TITLE: If $G$ is a group and $N$ is a nontrivial normal subgroup, can $G/N \cong G$?
QUESTION [17 upvotes]: 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 $G$.

REPLY [2 votes]: As mentionned by Derek Holt, the question is equivalent to find non-hopfian groups. In fact, it is easy to prove

Claim: Let $m,n \in \mathbb{Z} \backslash \{0, 1, -1\}$ be two coprime numbers. Then the Baumslag-Solitar group $BS(n,m)=\langle a,b \mid ab^ma^{-1}=b^n \rangle$ is non-hopfian.

Let $\varphi : BS(n,m) \to BS(n,m)$ be the morphism sending $a$ to $a$ and $b$ to $b^m$; $\varphi$ is well-defined since $\varphi(ab^ma^{-1}b^{-n})=1$.

$\varphi$ is surjective. Indeed, $a,b^m \in \mathrm{Im}(\varphi)$ and $b^n=ab^ma^{-1} \in \mathrm{Im}(\varphi)$. Because $m$ and $n$ are coprime, there exist $p,q \in \mathbb{Z}$ such that $pm+qn=1$, hence $$b= b^{pm+qn}=(b^m)^p \cdot (b^n)^q \in \mathrm{Im}(\varphi).$$
$\varphi$ is not injective. Indeed, $[aba^{-1},b] \neq 1$ (using Britton's lemma for example) but $\varphi([aba^{-1},b])=1$.

A necessary and sufficient condition is given here to know when $BS(n,m)$ is hopfian or not.