What is a (simple) example of a group $G$ with an endomorphism $G\rightarrow G$ That is surjective, but not injective?
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$\begingroup$ What could I have tried? I'm looking for an example, either I know one, or I don't $\endgroup$ – JimmyP Mar 23 '15 at 18:32
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3$\begingroup$ Yes, but it isn't as though the only options are "I know an example immediately" and "such an example will never occur to me no matter how long I think about it". What properties of $G$ most hold? For instance, could $G$ be finite? $\endgroup$ – BaronVT Mar 23 '15 at 18:39
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Let $G= H^\infty$ for some non-trivial group $H$ (like $\Bbb Z$ or $\Bbb Z_2$), and let $\phi:G\to G$ be given by. $$ \phi(h_1,h_2,h_3,\ldots )=(h_2,h_3,\ldots) $$
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$\begingroup$ @JimmyP for any $(h_2,h_3,\dots)\in G$ there is an element $u\in G$ such that $\phi(u)=(h_2,h_3,\dots)$. E.g. $u=(e,h_2,h_3,\dots)$ where $e$ is the identity of $H$. $\endgroup$ – drhab Mar 23 '15 at 19:04
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$\begingroup$ @drhab but isnt it also injective in that case? $\endgroup$ – JimmyP Mar 23 '15 at 19:07
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$\begingroup$ Not if $H$ is not trivial. If $x\in H$ with $x\neq e$ then you can also take $v=(x,h_2,h_3,\dots)\neq(e,h_2,h_3,\dots)=u$. $\endgroup$ – drhab Mar 23 '15 at 19:09
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There are no examples with $G$ finite. So you need $G$ to be infinite.
Consider the set of all polynomials with rational coefficients as an additive group.
Then the derivation map $p\mapsto p'$ is a surjective, but not injective, endomorphism.
Or consider the set of complex numbers of modulus $1$ as a multiplicative group.
Then the square map $z\mapsto z^2$ is a surjective, but not injective, endomorphism.
