Powers of finite simple groups

Question

I have heard about the following result: for each finite simple non-abelian group $S$ and each natural number $r\ge 2$ there exists a number $n=n(r,S)$ such that the power $S^n$ is $r$-generator but $S^{n+1}$ is not $r$-generator. What is known about the numbers $n(r,S)$? Could someone give me references to this, please?

(I have posted this already on mathstackexchange.com, but did not get a response.)

Edit: This question is in a sense a converse to Bounding from below the cardinality of a set of generators of the $n$-fold cartesian product of a finite group. There it is basically asked for a given (arbitrary, finite) group $G$ and a given number $n$, how small can a generating set for $G^n$ possibly (not) be. In my question the input parameters were a finite simple group $S$ and a number $r\ge 2$ and the question was how big a number $n$ can possibly be so that $r$ elements are sufficient to generate the power $S^n$. Also I was interested in how this number (the biggest such $n$) is actually computed in concrete examples (or whatever is known about the computation of these numbers).

Basically I wanted to know, given a finite simple non-abelian group $S$ and a number $r$, the product of how many copies of $S$ do I need to take to get the $r$-generated free object in the formation generated by $S$.

@Editors/moderators: please feel free to delete the question if it is inappropriate.

Answer 1

See Collins's thesis, Theorem 2.22, page 21.

Theorem 2.22. Let $S$ be a nonabelian simple group and $h_{n-1}(S) < k \le h_n(S)$. Then $r(S^k)=n$.

Here, $r(G)$ is the minimal number of generators of $G$ and $h_n(G)$ is the reduced Euler function i.e. the number of generator sequences of length $n$ of $G$ divided by $|{\rm Aut}\ G|$.

Answer 2

I have no reference for this problem, but let's at least write down the trivial bounds.

Let $s_1,\dots s_r\in S^n$ and suppose $n>|S|^r$. Associate to each index $i$ the element $$(\pi_i(s_1),\dots,\pi_i(s_r))\in S^r,$$ where $\pi_i:S^n\to S$ is the projection onto the $i$th factor. Since $n>|S|^r$ there are distinct indices $i$ and $j$ such that $\pi_i(s_k)=\pi_j(s_k)$ for each $k$. But this implies that $\pi_i\times\pi_j$ maps $\langle s_1,\dots,s_r\rangle$ into the diagonal subgroup of $S\times S$, so $s_1,\dots,s_r$ do not generate $S^n$.

Next we claim that if $S^n$ can be generated with $r$ elements then $S^{n+1}$ can be generated with $r+1$ elements. Indeed take $r$ generators $s_1,\dots,s_r$ of $S^n$ and consider the elements $$(s_1,\pi_1(s_1)),\dots,(s_r,\pi_1(s_r)),(e,x)\in S^n\times S.$$ Here $x$ is any nonidentity element of $S$. By conjugating the last element of this list by the first $r$ elements you see that the elements together generate $1\times S$ by simplicity, so indeed they generate $S^{n+1}$.

Finally recall that every finite simple group is $2$-generated. Thus the minimal number of generators of $S^n$ starts at $2$ and climbs to infinity never rising more than $1$ step at a time, so your function $n(r,S)$ is well defined, and the things we've said so far demonstrate the bounds $$r-1 \leq n(r,S)\leq |S|^r.$$

It seems to me that $n(3,S)$ should tend to infinity with $|S|$, but I don't see how to prove that right now.