Why Quotient group is not a subgroup? $G/N = \{ gN  | g \in G \}$ is factor group where $N$ is normal subgroup of $G$. My doubt is that is it subgroup of $G$ ? Though I found it is not a subset of $G$. But how, my claim is that if we take $x \in  G/N$ so $x= an$ for some $a \in G$ and $n \in N$. We know $n$ belongs to $G$ as it is element of normal subgroup $N$. So $x= an$ should belongs to $G$.
 A: Sometimes you can actually think of $G/N$ as a subgroup of $G$.
If $N$ is a normal subgroup of $G$, the elements of $G/N$ are cosets, i.e. subsets of $G$ of the form $xN$ for some $x \in G$.  The element $x$ is a representative of the coset $xN$, and the other representatives of the same coset as $xn : n \in N$.
In certain situations, there is, for each element $X$ of $G/N$ a choice $g_X \in G$ of a representative for $X$ such that the assignment $X \mapsto g_X$ defines an injective group homomorphism $G/N \rightarrow G$.  Then the image of $G/N$ under this homomorphism is a subgroup of $G$ which is isomorphic to $G/N$.  We say in this case that $N$ splits in $G$.
Example: Let $\sigma$ be a permutation on $n$ elements.  The $n$ by $n$ permutation matrix $g_{\sigma}$ associated to $\sigma$ is the matrix obtained from the $n$ by $n$ identity matrix by permuting the rows according to $\sigma$.  For example, if $n = 3$ and $\sigma = (1 \space 2)$, then
$$g_{\sigma} = \begin{pmatrix} & 1 \\ 1 \\ & & 1 \end{pmatrix}$$
Let $N$ be the group of invertible diagonal matrices with entries in $\mathbb{R}$, and $G$ the group generated by $N$ and all the $g_{\sigma}$.  Then $N$ is normal in $G$, and the quotient $G/N$ is isomorphic to the symmetric group, the elements of $G/N$ being the cosets $g_{\sigma}N : \sigma \in S_n$.  They are all distinct for distinct $\sigma$.  The choice of representatives $g_{\sigma}$ for the cosets $g_{\sigma}$ defines an injective group homomorphism $G/N \rightarrow G$, so we can think of $G/N$ as a subgroup of $G$.
Nonexample: let $G = \mathbb{Z}$, and $N = 2\mathbb{Z}$.  The quotient $G/N$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.  This is a group with two elements.  There is no way we can inject $G/N$ into $G$, because then there would exist an element of $G$ with order two.  But every element of $G$ except $0$ has infinite order.
A: No, $x=aN=\{an\mid n\in N\}\notin G$. There's where you have made a mistake.
A: It sounds like you're suggesting we choose a representative element $an$ for each $aN \in G / N$ to try and associate the quotient group with a subgroup of $G$.  The reason this doesn't work is that the set of representatives will not typically be closed under the group operation, thus won't form a subgroup.
For example, consider $\mathbb{Z} / 2\mathbb{Z}$.  The most obvious choices of representative elements in $\mathbb{Z}$ would be $0$ and $1$.  But $1+1=2$, not $0$, in $\mathbb{Z}$ (proof left as an exercise). Indeed, you can never find representatives for $0$ and $1$ in $\mathbb{Z}$ that match their properties in $\mathbb{Z} / 2\mathbb{Z}$ - for example, in $\mathbb{Z} / 2\mathbb{Z}$ both elements have the property that $x+x+x=x$, but $0$ is the only element with that property in $\mathbb{Z}$. 
A: Here $gN = \{gn\mid n\in N\}$ is an element of $G/N$, not a subset of it.
A: This is just one example of equivalence relation and equivalence classes determined by the relation.
The normal subgroup $N$ determines an equivalence relation on the elements of $G$ and $G/N$ is the set of equivalence classes which is also a group determined by the group operation of $G$. So, yes, the elements of the elements of $G/N$ are elements of $G$, but $G/N$ is not a subgroup of $G$.
