Clearly every finite group has a minimal generating subset.
Is there any formula for the number of minimal generating subsets of a finite group?
Is it known which groups have a unique minimal generating subset?
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Sign up to join this communityClearly every finite group has a minimal generating subset.
Is there any formula for the number of minimal generating subsets of a finite group?
Is it known which groups have a unique minimal generating subset?
1.No, this is a hard question in general. It could maybe be done for special classes of groups, say nilpotent groups.
2.The only (finitely-generated) groups which have a unique minimal generating subset are the trivial group and the cyclic group of order 2.
Let $G$ be a group with a unique minimal generating subset $S$. As Gerhard Paseman said in the comments, we can replace a non-involution by its inverse, so we can assume that every element in $S$ is an involution. Now, let $s$ and $t$ be distinct elements of $S$ and let $S^*=(S\setminus\{t\})\cup \{st\}$. Clearly, $S^*$ generates $G$. Since $G$ has a unique minimal generating set and $|S|=|S^*|$, $S^*$ must be minimal (otherwise we'd get a smaller generating set) and thus $st$ is an involution and $s$ and $t$ commute. Since $s$ and $t$ were arbitrary elements of $S$, $G$ is an elementary abelian $2$-group and it is easily seen that it must have order at most $2$.
In finite simple groups, most pairs of elements generate, so (at least asymptotically), the number of generating pairs is $\asymp |G|^2.$ See, for example:
Robert M. Guralnick, Martin W. Liebeck, Jan Saxl, and Aner Shalev, MR 1707675 Random generation of finite simple groups, J. Algebra 219 (1999), no. 1, 345--355.
If $G$ is a p-group and the Frattini quotient $G/\phi(G)$ has order $p^r$ (it's an elementary abelian group) then the number of minimal generating sets equals the number of bases of $\mathbb{F}_p^r$: $\frac{(p^r-1)(p^r-p)\cdots(p^r-p^{r-1})}{n!}$.
More generally: If $G$ is nilpotent, each gnerating set of $G/[G,G]$ lifts to a generating set of $G$ and the cardinality of a minimal generating set equals the rank of $G/[G,G]$. Hence, counting the minimal generating sets boils down to counting the minimal generating sets of an abelian group. But I don't know a formula for this number by heart.