Let $\sigma,\tau \in S_{13}$, with $\sigma=(1,7,12,10)(2,4,5,6,3)(8,11)(9,13)$ and $\tau^3=\sigma$
I would like to find such a $\tau$ or proof it doesn't exist in $S_{13}$.
In order to do so, I have calculated the order of $\sigma$ which is: $Ord(\sigma)=20$. (let the different cycles in σ respectively be $a_i$ then $Ord(\sigma)=lcm(a_1,a_2,a_3,a_4)=lcm(4,5,2,2)=20)$
Thus $\sigma^{20}=(1)$ which gives $\tau^{60}=(1)$ I don't think it exists since 3 is co-prime with, 2 & 5 but I'm not really sure what to do next in order to prove this.
For example: let $\varrho \in S_{13}, ϱ^4=\sigma$. Now I think it would be possible to create such a $\varrho$.
new info I understand that $ϱ^4=σ$ is not possible since this gives, $σ^{21}=σ=ϱ^{84}$, 84=2*2*3*7. This element would be in $S_{14}$. But $σ^{21}=σ=τ^{63}$, with 63=3*3*7. Since the cardinality of these cycles add up to 13 it seems possible to create such a τ. Could someone explain how to actually find this tau?
Thanks in advance!
Regards
Since the $Ord(\tau)=60$ we must find a permutation composed of $n$ cycles where each cycle cardinality summed up gives 13 (degree), and $lcm$ gives 60 (order).
Based on the factorisation of $60 = 2^2\cdot 3\cdot 5$ the corresponding cycle type is:
$type(\tau) = (1,2,1,0,1,0,0,0,0,0,0,0,0)$
Now prove that $\tau^3 = \sigma$
$type(\tau^3) = (4,2,0,0,1,0,0,0,0,0,0,0,0)$
But $type(\sigma) = (0,2,0,1,1,0,0,0,0,0,0,0,0) \neq type(\tau^3)$
You can try to give different representations of $\tau$ and see if you can get the same cycle type as $\sigma$