Compute $\sigma$, given $\sigma^{11}$ in $S_{10}.$ 
Let $\sigma \in S_{10}$ with $\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)$. How to compute $\sigma$?

I tried:
$\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
Now I tried to find two permutations such that 
$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\  &  &  &  &  &  & &  &  & \end{pmatrix}\cdot \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\  &  &  &  &  &  & &  &  & \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
But it's not working.
How to find $\sigma$ here?
 A: Let $\tau= (314)(529687a)$ (with $a=10$) & so $\tau^{21}=e$ & $\sigma^{21}=e$. We have $ \tau= \sigma^{11}$. Squaring gives $\sigma=\tau^2$. Which is easy to compute.
A: The first thing to note is that $11$ is $-1$ modulo $3$ and $4$ modulo $7$. In particular, we note that $(1,3,4)^{11} = (3,1,4)$, so if we can find an element $\tau$ of the symmetric group on $\{2,5,6,7,8,9,10\}$ whose eleventh power is $(5,2,9,6,8,7,10)$, we're done. Now, we note that the $n$th power of a $k$ cycle is a $k$ cycle whenever $k$ and $n$ are coprime, as in this case, so it's reasonable to guess that we're looking for a $7$-cycle, and we note that the eleventh power of a $7$-cycle is equal to its $4$th power, so we're looking for $(a,b,c,d,e,f,g)$ such that  $(a,b,c,d,e,f,g)^4 = (5,2,9,6,8,7,10)$. But $(a,b,c,d,e,f,g)^4 = (a,e,b,f,c,g,d)$, so we have $(a,b,c,d,e,f,g) = (5,9,8,10,2,6,7)$. Putting these together (since disjoint cycles commute), we have $$((1,3,4)(5,9,8,10,2,6,7))^{11} = (3,1,4)(5,2,9,6,8,7,10).$$
