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Suppose a cycle $\sigma\in S_n$ has decomposition a product of disjoint $p$-cycles, for $p$ a prime. So each cycle has order $p$, and thus $\sigma^p=id$, so $\sigma$ has order dividing $p$, and thus order $p$ when $\sigma\neq id$.

What if $\sigma$ has a decomposition into a product of disjoint $m$-cycles, for $m$ composite. I know $\sigma^m=id$, and thus has order dividing $m$. In every example I've tried, it seems $\sigma$ still has order $m$. Is it possible for $\sigma$ to have order smaller than $m$ (when not the identity of course)? If so, is there an example?

Much thanks,

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exercise: the order of a product of disjoint cycles is the lcm of the cycle lengths – yoyo Aug 31 '11 at 23:05
up vote 2 down vote accepted

A permutation that is the product of disjoint permutations that all have the same order, has the same order as its factors. So your $\sigma$ will have order $m$, guaranteed.

(It must have order at most $m$ because the disjoint factors commute, so $\sigma^m = (xy\cdots z)^m = x^m y^m \cdots z^m = 1\cdot 1\cdots 1 = 1$. On the other hand, by the same argument, if it had order less than $m$, each factor would also have that order, because the only way a product of disjoint permutations can be the identity is if each of the factors are).

(By the way $\sigma$ is not itself a "cycle" if it is the product of two or more disjoint permutations).

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