How to find the subgroups of $S_n$ which are isomorphic to $\mathbb{Z}_l$? This post is somewhat generalization of my previous question, number of subgroups in $S_5$ which are isomorphic to $\mathbb{Z}_5$
I want to generalize the previous post and want some procedure or known facts regarding this.
i.e.,

How to find the subgroups of $S_n$ which are isomorphic to $\mathbb{Z}_l$?

My naive idea is to express the explicit element of $S_n$ and see the cyclic order of $l$... But for me it seems it is not enough.
 A: Yes, every cyclic subgroup of $S_n$ of order $k$ (I'm using $k$ because I find $l$ difficult to read, too close to $1$ for my taste) is generated by an element of $S_n$ of order $k$ (and of course often the same subgroup can be generated by different elements, but you don't seem to be interested in counting subgroups in this question, only finding them - so that's another story - but the below may assist in counting them).
It helps to remember that the order of an element of $S_n$ can be determined from its cycle type - that is, the list of positive integers $(a_1, \dots, a_m)$ for which the element can be written as product of disjoint cycles of lengths $a_1, \dots, a_m$,  respectively.  In this event, fairly simple calculations can show you that the order of an element of cycle type $(a_1, \dots, a_m)$ is $\operatorname{lcm}(a_1, \dots, a_m)$.
So in terms of cycle types, you're looking for ones corresponding to tuples of positive integers $(a_1, \dots, a_m)$ with $\operatorname{lcm}(a_1, \dots, a_m) = k$.  And because the disjoint cycles that make up this cycle type have to fit in $S_n$ it is also required here that $a_1 + \dots + a_m = n$.  (There's no harm in thinking of fixed points of a permutation as cycles of length $1$, which is why I have equality here.)
Recall that tuples of positive integers summing to a positive integer $n$ are sometimes called partitions of $n$.  You're looking to see if there are any partitions of $n$, the $\operatorname{lcm}$ of which is $k$.  If there are any, then you just fill in cycles corresponding to that cycle type.  For many values of $n$ and $k$ computers would likely to be better at performing this computation than people.  I don't know of a way to simplify it, or a simpler way of approaching it.
For example if $n=10$ and $k=6$, I'm looking for partitions of $10$ with $\operatorname{lcm}(6)$.  Doublecheck, e.g. via your own reasoning or via http://oeis.org/wiki/10, that there are not so many such partitions, including e.g. $(3,2,1,1,1,1,1)$, $(3,2,2,2,1)$, and $(6,1,1,1,1)$ and others, but not so many.  If I've done this right that means the subgroups of $S_{10}$ of order $6$ must be generated permutations of these cycle types, e.g. those of the form $(abc)(de)$, or $(abc)(de)(fg)(hi)$, or $(abcdef)$, etc.  It's pretty easy to write down examples of these (and others), and from the above we know that those are all of them.
From there, maybe you can begin identifying when subgroups with these types of generators happen to be equal, and begin counting them.
