# When is a surjective homomorphism with a right-inverse an isomorphism?

Let $G$ be an abelian group (or more generally a module over a ring $R$) and $\phi:G\to G$ a surjective group homomorphism with a right-inverse $\psi: G\to G$, i.e. $\phi\circ \psi = id_G$. So, we have a $split$ short exact sequence $$0\to K:=\ker(\phi)\to G\overset{\phi}{\to} G\to 0$$ and hence $G\cong K\oplus G$.

Question: Under which conditions on $G$ can we conclude that $\phi$ is an isomorphism (i.e. that $K=0$)?

I suspect that there is a notion/name for such groups $G$ that I'm ignorant of. So, references are very welcome as well. Thanks in advance for any contribution.

• If $G$ is a finitely generated abelian group or module over a noetherian ring, every surjective endomorphism of $G$ is an isomorphism (the verification is easy). If the ring is not noetherian, the same is true, but the proof (though short) is not so obvious (it is a result by Vasconcelos, I think). On the other side, there are surjective endomorphisms of a free not finitely generated abelian group (and so split) that are not isomorphisms: take $F\oplus F\to F$ surjective not injective and sum to each side an infinite number of copies of $F$. – A.G Dec 16 '16 at 17:34
• math.stackexchange.com/questions/239364/… – A.G Dec 16 '16 at 17:41
• Have you heard of hopfian groups? Anyway if your abelian group $G$ is finitely generated (as a group) then your $K = 0$. – M.U. Dec 16 '16 at 20:06
• @A.G. Thanks for your comment. The groups I'm interested in (in particular $\mathbb{R}/\mathbb{Z}$) are typically not finitely generated. – Dave Dec 19 '16 at 8:46
• @M.U. Thanks for the reference to Hopfian Groups, I'll definitely look into that. – Dave Dec 19 '16 at 8:46