Permutations and cycles

**worc3247** · January 25th 2011, 11:15 AM

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

**roninpro** · January 25th 2011, 02:19 PM

Suppose we have a permutation $\rho$ . It can be broken up into a product of disjoint cycles: $\rho=\sigma_1 \sigma_2\cdots \sigma_k$ . If $n_1, n_2,\ldots, n_k$ are the lengths of each cycle, it can be shown that the order of $\rho$ is $\text{lcm}(n_1,n_2,\ldots n_k)$ .

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

I will leave it to you to conclude.

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