My Math Forum Proof on product of cycles (permutation groups)

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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.

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