[Math] How large can the smallest generating set of a group $G$ of order $n$ be

gr.group-theory

Let $n$ be a natural number. For every group $G$ of order $n$, denote

$d(G)$ : The number of elements of the smallest generating set of $G$

How large is the maximum possible value of $d(G)$ depending on $n$ ?

If $n$ is a cyclic number, we have $d(G)=1$ for every group of order $n$.
For $n=2p$ , $p$ an odd prime, there are two groups : the cyclic group and the dihedral group with $2$ generators, so in this case the maximum value is $2$.

But I wonder, if the maximal value for $d(G)$ can be determined in general, assuming the factorization of $n$ is known. Is the value known for $n=2048$, for example ?

Best Answer

By a Theorem of Guralnick and Lucchini (which does require CFSG), if each Sylow subgroup of $G$ (ranging over all primes) can be generated by $r$ or fewer elements, then $G$ can be generated by $r+1$ or fewer elements. As noted in comments, if $G$ has a Sylow $p$-subgroup $P$ of order $p^{a}$, then $P$ can be generated by $a$ or fewer elements (and $a$ are needed if and only if $P$ is elementary Abelian). Hence if $|G|$ has prime factorization $p_{1}^{a_{1}}p_{2}^{a_{2}} \ldots p_{r}^{a_{r}}$ with the $p_{i}$ distinct primes, and the $a_{i}$ positive integers, then $G$ can be generated by $1 + {\rm max}(a_{i})$ or fewer elements.

(The result attributed to Guralnick and Lucchini was not a joint paper, rather a result proved independently at around the same time: references:

R. Guralnick, "A bound for the number of generators of a finite group, Arch. Math. 53 (1989), 521-523.

A Lucchini: "A bound on the number of generators of a finite group", Arch. Math 53, (1989), 313-317).