Find a permutation $\sigma \in S_9$ such that $\sigma^2=(13579)(268).$

So I know that $\sigma^{10}=\sigma.$ But I don't know $\sigma^5$..... Is $\sigma^{10}=\sigma^4\sigma^6$? I doubt this is the case because I am not getting the correct answer.

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    $\begingroup$ You don't have $\sigma^{10} = \sigma$ for that permutation. $9$ is not an exponent of $S_9$. What is the order of the product of two disjoint cycles? $\endgroup$ – Daniel Fischer Mar 9 '14 at 21:13
  • $\begingroup$ Why don't I have a $\sigma^{10}=\sigma$? Isn't $\sigma^9=\epsilon$, with $\epsilon$ being the identity. Is the order 15? $\endgroup$ – allie Mar 9 '14 at 21:20
  • $\begingroup$ Because $lcm(3,5)=15$, you have $\sigma^{30}=\epsilon$. $\endgroup$ – vadim123 Mar 9 '14 at 21:24
  • $\begingroup$ In this case, the order is indeed $5\cdot 3 = 15$. $\endgroup$ – Daniel Fischer Mar 9 '14 at 21:25
  • $\begingroup$ @vadim: You mean $\sigma^{15}=\epsilon$, right? (Not that what you said is wrong...) $\endgroup$ – TonyK Mar 9 '14 at 21:26

A "square root" of $(13579)$ is $(17395)$. A "square root" of $(268)$ is $(286)$.

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  • $\begingroup$ How did you figure that? $\endgroup$ – allie Mar 9 '14 at 21:38
  • $\begingroup$ Well, it just sort of came to me in a flash. But you can see that it works, right? In fact, any odd-length cycle $(s_1s_2\ldots s_{2n-1})$ has the square root $(s_1s_{n+1}s_2s_{n+2}\ldots s_{2n-2}s_{n-1}s_{2n-1}s_n)$. $\endgroup$ – TonyK Mar 9 '14 at 21:52

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