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Symmetric groups proof

  1. Mar 19, 2015 #1
    • Member warned that the template is required
    If σ is a k-cycle with k odd, prove that there is a cycle τ such that τ^2=σ.

    I know that every cycle in Sn is the product of disjoint cycles as well as the product of transpositions; however, I'm not sure if using these facts would help me with this proof. Could anyone point me in the right direction?
     
  2. jcsd
  3. Mar 19, 2015 #2

    Dick

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    Here's a general strategy for problems like this. Start with simple cases. Take (123). That's (132)^2. Easy enough. Now try (12345). Just use trial and error. Can you see any pattern emerging you can use to solve the general case?
     
  4. Mar 19, 2015 #3
    If k is odd, then k=2n+1 for some n. That would be a good place to start.

    (edited to change order to parity)

    (edited again to write out what k is)

    (doh)
     
    Last edited: Mar 19, 2015
  5. Mar 19, 2015 #4

    Dick

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    Hah. That's a better solution. I like it.
     
  6. Mar 19, 2015 #5

    Dick

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    Doh, here too. But I know what you are onto. The parity of the group element is even but the order of the group element is odd. That's all you need. You could also discover a form for the square root by using my route.
     
  7. Mar 20, 2015 #6
    Okay, I got it now. Thanks! There's another one I can't figure out.

    Prove that every element of An is a product of n-cycles.

    I always think of induction when I see a "for every element of such and such", but I don't think induction would be the right direction for this one. Instead, I would let α∈An where α is arbitrary and try to show that this can be written as the product of n-cycles. How could I get this to work?
     
  8. Mar 20, 2015 #7
    I haven't thought very hard about it but I think parity might be important in that one because An is the subgroup of even permutations. So any element of An can be written as a product of an even number of transpositions, & so would any product of any number of elements. So whatever you get will be another even permutation. I don't know if that matters & I'm not sure off the top of my head why they have to be n-cycles (although I can believe it) so maybe that doesn't help, that's just the first thing I thought of.
     
    Last edited: Mar 20, 2015
  9. Mar 21, 2015 #8
    So I noticed that if n is odd it seems a lot easier because n-cycles are already even. If n is even there are no n-cycles in An so I had to make some & I found that (fingers crossed) the following product of n-cycles works to create some (n-1)-cycles which will be in An:
    $$(x_{1},\,x_{n},\, x_{n-1},\, ...,\,x_{2})(x_{1},\,x_{3},\,x_{5},\,...,\,x_{n-1},\,x_{2},\,x_{4},\,...,\,x_{n}) = (x_{1},\, x_{2},\, x_{3},\,...,\,x_{n-1})(x_{n})$$

    For example if n is 4 I have (1432)(1324) = (123)(4) & if n is 6, (165432)(135246) = (12345)(6) etc just relabel the ##x_{k}## to get whatever elements are needed & they're products of n-cycles. I think that's one way it could work.... :olduhh:
     
  10. Mar 21, 2015 #9

    Dick

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    We are probably getting close to doing too much thinking for tropian1 with too little feedback, but think about generators for ##A_n##. ##S_n## is generated by all transpositions (2-cycles). ##A_n## is generated by all products of pairs of transpositions, yes? Wouldn't it be nice if you could make a pattern where the product of two n-cycles could create an arbitrary product of a pair of disjoint 2-cycles? You are going to need n>3, of course.
     
    Last edited: Mar 21, 2015
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