# Simple example of non-injective group endomorphism?

I'm not a mathematician, but I'm teaching a bit of algebra to some budding logicians, and introducing them to/reminding them of the notions of isomorphism, homomorphism, etc. I'd like to give them an example of an endomorphism which isn't an automorphism, so that they can see the point of there being a name for these separate concepts. I'd also like the example to be as simple as possible, ideally just with some infinite group. But it's proving to be harder than I expected to do this. Every candidate I've come up with turns out to be either an automorphism or not really a homomorphism in the first place.

EDIT: Should have said explicitly from the beginning, I'm hoping for an example where the homomorphism is surjective.

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eh what is about $\mathbb{Z}/2 \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ which maps everything to zero ? – Dominic Michaelis Sep 11 '13 at 20:00
$\rho\colon S^1\rightarrow S^1\colon e^{i\theta}\mapsto e^{ni\theta}$, $n\geq 2$. In this case $|\ker\rho|=n$. – Dan Rust Sep 11 '13 at 20:06
Or the endomorphism of $\mathbb{ZZ/4Z}$ which makes to the subgroup of order two? If you want to blow their minds, look at the answers to this question, which gives surjective, non-injective endomorphisms. (And for anon's comment, an easy example is $\mathbb{Z}: i\mapsto 2i$.) – user1729 Sep 11 '13 at 20:07
@dubiousjim Take a circle of elastic. Cut it, and stretch it so that it traces the same circle twice (for $n=2$). – user1729 Sep 11 '13 at 20:16
Let $S^1=\{e^{i\theta}\in\mathbb{C}\mid \theta\in[0,2\pi)\}$ be a unit circle with group operation addition of angles modulo 1 (so take the fractional part of the sum of angles). $\rho$ is then just the map which 'wraps' the circle around itself twice. – Dan Rust Sep 11 '13 at 20:17

Consider $G=\mathbb{R}\setminus 0$ (real numbers without zero) as a group with respect to the multiplication. Then the map $a\to |a|$ is an endomorphism, but not an isomorphism.