TITLE: Does $G\cong G/H$ imply that $H$ is trivial?
QUESTION [81 upvotes]: Let $G$ be any group such that
$$G\cong G/H$$
where $H$ is a normal subgroup of $G$.
If $G$ is finite, then $H$ is the trivial subgroup $\{e\}$. Does the result still hold when $G$ is infinite ? In what kind of group could I search for a counterexample ?

REPLY [2 votes]: Proving Baumslag-Solitar groups is a counter-example is not at all tough, so I will write this proof for $BS(m,n)$
Choose any two integers $m$ and $n$ such that $(m,n)=1$ and $m,n \neq -1,0,1$.
Let $G=BS(m,n)=\langle a,b\ |\ b^{-1}a^mb=a^n \rangle$.
Consider $\phi: G \to G$ where $\phi(a)=a^m$ and $\phi(b)=b$, it is easily checked that $\phi$ is well defined. 
Claim-1 $\phi$ is surjective.
We see that $a^m,b \in \text{Im}(G)$, so $a^n\in \text{Im}(G)$ also. Now as $(m,n)=1$, $k_1m+k_2n=1$ for some $k_1,k_2\in \Bbb{Z}$, and  as $\text{Im}(G)$ is a subgroup, $a=a^1\in \text{Im}(G)$, thus $\phi$ is an epimorphism.
Claim-2 Ker$(\phi)$ is non-trivial.
$[b^{-1}ab,a]=b^{-1}abab^{-1}a^{-1}ba^{-1}\neq 1$ by Britton's lemma, but $\phi([b^{-1}ab,a])=\phi([b^{-1}a^mb,a^m])=[b^n,b^m]=1$, so kernel is non-trivial. $\hspace{4cm}\blacksquare$