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 $G$.

Yes. Let $G$ be the additive group of the complex numbers, and let $N$ be the subgroup consisting of the real numbers.
Edit in response to comment by @GA316:
$(\mathbb C,+)/\mathbb R$ is clearly isomorphic to $(\mathbb R, +)$, and it is well known (but this requires the Axiom of Choice) that $(\mathbb C,+)\cong(\mathbb R,+)$.
My answer is a special case of the answer posted simultaneously by Asaf Karagila, considering $\mathbb C$ as a vector space over the field of rational numbers.
