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    • Thread Starter

    Let \rho \inSym(n), p be prime, r be the remainder when n is divided by p (so 0\leqr<p and n=qp+r for some integer q).

    1. Show that \rho^p = \iota iff the cycles of \rho all have lengths 1 or p.

    2. Show that if \rho^p = \iota then |Supp(\rho)| is a multiple of p and |Fix(\rho)|\equiv r(mod p).

    I really don't have many ideas on these at all.

    Fix(\rho) := {x|x\rho = x}
    Supp(\rho) := {x|x\rho \neq x}

    1) If all cycles have length 1 then it is clear that \rho^p is the identity.
    I don't know what I can deduce from all cycles having length p. The other way around, I can see if we have the identity that all cycles could be length 1, but I don't know how to go about getting length p.

    2) I have no idea how to start this.

    • PS Helper

    PS Helper
    For (i), you should know the result that a permutation \sigma = \sigma_1 \sigma_2 ... \sigma_n (where the \sigma_i are disjoint permutations) has order the lowest common multiple of the orders of all the \sigma_i. If you haven't already proved this result it's maybe worth proving it (it's not too difficult), because it makes the answer fall out fairly fast.

    Unfortunately my memory fails me for (ii), but it's worth looking more closely at the definitions given to try and work out how they relate to p and r.
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Updated: January 24, 2010
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