What is the general formulae for calculating the number of conjugate subgroups in Sn for a particular cycle?
And what is the relation between number of commuting elements with a particular cycle of Sn and conjugating elements? Please explain with an example.
Any help would be greatly appreciated.
suppose you have a cycle $c=(a_1,\dots,a_k)$. For any permutation $\sigma$ of $S_n$, you´ll have that $\sigma^{-1} c\sigma$ is the cycle $(\sigma(a_1),\dots,\sigma(a_n))$ of $S_n$. Thus, the conjugates of a cycle of length $k$ in $S_n$ are all the cycles of length $k$. And the conjugate subgroups of the one generated by the cycle are the subgroups generated by cycles of length $k$.
How many of them are there? Well, there are $\frac{n!}{(n-k)!}$ ways of picking $k$ ordered elements out of $n$, and each cycle can be represented in $k$ different ways (for example $(123)=(231)=(312)$), you have $\frac{n!}{k\cdot(n-k)!}$ conjugate elements of the cycle.
However, in the subgroup generated by a cycle of order $k$ there are $\phi(k)$ different $k$-cycles which generate it (where $\phi(k)$ is the Euler Phi function). That means there will be just $\frac{n!}{k\phi(k)(n-k)!}$ conjugate subgroups.
As for the second question, the elements of the group which commute with a given elements for a subgroup called the centralizer. To work out its order, you can consider the action of $S_n$ on itself by conjugation. Then the orbit-stabilizer theorem tells us that the order of the group is the order of the centralizer of an element times the size of its conjugation class.
For example, if you consider the element $(123)$ in $S_5$, it has $\frac{5!}{3(5-3)!}=20$ conjugate elements (all the $3$-cycles), so its centralizer will have order $\frac{|S_5|}{20}=6$. (In this case it will be the subgroup generated by $(123)$ and $(45)$)