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 $|S|\geq n$?



Bounding from below the cardinality of a set of generators of the $n$-fold cartesian product of a finite group
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"](http://www.math.ucla.edu/~pak/papers/sim.pdf), page 4.


