Is it necessary that normal subgroups must have an index of 2 always? Can there exist normal subgroup having index greater than two.... It's just a question that occurred in my head while doing problems , I need no proof as such.
 A: Yes, it is possible. For abelian groups you can take any subgroup of index greater than two. It is normal. There are also many examples for non-abelian groups.
The symmetric group $S_4$ has many normal subgroups of index greater than $2$, see here.
A: An even simpler answer: the trivial subgroup is always a normal subgroup. So your question is equivalent to the question "Does there exist a group of order greater than 2?"
Of course the answer is yes. Consider $G = \mathbb Z/3\mathbb Z$.
A: Let $G=H \times C_n$, where $H$ is any group and $C_n$ is the cyclic group of order $n$. Then $H$ is normal and $|G:H|=n$.
A: You might recall that the index of a normal subgroup $N < G$ equals the order of the quotient group $G/N$. Also, for any surjective homomorphism $G \to Q$ with kernel $N$, the group $Q$ is isomorphic to $G/N$, and so the index of $N$ equals the order of $Q$. 
So your question is equivalent to asking: 


*

*Can there exist surjective homomorphisms $G \to Q$ such that $Q$ has order greater than $2$?


Yes there can. For example, the map $f : \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y)=y$ is a surjective homomorphism, with respect to the group operations of vector addition on $\mathbb{R}^2$ and ordinary addition on $\mathbb{R}$. The kernel of $f$ is the $x$-axis, and its index equals the order of the image group $\mathbb{R}$, which is uncountably infinite. 
A: You're probably confused by the fact that an index $2$ subgroup is automatically normal; however this is the exception, rather than the norm. 
A famous theorem by Feit and Thompson implies that every group with odd order has a proper and non trivial normal subgroup and, of course, such a subgroup cannot have index $2$.
More easily, an abelian group has subgroup of every possible order (taking Lagrange's theorem into account, of course):

If $G$ is a finite abelian group of order $n$ and $d$ is a divisor of $n$, then $G$ has a subgroup $H$ of order $d$ (and index $n/d$).

If $G$ is cyclic, the proof is very simple: take a generator $x$ of $G$; then the subgroup $\langle x^{n/d}\rangle$ generated by $x^{n/d}$ has order $d$ (and index $n/d$).
