# Surjective, but not injective endomorphism

What is a (simple) example of a group $G$ with an endomorphism $G\rightarrow G$ That is surjective, but not injective?

• What could I have tried? I'm looking for an example, either I know one, or I don't – JimmyP Mar 23 '15 at 18:32
• 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? – BaronVT Mar 23 '15 at 18:39

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. $$\phi(h_1,h_2,h_3,\ldots )=(h_2,h_3,\ldots)$$

• Why is this surjective? – JimmyP Mar 23 '15 at 18:46
• @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 $H$. – drhab Mar 23 '15 at 19:04
• @drhab but isnt it also injective in that case? – JimmyP Mar 23 '15 at 19:07
• Not if $H$ is not trivial. If $x\in H$ with $x\neq e$ then you can also take $v=(x,h_2,h_3,\dots)\neq(e,h_2,h_3,\dots)=u$. – drhab Mar 23 '15 at 19:09
• @drhab I get it now, thanks! – JimmyP Mar 23 '15 at 19:11

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.