Why in Quotient Group we need normal subgroup?

Let $G$ be a group and let $e$ denote its identity.

Forming a quotient $G/\sim$ of group $G$ is actually forming a partition on $G$ such that $[a][b]=[ab]$ is a well defined multiplication. Here $[a]$ and $[b]$ are elements of the partition that are represented by $a\in[a]$ and $b\in[b]$ respectively.

The multiplication is well defined if $[a'b']=[ab]$ whenever $a'\in[a]$ and $b'\in[b]$. Further it is not difficult to prove that $[e]$ will be a subgroup and will work as identity of the quotient.

Now if $a\in[e]$ and $g\in G$ then we find $$[gag^{-1}]=[g][a][g^{-1}]=[g][e][g^{-1}]=[geg^{-1}]=[e]$$This proves the necessity of: $$a\in[e]\implies gag^{-1}\in[e]$$ So $[e]$ is bound to be a normal subgroup.