# Finitely generated subgroup [closed]

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)

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## closed as too broad by YCor, Stefan Kohl, Ryan Budney, Wolfgang, Frieder LadischDec 4 '15 at 11:49

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

This question is absurdly broad. As it clearly has no single answer, it should certainly be community wiki. – HJRW Jul 12 '11 at 9:15
related question: mathoverflow.net/questions/26059/… – Ian Agol Jul 12 '11 at 18:20
The class of slender groups is closed under taking subgroups, quotients, and finite products. If it were closed under taking direct products, then it would form a variety, but this is not the case. en.wikipedia.org/wiki/Variety_(universal_algebra) It's also closed under taking extensions. This is very closely related to an extension closed variety of finite groups, but I'm not sure what's been considered in the infinite group case. – Ian Agol Jul 12 '11 at 19:21
Let $Z=\gamma_s(G)\cap Z(G)$. Why if $\gamma_s(G)/Z$ is finite then there exists a finitely generated subgroup $U$ of $G$ such that $\gamma_s(G)=\gamma_s(U)Z$? – user83632 Dec 4 '15 at 9:37
@user83632 BTW if you ask a question in these terms, it will be put on hold as unclear (you have to explain your notation). – YCor Dec 4 '15 at 12:51

A consequence of the Nielsen-Schreier 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.

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Also, if $G$ is finitely related your finite-index subgroup will be finitely related also. – ADL Jul 12 '11 at 7:18
@Alan: Interesting. Can you add a sketch of proof? – Martin Brandenburg Jul 12 '11 at 20:33
Martin - realise G as the fundamental group of a finite complex. The subgroup is the fundamental group of a finite-sheeted covering space, which is also a finite complex, and hence is finitely presented. – HJRW Jul 12 '11 at 21:09

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.

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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.

Is every finitely presented slender group virtually polycylic?

The existence of continuum-many Tarski monsters is strong evidence that there is no reasonable classification of finitely generated slender groups.

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For context, "slender" here means noetherian, i.e. every subgroup is f.g., or equivalently there no properly ascending infinite chain of subgroups. (Slender groups in wikipedia points to an unrelated definition.) [I subsequently found the definition of slender in Max' answer but I leave the comment] – YCor Dec 4 '15 at 12:54

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 Zassenhous-Lazard filtration) is nilpotent.

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