If $G$ is a group and $N$ is a nontrivial normal subgroup, can $G/N \cong G$? I know $G/N$ is isomorphic to a proper subgroup of $G$ in this case, so the gut instinct I had was 'no'. But there are examples of groups that are isomorphic to proper subgroups, such as the integers being isomorphic to the even integers, so that reasoning doesn't work. However in this case the even integers are not a quotient of the integers.
edit: I realize now that $G/N$ is not necessarily isomorphic to a proper subgroup of $G$, just a subgroup of $G$.

Another example: Fix a prime $p$ and let $T_p=\{z\in\mathbb{C}|\exists n\geq0\hspace{5pt} z^{p^n}=1 \}$.
Consider the map $\varphi: T_p\to T_p$ defined by $\varphi(z)=z^p$. Then $\varphi$ is a homomorphism with non-trivial kernel, $N=\{z\in \mathbb{C}|z^p=1\}$ and $\operatorname{Im}(\varphi)=T_p$, so $T_p/N\cong T_p$.
