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Please help me to find an example of a group epimorphism $f:G\to G$ which is not an isomorphism.

I understand that $G$ should an infinite group.

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marked as duplicate by user1729, Hagen von Eitzen, Dan Rust, Vedran Šego, TZakrevskiy Sep 23 '13 at 18:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Such a group is called non-Hopfian. See this old question for examples. – user1729 Sep 23 '13 at 16:22
You need a non-hopfian group, which are not that easy to spot and are rather nasty in a way: try $$BS(2,3):=\langle\;a,b\;;\;b^{-1}a^2b=a^3\;\rangle$$ and the map $\;a\mapsto a^2\;,\;b\mapsto b\;$ – DonAntonio Sep 23 '13 at 16:25
@user1729: I am so sorry for my self. This is the second duplicate question I have seen. – Babak S. Sep 23 '13 at 16:25
(It is interesting to note that Don Antonio's group is finitely presented, but neither of the two in the answers here are.) – user1729 Sep 23 '13 at 16:27
Indeed @user1729...and as such it must be non-residually finite since it is non-hopfian. – DonAntonio Sep 23 '13 at 16:28

Hint: consider infinite direct sums (of copies of a given group)...

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Let $p$ is prime and let the group $G=\mathbb Z(p^{\infty})=\{\frac{a}{p^k}+\mathbb Z\mid (a,p)=1,k\in\mathbb N\}$. I hope you know this fact that for any subgroup $H$ of the group above; $G/H\cong G$. Now I am thinking of the natural epimorphism $\pi: G\to G$.

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but domain and codomain should be same – Anupam Sep 23 '13 at 16:21
@SucharitaDeb: They are the same. In fact, group $G$ is abelian and maybe you wanted the group to be non-abelian. – Babak S. Sep 23 '13 at 16:23
Well done, @Babak...but in the last line I think you may have meant $\;\pi :G\to G\;$ +1 – DonAntonio Sep 23 '13 at 16:26
@DonAntonio: Yes I have, Don. – Babak S. Sep 23 '13 at 16:28
Babak gives here what is perhaps the easiest, or maybe the best well-known, example of non-hopfian group: the Prüfer $\;p-$group $\,\Bbb Z_{p^\infty}\;$ – DonAntonio Sep 23 '13 at 16:31

Consider $G = \mathbb{Z}[x]$ under addition and the map sending $a_0 + a_1x + ... + a_n x^n$ to $a_1 + a_2 x +...+ a_n x^{n-1}$. This is clearly an epimorphism yet isn't injective.

Also for a contextual example the derivative map on $\mathbb{Q}[x]$.

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Isn't this just a different way of viewing anon's example? (I ask to verify my thinking, not to criticise.) – user1729 Sep 23 '13 at 18:50
Yes...however anon didn't really give an example! Just gave a group which you can construct such a map on. – fretty Sep 23 '13 at 18:53

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