Bounding from below the cardinality of a set of generators of the $n$-fold cartesian product of a finite group

Question

Let $G$ be a nontrivial finite group. Given $n\in\mathbb{Z}_{\geq 1}$, let $G^n$ be the cartesian product of $n$ copies of $G$. Further let $S\subseteq G^n$ be a generating set of $G^n$.

Question: Do we always have $|S|\geq n$?

Answer 1

The answer is no. -- For example, ${{\rm A}_5}^3$ is $2$-generated. -- We have e.g. $$ \langle (2,5)(3,4)(6,7,8,9,10)(11,12,15), (1,3,2,4,5)(6,7,9,10,8)(11,13,12,14,15) \rangle \ \cong \ {{\rm A}_5}^3. $$ Finding such generator pair is easy -- just pick two random elements until you find some which generate.

Actually, according to Hall, even ${\rm A}_5^{19}$ is $2$-generated (while ${\rm A}_5^{20}$ is not). Cf. e.g. Pak's "On probability of generating a finite group", page 4.

Answer 2

It is proved in

J. Wiegold, Growth sequences of finite groups, III, J. Austral. Math. Soc. Ser. A 25 (1978) 142–144

that a direct product of $n$ nonabelian simple (or quasisimple) groups can be generated by at most $2 + {\rm Ceiling}( \log_{60} n)$ elements.

On the other hand, given $r$ elements of $G^n$, there are at most $|G|^r$ possible projections on to the components. So, if they generate $G^n$, then no two of the projections can be the same, and hence $n \le |G|^r$ and we have the lower bound $r \ge \log_{|G|} n$.

So the best possible bound is logarithmic in $n$.

Answer 3

For an alternative proof of Wiegold's results, and in particular for finite groups G such that the number of generators of G^n is comparable to log n, see Section 3 in Jacques Thévenaz, Maximal subgroups of direct product groups, Journal of Algebra 198 (1997) 352-361.

Pierre de la Harpe.