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I saw the following statement in a book: If $\sigma, \tau\in S_n$ such that $\sigma\tau=\tau\sigma$, then the order of $\phi=\sigma\tau$ is the least common multiple of the orders of the permutations $\sigma$ and $\tau$. Is the statement true? A similar result which I have proved holds for a permutation written as a product of disjoint cycles: if a permutation is written as a product of disjoint cycles, then the order of the permutation is the least common multiple of the orders of the cycles. There is no condition for the permutations $\sigma$ and $\tau$ here. Thank you!

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    $\begingroup$ If $\tau =\sigma$ has order $2$, then $\phi $ has order $1$ $\endgroup$ – HK Lee Dec 29 '14 at 13:15
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If $\sigma\tau = \tau\sigma$, then $$(\sigma\tau)^k = \sigma^k \tau^k$$ for any $k \in \mathbb N$. Therefore we can say that the order of $\phi = \sigma\tau$ divides the lowest common multiple of the orders of $\sigma$ and $\tau$.

However, it is false in general the the order of $\phi$ will equal this lowest common multiple.

For example, if $\sigma \in \langle \tau \rangle$, then $\tau$ and $\sigma$ will always commute, but the order of $\sigma\tau$ could be less than the lowest common multiple. In $S_3$, $(123)$ and $(132)$ both commute and have order $3$, but their product is $e$ which has order $1$.

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This is not true in general. $\tau = \sigma^{-1}$ has the same order as $\sigma$.

Also suppose $\sigma$ has order $9$, then if $\tau = \sigma^2$ then $\tau$ has order $9$, but $\sigma \tau$ has order $3$.

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