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This is probably already well-known or too big to answer. Let $G$ be a finite group and $G^{ab}$ be the abelianization of the group G. Is there any bound on $d(G)=\min\{\#S\mid G=\langle S\rangle\}$ by using $d(G^{ab})$ without considering the order of $G$?

Thanks for any answer and comments.

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    $\begingroup$ $G$ can be perfect in which case the abelianization is trivial and you don't learn anything about $G$ from its abelianization. In particular $G$ can require arbitrarily many generators to generate; to be explicit take $G$ to be a direct product of a bunch of copies of $A_5$. In a positive direction there's Burnside's basis theorem: groupprops.subwiki.org/wiki/Burnside%27s_basis_theorem $\endgroup$ – Qiaochu Yuan Sep 28 at 2:52
  • $\begingroup$ Actually I don't know how to prove that $A_5^n$ has unbounded rank as $n \to \infty$. Seems plausible though. $\endgroup$ – Qiaochu Yuan Sep 28 at 3:20
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    $\begingroup$ @QiaochuYuan: pigeon says there is a large set of coordinates where each generator restricts to a diagonal action $(g,g,g,...,g)$. $\endgroup$ – Ville Salo Sep 28 at 4:55
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    $\begingroup$ Wiegold actually proved that for $G$ finite perfect nontrivial, the generating rank of $G^n$ grows logarithmically. $\endgroup$ – YCor Sep 28 at 6:26
  • $\begingroup$ Thanks Qiaochu and Ville. My original question is to show the minimal number of generating set of the following group stays bounded. Let $E_1=S_d$ a symmetric group of an odd degree $d$. Let $E_2=E_{1}\wr E_1\cap \ker(\sgn)$ where the $\sgn$ is the natural sign function by embedding $E_2$ to a symmetric group. For n>2 we define $E_n$ recursively by doing wreath product and intersect with the kernel of $\sgn\circ\res_2$ where $\res_2$ means we restrict the entire wreath product to the first two "coordinate". It can be shown that $E_n/E_n'\cong C_2$, and it is known that ...(continuous) $\endgroup$ – Wayne Peng Sep 28 at 17:39
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Okay, so let's fill in the details on Ville's nice argument in the comments: there is no such bound, and to prove this it suffices to exhibit a sequence of finite perfect groups whose ranks are unbounded. We'll take the sequence $A_5^n$ to be concrete although the argument applies to powers of any finite perfect group. If we take $k$ elements $\{ g_1, \dots g_k \}$ of $G_n$, their projections to each copy of $A_5$ can take at most $60^k$ possible values, so by pigeonhole there's at least one subset of the indices $S \subseteq \{ 1, 2, \dots n \}$ of size at least $\left\lfloor \frac{n}{60^k} \right\rfloor$ such that the projections of the $g_i$ to each copy of $A_5$ indexed by each $i \in S$ are the same. If $|S| \ge 2$ it follows that $\{ g_1, \dots g_k \}$ can't generate $A_5^n$, hence

$$\text{rank}(A_5^n) > \log_{60} \frac{n}{2}.$$

In the positive direction, Burnside's basis theorem implies that $\text{rank}(G) = \text{rank}(G^{ab})$ if $G$ is a finite $p$-group.

Edit: It's maybe also worth mentioning that we needed to do this construction because it doesn't suffice to just take, say, arbitrarily large finite simple groups; it's known that all nonabelian finite simple groups have rank exactly $2$.

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