# Permutation of order 12 and 30 in $S_{9}$

If I have the group $S_{9}$ and $\sigma, \tau \in S_{9}$ where $\vert \sigma \vert = 5$ and $\vert \tau \vert = 6$ is it then possible to have $\vert \sigma \tau \vert = 30$ and $\vert \sigma \tau \vert = 12$ why/why not?

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No element of $S_9$ has order $30$ (think about cycle types). – Chris Eagle Mar 31 '12 at 9:33
Not possible because $30\ne 12$... – lhf Apr 1 '12 at 2:59
@lhf, I took it to be two separate questions, one about order 30, another about order 12. But, who knows? – Gerry Myerson Apr 1 '12 at 6:27

$(12345)(234657)=(1365)(247)$.