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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Such a group is called non-Hopfian. See this old question for examples. – user1729 Sep 23 '13 at 16:22
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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
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@user1729: I am so sorry for my self. This is the second duplicate question I have seen. – Mikasa Sep 23 '13 at 16:25
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(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
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Indeed @user1729...and as such it must be non-residually finite since it is non-hopfian. – DonAntonio Sep 23 '13 at 16:28
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@DonAntonio Indeed! Have you ever read Higman's paper which gives a 3-generator, 2-relator example? It is actually really easy to see that it is non-Hopfian - easier than Baumslag-Solitar's example. I feel that this is much more pleasing than the BS-example because I can see where it comes from... Take $A\leq H$ and suppose there exists an automorphism $\phi$ of $H$ such that $A\phi=B\lneq A$. Then take $G_A=H_1\ast_{A_1=A_2}H_2$ (H_i=H, etc.). This is isomorphic to $G_B=H_1\ast_{B_1=B_2}H_2$, but the natural map $G_B\rightarrow G_A$ got by adding in the rest of $A$ has non-trivial kernel. – user1729 Sep 23 '13 at 18:47
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(@DonAntonio Also, in the (specific) example Higman gives, the non-injective endomorphism an a Nielsen transformation. I find this incredibly disturbing. Although...when I think about it, this is probably true always - every automorphism of a group can be viewed as a Nielsen transformation over some generating set, so I presume the same holds even if we relax the injectivity. I think I might need to lie down now.) – user1729 Sep 23 '13 at 19:03
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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 $\pi: G\to G$.
Mikasa
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@SucharitaDeb: They are the same. In fact, group $G$ is abelian and maybe you wanted the group to be non-abelian. – Mikasa Sep 23 '13 at 16:23
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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
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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
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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 $\mathbb{Q}[x]$.
fretty
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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
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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