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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.

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    $\begingroup$ 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 $19$, $7135$, $161420$. $\endgroup$ – Derek Holt Nov 22 '14 at 4:04
  • $\begingroup$ Thanks for this reference and also for the one pointing to Wiegold's paper in your answer to an earlier question! $\endgroup$ – user 59363 Nov 23 '14 at 18:35
  • $\begingroup$ see mathoverflow.net/questions/198785/… $\endgroup$ – Andreas Thom Mar 2 '15 at 8:03
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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 $|{\rm Aut}\ G|$.

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    $\begingroup$ But what is known about the computation of the numbers $h_r(S)$, for given $S$ and $r$? $\endgroup$ – user 59363 Nov 23 '14 at 19:45
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    $\begingroup$ 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. $h_2(A_5)=19$). $\endgroup$ – Anton Klyachko Nov 23 '14 at 21:07
  • $\begingroup$ Thanks for the reference! It contains several other things which are also intersting for me. $\endgroup$ – user 59363 Nov 25 '14 at 20:44
  • $\begingroup$ Oh, I see: mathoverflow.net/a/53162/24165 $\endgroup$ – Anton Klyachko Nov 27 '14 at 10:42
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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.

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  • $\begingroup$ 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.) $\endgroup$ – user 59363 Nov 21 '14 at 21:02

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