# Bound the number of the minimal generating set of group G by its abelianization

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 $$G$$?

• $G$ 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 – Qiaochu Yuan Sep 28 at 2:52
• Actually I don't know how to prove that $A_5^n$ has unbounded rank as $n \to \infty$. Seems plausible though. – Qiaochu Yuan Sep 28 at 3:20
• @QiaochuYuan: pigeon says there is a large set of coordinates where each generator restricts to a diagonal action $(g,g,g,...,g)$. – Ville Salo Sep 28 at 4:55
• Wiegold actually proved that for $G$ finite perfect nontrivial, the generating rank of $G^n$ grows logarithmically. – YCor Sep 28 at 6:26
• 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) – Wayne Peng Sep 28 at 17:39

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 $$2$$.