Is a factor group an isomorphism of the group it's in?

Question

The definition for factor groups that I was given in my textbook (A First Course In Abstract Algebra by John B. Fraleigh) was:

If $N$ is a normal subgroup of a group $G$, the group of cosets of $N$ under the induced operation is the factor group of $G$ modulo $N$, and is denoted $G/N$. The cosets are residue classes of $G$ modulo $N$.

After reading the properties section from the wikipedia article on factor groups, I know that the $G/\{e\}$ is isomorphic to $G$, but I am wondering why other factor groups are not isomorphic to $G$? Since the factor group is the group of cosets for some subgroup $N$ and every element of $G$ belongs to some coset of $N$ (by Lagrange's Theorem), shouldn't there be an isomorphism between the factor group $G/N$ and the group $G$?

If I am thinking about this the wrong way, any clarification would be welcome. Thanks in advance for any help.

Answer 1

Suppose $G$ is a finite group of order $n$. If $G$ and $G/N$ are isomorphic then $|G|=|G/N|=\frac{|G}{|N|}$. This implies $|N|=1$. Hence, $N=\{e\}$.

Answer 2

Even in infinite case , this quotient might not be isomorphic to the original group. Consider $\mathbb{Q}/\mathbb{Z}$. This is not isomorphic to $\mathbb{Q}$ (why?).