G is a finitely generated group and F is its subgroup.
Q: Under what known sufficient conditions can we guarantee that F is finitely generated? (e.g. G is Abelian)
G is a finitely generated group and F is its subgroup. Q: Under what known sufficient conditions can we guarantee that F is finitely generated? (e.g. G is Abelian) 


A consequence of the NielsenSchreier theorem is the following: If a group generated by $n$ elements, then every subgroup of finite index $k$ is generated by $kn−k+1$ elements. See also this aops discussion; there jmerry gives a direct algebraic proof. 


If $G$ is polycyclic, then every subgroup is finitely generated. In fact, one of several ways to define a polycyclic group is to demand that it is a solvable group for which all subgroups are finitely generated. So, this might seem kind of tautological, but polycyclic groups have other definitions and they come up quite a bit in various areas. This is a special case of the class of Noetherian groups, also known as slender groups, in which are defined by having the property that every subgroup is finitely generated. You can find some more information on this subject in this MO question. 


Further to Max's answer, and in an effort to make this very general question more specific, I have a feeling that the following question is open.
The existence of continuummany Tarski monsters is strong evidence that there is no reasonable classification of finitely generated slender groups. 


If you look at the toplogical case, then a profinite group has finite rank if all its closed subgroups are finitely generated (toplogically). Now a pro$p$ group of finite rank can be characterized in many ways. For instance, a pro$p$ group has finite rank if and only if it is a Lie group over the $p$adic integers if and only if it is linear over the $p$adics if and only if it has polynomial subgroup growth if and only if the associtaed graded Lie algebra (with respect to the ZassenhousLazard filtration) is nilpotent. 

