1

Given a (finite) group $G$. Is there any bounds on the minimum number of generators $d(G)$?

For example, it is clear that $d(G) \geqslant d(G^{ab})$. Where the right hand side can be easily computed. Unfortunately, this does not give much information, e.g. $A_n^{ab}$ is trivial, but $A_n$ has two generators itself.

UPD: Actually, I am more interested in lower bounds.

  • 4
    The question is quite too general as now stated. Some particular but interesting cases are treated in this question about powers of finite simple groups. – YCor Dec 12 '21 at 17:51
  • 2
    See http://www.numdam.org/article/RSMUP_1990__83__201_0.pdf – Benjamin Steinberg Dec 12 '21 at 17:53
  • @YCor partial answers are acceptable too. –  Dec 12 '21 at 17:54
  • 3
    @SemyonAbramyan I hope so :) the point is that for a too broad/open-ended question, too many answers are acceptable. Since there has already been more focussed questions on the subject, you might have a more focussed one, at least taking into account existing ones. – YCor Dec 12 '21 at 18:02
  • 1
    Lots of bounds have been proved for various specific types of finite groups, such as transitive/primitive subgroups of $S_n$, completely reducible/reducible/primitive subgroups of ${\rm GL}(n,K)$. – Derek Holt Dec 12 '21 at 18:26
  • 1
    The title of @BenjaminSteinberg's reference: Lucchini - Some questions on the number of generators of a finite group. – LSpice Dec 12 '21 at 18:30
  • BTW, what you are calling $n(G)$ is far more commonly denoted by $d(G)$ (I'm not sure why). – Derek Holt Dec 12 '21 at 18:32
  • @DerekHolt thank you! I have changed the notation. –  Dec 12 '21 at 21:07
  • 1
    $d(G)\le\lfloor\log_2|G|\rfloor$. More precisely, $d(G)\le\Omega(|G|)$ (i.e., the number of prime divisors of $|G|$, with multiplicity). – Emil Jeřábek Dec 12 '21 at 21:44
  • 2
    Example of lower bound: If a group has rank $\le k$ then every subgroup of index $m$ has rank $\le 1+m(k-1)$. Application: let $F$ be a finite group and $V$ a finite $F$-module, whose underlying abelian group has rank $m$. Then $F\ltimes V^n$ has rank $\ge 1+(nm-1)/|F|$ (while if $F$ is perfect and $V$ has trivial $F$-coinvariants, $F\ltimes V^n$ is a perfect group). Example: for prime power $q\ge 4$, the perfect group $\mathrm{SL}_2(q)\ltimes (q^2)^n$ has rank $\ge 1+(2n-1)/(q^3-q)$. – YCor Dec 13 '21 at 10:43
  • There is a link between $d(G)$ and the subgroup lattice of $G$, see Corollary 7.3 in this paper. – Sebastien Palcoux Dec 15 '21 at 02:59

0 Answers0