Let $G$ be a finite group, and let $M(G)=H_2(G,\mathbb{Z})$ be its Schur multiplier. Are there any known bounds on the number of generators of $M(G)$ in terms of $G$? For example, if $G$ is abelian of rank $r$, then $M(G)$ can be generated by $r(r-1)/2$ elements. Is there anything that can be said for more general finite groups?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
If a finite group has a presentation with $g$ generators and $r$ relations, then the Schur multiplier is generated by $r-g$ elements. There's been lots of work studying groups and presentations where this is minimal ("efficient" presentations and groups).
answered Dec 7, 2017 at 11:11
-
1$\begingroup$ Would you please give the reference to this result? $\endgroup$ Commented Dec 7, 2017 at 11:26
-
1$\begingroup$ @M.FarrokhiD.G. It's very old and well-known, and I think it goes back to Schur. A slightly more modern reference: D.L. Johnson and E.F. Roberston, "Finite groups of deficiency zero" in Homological Group Theory (ed. C.T.C. Wall), C.U.P., 1979. But Googling "deficiency efficient Schur" will give lots of other references. $\endgroup$ Commented Dec 7, 2017 at 11:47
-
1$\begingroup$ One should read "If a finite group has a presentation with ...". A general bound for an arbitrary finitely presented group has been obtained by D. Epstein (1961) who gives much of the credits to P. Hall (the bound on the number of generators is enlarged by adding the free rank of the abelianization of $G$, see lemma 1.2 of "Finite presentations of groups and 3-manifolds"). Epstein cites I. Schur (1904) and B. H Neumann (1956) as the authors/finders of early work on the relation between the Schur multiplier and the deficiency of a group. $\endgroup$ Commented Aug 31, 2019 at 16:57
-
1$\begingroup$ @LucGuyot Thanks. I've edited my answer to correct that. $\endgroup$ Commented Aug 31, 2019 at 22:00