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.

There are some references in the answers to the earlier question, but if you are interested in some exact computations, in a paper by myself and Murray Macbeath, "Certain maximal characteristic subgroups of the free group of rank 2", COMMUNICATIONS IN ALGEBRA, 25(4), 1047- 1077 (1997), we compute $n(2,S)$ with $S={\rm PSL}(2,q)$ for small $q$ . For example with $q=5,31,125$ , $n(2,S)$ is respectively $19$ , $7135$ , $161420$ . — Derek Holt, Nov 22 '14 at 04:04
Thanks for this reference and also for the one pointing to Wiegold's paper in your answer to an earlier question! — user 59363, Nov 23 '14 at 18:35
see http://mathoverflow.net/questions/198785/what-is-the-growth-of-the-rank-of-a-power-of-a-finite-simple-group — Andreas Thom, Mar 02 '15 at 08:03

Anton Klyachko · Accepted Answer · 2014-11-22T07:03:54.467

7

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 $|{\rm Aut}\ G|$ .

edited Nov 22 '14 at 07:03

answered Nov 22 '14 at 04:03

Anton Klyachko

3,912

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But what is known about the computation of the numbers $h_r(S)$ , for given $S$ and $r$ ? – user 59363 Nov 23 '14 at 19:45
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The automorphism groups of finite simple groups are well known. So, we have to calculate the (non-reduced) Euler function $\phi_n(G)$ (ie. the number of generating n-tuples). In Section 1.1, Collins describes a technique of such calculations that allowed Hall (in 1936) to calculate, e.g., $\phi_2(A_5)=19\cdot 120$ (ie. $h_2(A_5)=19$ ). – Anton Klyachko Nov 23 '14 at 21:07
Thanks for the reference! It contains several other things which are also intersting for me. – user 59363 Nov 25 '14 at 20:44
Oh, I see: http://mathoverflow.net/a/53162/24165 – Anton Klyachko Nov 27 '14 at 10:42

score 3 · Answer 2 · answered Nov 21 '14 at 18:47

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.

Thanks a lot for your effort. The trivial upper bound can be even chosen to be at most $\vert S\vert^r-1$ : the $r$ -generator free object $F$ in the variety generated by $S$ is sitting inside $S^{\vert S\vert ^r}$ (for universal algebraic reasons) and each $r$ -generator power of $S$ must be a quotient of $F$ --- it must be a proper quotient since $F$ has also solvable quotients, for example, so they cannot be isomorphic. I was more interested in results of the kind: for which $(r,S)$ has the number $n(r,S)$ been computed? (E.g. $n(2,A_5)=19$ , but again I have no reference for this.) — user 59363, Nov 21 '14 at 21:02

Powers of finite simple groups

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