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Math Help - Another Cycle Decomposition Question

  1. #1
    Super Member craig's Avatar
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    Another Cycle Decomposition Question

    In S_4 let \sigma be a cycle of length 4, and \tau be a transposition.

    Show \sigma \tau is even, and briefly explain why \sigma \tau\neq id

    Let \sigma = (abcd), then the decomposition would be (ab)(ac)(ad), and let \tau = (ef). This means that \sigma \tau would be (ab)(ac)(ad)(ef), which is even. Think I've gone about this in the correct way?

    For any transposition (ij), (ij)^2 = id, this means that (ij) = (ij).

    For our transpositions above, (ab)\neq(ac)\neq(ad), so even if (ef) = (ab) or (ac) or (ad), it would still not equal the identity transposition. Would this be enough to briefly explain?

    I managed to attempt the first couple of sections of the question, but this last part has got me very confused:

    Deduce that \sigma \tau is either a cycle of length 3, or a product of 2 disjoint transpositions. Conclude that \sigma = \gamma \lambda \tau for some not necessarily disjoint permutations \gamma, \lambda \in S_4

    Would greatly appreciate some help in how to go about this problem.

    Thanks in advance
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  2. #2
    Senior Member Tinyboss's Avatar
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    You're definitely on the right track. In general, an n-cycle is even for odd n, and vice-versa, so the first result is immediate. For the second part, I'd just note that the 4-cycle moves 4 elements, and any transposition can only restore at most 2 to their original place.
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by Tinyboss View Post
    You're definitely on the right track. In general, an n-cycle is even for odd n, and vice-versa, so the first result is immediate. For the second part, I'd just note that the 4-cycle moves 4 elements, and any transposition can only restore at most 2 to their original place.
    Thanks for the reply. That's a good idea about only restoring 2, didn't think of that.

    Any idea how to approach the last section?
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  4. #4
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    cycle decompositoions

    we know there is 12 even permutations in s4, we also know that id is not one of the solutions. (as you have said). solution have to be one of the other 11 even permutation, which are 8x3-cycles and 2xproduct of two disjoint transpositions.

    hope it will help
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  5. #5
    Super Member craig's Avatar
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    Yes that helps, cheers for the reply!
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