Is a finite index subgroup of a finitely presented subgroup finitely presented?

Question

I do know Schreier's theorem, which states that a finite index subgroup of a finitely generated group is finitely generated. Other than this, I have no reason to suspect a positive answer to my question, other than it would be nice.

Answer 1

This is a consequence of the Reidemeister-Schreier algorithm. See Chapter 2, Proposition 4.2 in [1] or Chapter 2, Corollary 2.8 in [2]

[1] Lyndon and Schupp, Combinatorial group theory. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 89. Springer-Verlag, Berlin-New York, 1977. xiv+339 pp.

[2] Magnus, Karrass and Solitar, Combinatorial group theory. Presentations of groups in terms of generators and relations. Second revised edition. Dover Publications, Inc., New York, 1976. xii+444 pp.

Answer 2

It is indeed true. Because being finitely presentable is a geometric property. More precisely "being finitely presentable" is preserved under quasi-isometries and a subgroup of finite index is quasi-isometric to the big group.

Answer 3

Yes they are, finite index subgroups of finitely presented subgroups are finitely presented. In fact it is known that if a group is quasi-isometric to a finitely presented group, then it is finitely presented (finite index subgroups are quasi-isometric).

This is proved in Lectures on geometric group theory, by Cornelia Druţu and Michael Kapovich, which is still in the works, but as it is now, it is Cor 6.40. The idea is sketched in Topics in geometric group theory by de la Harpe too, V.A.3.