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I have the following question : $$\alpha=(714)(3925)$$ Find $\beta \in S_9$ such that : $\beta^5=\alpha$

I know that the $o(\alpha)=12$

I do have the answer, yet I don't understand the way, I tried to "play" with $\beta$ and $\alpha$ since we know that $\alpha=\alpha^{13}$ yet were unsuccessful with achieving result.

Any ideas? (If you could please go to detail so I understand how to approach to such questions I'll be grateful).

Thank you!

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    you might find it helpful to look at the powers of $(7,1,4)$ and $(3,9,2,5)$ – 2017-01-27
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    Note that $5\cdot 5\equiv 1 \bmod 12$. – 2017-01-27

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Hint: $\alpha^{12}=1$ so that $\alpha=\alpha ^{13}=\alpha^{25}=\alpha^{37}=\dots $

Now can you spot a fifth power?

And can you see how you would generalise this?

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    Fifth power is $61$, generalise maybe $k\in \mathbb{Z}$ so that $12k+1$? – 2017-01-27
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    @JaVaPG Note $(\alpha^5)^5$. If $\alpha^r=1$ then $\alpha$ is an $q$-th power if you can solve $nr+1=mq$ because then $\alpha=\alpha^{nr+1}=\alpha^{mq}$. This can be done, if it is possible, using Euclid's algorithm. If $r$ is the order of $\alpha$ then this can be done iff $p$ is coprime to $r$. – 2017-01-27
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    I'm still not sure about it this is what I did : $\beta^5=\alpha \rightarrow \beta^5=(\alpha^5)^5 \rightarrow \beta^5=(\alpha^{12})^2*\alpha$ so I managed to reach $\alpha^{nr+1}$ yet I don't really understand what I suppose to now, why did you choose $(\alpha^5)^5$ – 2017-01-28
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    @JaVaPG Suppose $\beta = \alpha^5$, you then have $\alpha=\alpha^{25}=\beta^5$ – 2017-01-28
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    How did you know to suppose $\beta = \alpha^5$? We are looking for $\beta^5=\alpha$ – 2017-01-28
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    @JaVaPG I was looking for $\alpha=\alpha^{12k+1}$ with $12k+1=5m$. I then put $\beta=\alpha^m$ – 2017-01-28