Every prime ordered group is simple, its because it doesn't admit any subgroups. But where comes the normal subgroup, why cant the people use just subgroups instead of normal subgroups in the definition of simple group. Does the normal subgroup has any influence when we are dealing with the structure of groups ???
In addition help me with an example of a group that admits subgroup but it doesn't have normal subgroups ???
If you demand that "$G$ be a non-trivial group with no non-trivial subgroups." then you would be limiting yourself to just the groups of prime order (i.e. isomorphic to the integers mod a prime). This isn't very interesting.
[Quick sketch of a proof: Let $G$ be a non-trivial group of prime order. Then $G$ is cyclic and thus isomorphic to integers mod a prime. Otherwise, suppose $G$ is a non-trivial group of composite order. Suppose $p$ is a prime divisor of the group's order. Cauchy's theorem then states that $G$ has an element of order $p$ and thus has a non-trivial (proper) subgroup of order $p$. So all non-trivial groups of composite order have non-trivial proper subgroups.]
Also, being simple is all about not having "interesting" quotients. You can only quotient by normal subgroups.
A simple group is a non-trivial group whose quotients are just itself and the trivial group (up to isomorphism). This parallels the definition of a prime number whose (positive integer) quotients are just itself and $1$.
In fact, the Jordan-Holder theorem states that every non-trivial finite group in some sense "factors" uniquely into simple groups (although "factors" isn't exactly the right term). In fact when you apply the Jordan-Holder theorem to the $\mathbb{Z}_n$ (integers mod $n$) groups, you'll recover the fundamental theorem of arithmetic. :)
EDIT: Note there is no need to assume $G$ is finite. Suppose $G$ is non-trivial has no non-trivial proper subgroups. Pick any non-identity element. This will generate a non-trivial subgroup. Since $G$ has no non-trivial proper subgroups, it must be generated by that element. Thus $G$ is cyclic. So either $G$ is isomorphic to the integers or integers mod some $n$. In the former case, we have many non-trivial proper subgroups (e.g., the even integers). Thus $G$ must be finite.