TITLE: Finding all $23$rd roots of a given element of $S_{10}$
QUESTION [5 upvotes]: Given the permutation 
  $$\sigma = 
\begin{pmatrix} 
1 &2 &3 &4 &5 &6 &7 &8 &9 &10\\
1 &2 &3 &7 &5 &6 &4 &8 &9 &10
\end{pmatrix}$$
  solve the equation $$\gamma^{23} = \sigma ,$$
  where $\sigma,\gamma \in S_{10}$ 

What I did: $\sigma $ in cycle notation is $(47)$ i.e a transposition.The order of $\sigma$ is 2 because the length of the cycle is 2, so that means $\sigma$ his it's own inverse (I think) i.e $\sigma \circ\sigma = e$.
Taking the equation $\gamma^{23} = \sigma$ and multiplying on the right by $\sigma$ I get  $\gamma^{23} \circ \sigma = e$ , so that means that $\gamma^{23}$ and $\sigma$ are inverses , that implies that $\gamma$ must be  equal to $\sigma$ since the order of $\sigma$ is 2 and $\gamma^{22} = \gamma^{24} = e$ since 22 and 24 are divided by the order of $\sigma$ , so the only option is $\gamma = \sigma$. 
Is that correct?Is there another way of solving this type of equations? I doubt it's right.

REPLY [8 votes]: Since $\sigma$ is a transposition, any solution $\gamma$ satisfies $e = \sigma^2 = (\gamma^{23})^2 = \gamma^{46} .$ The prime factorization of $46$ is $2 \cdot 23$, so $\gamma$ has order $1$ (not possible, since $e^{23} = e \neq \sigma$), $2$, $23$ or $46$.
Every element of $S_{10}$ has order dividing $|S_{10}| = 10!$. Now, $23$, being a prime larger than $10$, does not divide $10!$, so...

...$\gamma$ cannot have order $23$ or $46 = 2 \cdot 23$. Thus, it has order $2$, and so $$\sigma = \gamma^{23} = (\gamma^2)^{11} \gamma = e^{11} \gamma = \gamma .$$