I learned here that if the outer groups in a short exact sequence are finitely-generated, then the middle group is, too.
Question. Is there a generalization of this to long-exact sequences?
Julian Kuelshammer mentioned the horseshoe lemma but I'm not really sure how to apply it.
Well, if you have an exact sequence $$A\stackrel{f}\to B\stackrel{g}\to C$$ where $A$ and $C$ are finitely generated, then so is $B$, This follows from the short exact sequence $$0\to \ker(g) \to B\to \operatorname{im}(g)\to 0$$ where $\ker(g)=\operatorname{im}(f)$ and $\operatorname{im}(g)$ are finitely generated because $A$ and $C$ are. So you don't need to start with a short exact sequence: in an exact sequence of any length, any term trapped between two finitely generated terms is finitely generated.
However, you cannot separate the finitely generated terms by more than one term in between. For instance, for any $A$, there is an exact sequence $$0\to A\to A\to 0,$$ where $0$ is finitely generated by $A$ need not be.
One always has an exact sequence $$0\to A\to A\oplus C\to B\oplus C\to B\to 0.$$ So given $A$ and $B$ there is a four-term exact sequences where the middle terms can be arbitrarily "large".
