I have tried to solve it for the case when $n=1,2,3,4$ which i found that there is only one group for $n=1$, one group for $n=2,$ one group for$ n=3 $ and two groups for $ n=4$, then my question come, if i were given a number $n$ which is larger, how do i know how many different groups are there?
(I treat two isomorphic groups as the same group in this question.)
While anon's comment is correct, there is one thing we can do - count the number (when you see "number", read "number of isomorphism classes") of abelian groups of order $n$. While I'll briefly explain how we can do so, in some respects this is fairly useless for giving a lower bound on the number of groups of order $n$. For example, for $n = 1024 = 2^{10}$, it has been tediously calculated by a computer that there are $49487365422$ groups of order $2^{10}$ - yet only $42$ of these are abelian.
To do this, we use the following result:
If $G$ is a finite abelian $p$-group with $|G|=p^n$, then $$G \cong C_{p^{a_1}} \oplus \ldots \oplus C_{p^{a_n}} $$ where $0 < a_1, \ldots, a_n \leq n$, $a_1 + \ldots + a_n = n$, and $C_k$ denotes the cyclic group of order $k$.
Note that as $C_{pq} \cong C_p \oplus C_q$ if and only if $p$ and $q$ are coprime, this allows us to deal with all finite abelian $G$ is full generality. (A proof of the above result can be found in pretty much any textbook on abstract algebra). From the result in blockquotes, it follows that the number of abelian groups of order $p^n$ is equal to the number of integer partitions of $n$. Generally, we get:
The number of abelian isomorphism classes $A(n)$ of a finite group $G$ are, given that $|G| = p_1^{a_1} \ldots p_n^{a_n}$, and denoting the partition function as $P(n)$, are $$A(n) = P(a_1) \ldots P(a_n)$$
