(j)--->t^-1(j)--->j to see an in-depth example, let's see how (1 2 3)(1 2 4)(1 3 2) = (2 3 4). our first case applies to j = 2,3 or 4. if t = (1 2 3), we have: 2 = t(1) 3 = t(2) 4 = t(4) so (1 2 3)(1 2 3 4)(1 3 2) acts on 2 like so: 2 = t(1), so t^-1(2) = t^-1(t(1)) = 1 (pretty obvious that (1 3 2) = t^-1 takes 2 to 1, right?). now s (which is (1 2 4) in this case) acts on 1 by moving it to 2: s(1) = 2. finally, t(2) = 3, so tst^-1(2) = 3, tst^-1 moves 2 to 3. similarly, we get: 3--->2 (since t^-1(3) = t^-1(t(2)) = 2) --->4 (since s(2) = 4) --->t(4) = 4, that is: tst^-1(3) = 4, tst^-1 moves 3 (t(2)) to 4 (t(4)). and very quickly: 4--->4--->1--->2. the interesting part is what happens for j = 1: 1 = t(3), and 3 isn't affected by s at all. so when we take t^-1 of 1, we get 3, which s passes along unchanged, only to be "undone" by t back to 1. and tst^-1 = (2 3 4) indeed leaves 1 unscathed. Tags cycles, groups, permutation, product, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jacob Abstract Algebra 0 August 24th, 2013 12:23 AM Solarmew Applied Math 1 April 12th, 2012 09:01 AM samyuktha Abstract Algebra 35 February 14th, 2012 09:16 AM MarkyMark Abstract Algebra 2 January 15th, 2009 08:34 PM Frazier001 Abstract Algebra 1 October 23rd, 2007 12:22 AM

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