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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.

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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Dec 12, 2021 at 17:51
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    $\begingroup$ See numdam.org/article/RSMUP_1990__83__201_0.pdf $\endgroup$
    – Benjamin Steinberg
    Commented Dec 12, 2021 at 17:53
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    $\begingroup$ @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. $\endgroup$
    – YCor
    Commented Dec 12, 2021 at 18:02
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    $\begingroup$ 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 ${\rm GL}(n,K)$. $\endgroup$
    – Derek Holt
    Commented Dec 12, 2021 at 18:26
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    $\begingroup$ 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 $\ge 1+(2n-1)/(q^3-q)$. $\endgroup$
    – YCor
    Commented Dec 13, 2021 at 10:43

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