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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?

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user1868607
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    Such groups are called Hopfian groups. Trivially every finite group is Hopfian, but there are examples of non-Hopfian groups. In particular, you are asking us a proof which does not exist, because this is a false statement. – Crostul Sep 03 '16 at 22:01

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This is not true for infinite groups. For instance, there are infinite groups $G$ which are isomorphic to $G\times G$, see this question.

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Arthur
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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.

DonAntonio
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  • @Arthur The mentioning of "basis" should be enough...or else it is unneccessary if the OP hasn't yet heard of this. – DonAntonio Sep 03 '16 at 22:03
  • What do you do with that $T$ or what does it show? Would you conisder a non-injective epimorphism I'd get the point. But an automophism beats me. – quid Sep 03 '16 at 22:11
  • @quid I misunderstood at first the question, and when I wanted to fix the answer the site began acting funny and I couldn't see MathJax clearly until a few minutes ago (not the first time this happens, to my recollection). You're right: automorphisms don't work, but now there's an example of a non-injective epimorphism by means of the structure of $;\Bbb Q,-$ linear space of $;\Bbb R;$ , and this is fine now, as far as I can see. Thanks. – DonAntonio Sep 04 '16 at 07:41
  • @Rodrigo Check the fixed answer, please. – DonAntonio Sep 04 '16 at 07:42
  • Yes, I agree this now shows the reals are non-Hopfian. – quid Sep 04 '16 at 12:34
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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 $G/N\cong\mathbf{RP^1}\cong S^1$.

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