Is it possible to find all finite groups which has subgroup of order $n$? Is it possible to find all finite group which has a subgroup of certain order $n$ (here say for example $2$) ? I am thinking about this problem for few days, but not getting how to approach.
I am thinking that the theorem - 

Every finite group is isomorphic to a subgroup of $S_n$.

can give something about it. Can anyone answer please? I have searched for it, but it is not posted here and not available in internet.
 A: First, the result regarding $S_n$ that you are referring to is known as "Cayley's theorem".
In fact, a necessary condition is $n$ divides $|G|$, the order of the group $G$.
Indeed, if there exists a subgroup of $G$ that has order $n$, then $n$ will divide $|G|$ by Lagrange's theorem.
Under a certain condition, a converse holds.
Suppose that $n$ divides $|G|$, and pick Sylow groups $P_1, \ldots, P_r$, where we assume that $n = p_1^{k_1} \cdots p_r^{k_r}$ and $P_j$ is a $p_j$-Sylow group.
Now if $P$ is any $p$-group, then it has a normal subgroup of order $p^m$, whenever $p^m$ divides $|P|$. For $m=1$ this is Cauchy's theorem, and for general $m$ use Cauchy's theorem and divide out the corresponding cyclic group and then use induction.
Thus, pick for each $j$ a subgroup $Q_j \le P_j$ of order $p_j^{k_j}$. Then the $Q_j$ are pairwise disjoint for order reasons, so that they define a subgroup
$$
H := \langle Q_1, \ldots, Q_r \rangle
$$
that is a direct product of $Q_1, \ldots, Q_r$ (if each $Q_j$ is a normal subgroup of $H$) and hence has order $n$.
A: As noted in the comments by Display name and Orat, it is (as of today) improbable for general or even for a given $n$ to find all groups containing a subgroup of order $n$ in the sense of classification - to formulate a list of groups such as that any group is isomorphic to one in the list.
On the other hand, sometimes necessary and/or sufficient conditions can be given. One simple necessary condition is that $n$ divides the order of $G$, and one simple sufficient condition is that $G$ has order $n$.
For specific $n$, better conditions can be given. If $n = p$ is a prime, then Cauchy's theorem states that a necessary and sufficient condition is that $p$ divides the order of $G$. This is strengthened by Sylow's theorems: a necessary and sufficient condition for a finite group $G$ to have a subgroup of order $p^a$ is that $p^a$ divides the order of $G$. In terms of classification, even $p$-groups are hard to tell, let alone all groups who contain them.
When $n$ is no longer a prime, things get complicated. If we factor $n = p_1^{a_1}\cdots p_k^{a_k}$, then for a group $G$ whose order is a multiple of $n$ we can easily find subgroups $H_i$ with order $p_i^{a_i}$ for every $1 \leq i \leq k$. But we cannot guarantee in general that their product is again a subgroup (altough it will have the correct order as a set and every subgroup of order $n$ can be written in such way). I don't have a construction out of my pocket, but I believe that arbitrarily bad behaved examples can be found.
If your group $G$ is nilpotent, than each of the $H_i$ in the above construction are normal in $G$ and you can guarantee that the product $H = H_1\cdots H_k$ is again a subgroup of $G$ of order $n$. Classify all nilpotent finite groups - that's also a hard problem. 
According to GroupWiki, your property lies somewhere between supersolvable (which is weaker than nilpotency) and solvable, as the alternating group on four letters $A_4$ is a solvable group that does not satisty this property. It also states that any group satisfying it must be solvable. Note that I'm talking about an even stronger property: a group $G$ having subgroups of every order possible (at least one subgroup for each divisor of it's order). If we talk about a fixed $n$, then we go back to the original problem: if $H$ has order $n$ and $G$ is any finite group, $G\times H$ is a group containing a subgroup of order $n$, and their classification involves classifying all finite groups.
