$G/N \cong G$ then $N = \{1\}$ with $G$ infinite. This question similar to others posted before, but it is a particular case which I think can be solved differently. It was asked here or here. 
I need to proof the following:

If $G/N \cong G$ then $N = \{1\}$

Note that the finite case is very simple but what happens when $G$ is infinite? 
 A: This is not true for infinite groups. For instance, there are infinite groups $G$ which are isomorphic to $G\times G$, see this question.
A: The reals $\;\Bbb R\;$ is an example, just as any other non-hopfian group is, but to construct a explicit example of epimorphism is kind of ...er, ugly and non elementary.
First, choose a  basis of $\;\Bbb R\;$ as a linear space over the rationals (Hamel basis and thus AC kicks in strongly here), say $\;B:=\{r_i\}_{i\in I}\;$  , and now choose an infinite countable $\;J\subset I\;$ and let $\;A:=\{a_j\}_{j\in J}\;$. For simplicity, say $\;J=\Bbb N\;$ so that we can comfortably ennumerate the indexes: $\;J=\{1,2,3,...\}\;$ . Define now
$$T:\Bbb R_{\Bbb Q}\to\Bbb R_{\Bbb Q}\;,\;\;Tr_k=\begin{cases}0,&k=1\\r_{k-1},&2\le k\in J\\r_k,&k\in B\setminus A\end{cases}$$
and extend the definition by linearity. Prove the linear trasnformation we get is surjective but definitely not injective and thus $\;\Bbb R/\ker T\cong \Bbb R \;$, and $\;\ker T\neq \{0\}\;$
Other examples are the Baumslag-Solitar group $\;B(2,3)\;$ , or quasi-cyclic groups.
A: The claim is not true when $G$ is infinite.  Consider $G=S^1\subset\mathbf{C}$ under complex multiplication, and let $N\subset G$ be the subgroup generated by $-1\in G$.  Then $G/N\cong\mathbf{RP^1}\cong S^1$.
