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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.

Suggestions, please?

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

dubiousjim
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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
  • The trivial endomorphism is always available. Slightly more interesting is the question of injective non-surjective endomorphisms, and for this we need infinite groups $-$ take e.g. the coordinate-shift-forward map on any countably infinite direct sum of nontrivial groups. – anon Sep 11 '13 at 20:04
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    $\rho\colon S^1\rightarrow S^1\colon e^{i\theta}\mapsto e^{ni\theta}$, $n\geq 2$. In this case $|\ker\rho|=n$. – Dan Rust Sep 11 '13 at 20:06
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    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
  • Thanks for the quick comments. I was hoping for a surjective but non-injective endomorphism. (Doesn't mean I can get it.) – dubiousjim Sep 11 '13 at 20:09
  • @dubiousjim That is what the link in my comment gives you. The easiest example is simply $\mathbb{Z\times Z\times}\ldots$ then kill the left-most $\mathbb{Z}$-term, and shift the rest to the left by one. – user1729 Sep 11 '13 at 20:11
  • Right, thanks very much, looks like what I was looking for! – dubiousjim Sep 11 '13 at 20:12
  • @DanielRust, I'm going to have to work for a bit to understand your suggestion, can you explain it to me like I'm dumber? – dubiousjim Sep 11 '13 at 20:15
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    @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
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    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
  • Terrific, thanks much for all the help. – dubiousjim Sep 11 '13 at 20:19
  • @DanielRust You should give that example as an answer to the question I linked too. It is rather neat, and I find it helpful when these things are all in the same place (MSE is meant to be a sort of searchable information database thingy, after all). – user1729 Sep 11 '13 at 20:30
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    @user1729 Good idea. I've added an answer. – Dan Rust Sep 11 '13 at 21:00

1 Answers1

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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.

Boris Novikov
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