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Thread: Permutations and cycles

  1. #1
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    Permutations and cycles

    Let $\displaystyle \rho \in Sym(n)$ and let p be a prime number. Show that $\displaystyle \rho^p = \iota$ if and only if the cycles of $\displaystyle \rho$ have lengths 1 or p.

    I realise that this is an if and only if question, so it will need to be shown in the two directions.
    For showing that if the cycles all have lengths 1 or p, then $\displaystyle \rho^p = \iota$ I have have shown it for length 1 but can't see how to show it for p. I can't think of where to start for the proof in the other direction. Help anyone?
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  2. #2
    Senior Member roninpro's Avatar
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    Suppose we have a permutation $\displaystyle \rho$. It can be broken up into a product of disjoint cycles: $\displaystyle \rho=\sigma_1 \sigma_2\cdots \sigma_k$. If $\displaystyle n_1, n_2,\ldots, n_k$ are the lengths of each cycle, it can be shown that the order of $\displaystyle \rho$ is $\displaystyle \text{lcm}(n_1,n_2,\ldots n_k)$.

    So, if the order of $\displaystyle \rho$ is a prime $\displaystyle p$, then $\displaystyle p=\text{lcm}(n_1,n_2,\ldots n_k)$.

    I will leave it to you to conclude.
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