Recently, while studying for my group theory class, I tried to solve the following problem
Find all $7$-Sylow subgroups of $S_{14}$.
Since the order of $S_{14}$ is $14!$ I know that $P$, a $7$-Sylow subgroup of $S_{14}$, has order $|P|=7^2$. By a previous result this implies that $P$ is abelian. Hence $P$ must be isomorphic to $\mathbb{Z}_{49}$ or $\mathbb{Z}_{7}\times \mathbb{Z}_{7}$. But given an element $\sigma \in S_{14}$ and its decomposition into disjoint cycles $\sigma=\sigma_1 \cdots \sigma_n$ $$ |\sigma|={\rm lcm}(|\sigma_1|, \cdots, |\sigma_n|). $$ Since the order of a cycle equals the length of the cycle itself I know that in $S_{14}$, $|\sigma_i| \leq 14$. If $|\sigma|=49$ then $|\sigma_i|$ divides $49$. Since $|\sigma_i| \leq 14$ we must have $|\sigma_i| =7$. But then ${\rm lcm}(|\sigma_1|, \cdots, |\sigma_n|)=7$. So I conclude that there is no element of order $49$ in $S_{14}$. Hence $P \cong \mathbb{Z}_{49}$ is not possible and $P$ must be isomorphic to $\mathbb{Z}_{7}\times \mathbb{Z}_{7}$. In this case I must find all subgroups of $S_{14}$ isomorphic to $\mathbb{Z}_{7}\times \mathbb{Z}_{7}$.
So my questions are:
Are there any mistakes in what I have done so far?
Could someone give me a hint on how to find all the isomorphic subgroups for $\mathbb{Z}_{7}\times \mathbb{Z}_{7}$?
I know that all its elements are of order $7$ and that it is generated by the elements $(1,0)$ and $(0,1)$. So I tried to find all elements of order $7$ in $S_{14}$ which are the $7$-cycles and the product of two disjoint $7$-cycles (correct?). But I don't know how to find which of these pairs are going to generate the subgroups I am looking for.
Any help is appreciated.