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Let $G$ be a direct product of nonabelian simple groups $T_1,T_2,\dots,T_d$ with $d>1$. Can $G$ be generated by two elements of $G$?

Jiangmin Pan
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    Hallo! On this forum, posters are discouraged from simply asking questions with no context. You are expected to express your own thoughts on the problem, and say what results you think may be relevant. I can tell you that the answer is yes (provided that $d$ is not too big). You could try and prove for yourself that $A_5 \times A_5$ can be generated by two elements. – Derek Holt Jan 10 '21 at 08:33
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    For each non-abelian simple group $S$, let $n_{S,d}$ be the number of orbits of $\mathrm{Aut}(S)$ on the subset of $S^d$ consisting of generating $d$-tuples. If I'm correct, the answer is: such $G$ is generated by $\le d$ elements iff for each $S$, there are at most $n_{S,d}$ summands isomorphic to $S$. – YCor Jan 10 '21 at 09:10
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    J. Wiegold proved some precise results about the minimum number of generators of a direct product of non-Abelian simple groups. – Geoff Robinson Jan 10 '21 at 10:26
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    This post is related, and this one. – Derek Holt Jan 10 '21 at 12:07
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    As a complement to my previous comment: if the $T_i$ are pairwise non-isomorphic, then the product is always 2-generated. In particular, for every $d$ there exists such a product that is 2-generated. Also for $d>19$, the group $\mathrm{Alt}_5^d$ is not 2-generated. Does this answer your question? As Derek Holt says, giving more context would be helpful. – YCor Jan 10 '21 at 14:47
  • The introduction to this paper -- https://arxiv.org/pdf/1309.5219.pdf -- addresses this question in terms of dessins d'enfants. I think it includes @YCor's observation above about orbits of $Aut(S)$. It also surveys some analogous results to the result about 19 copies of $A_5$ mentioned by Derek, for other simple groups. – Nick Gill Jan 11 '21 at 15:04
  • @YCor It seems to me your latest comment is a complete and good answer; why don't you post it? I'll also say that, as someone who knows very little finite group theory, this seems to me to be a perfectly interesting and reasonable question. – David E Speyer Jan 12 '21 at 14:53
  • On second thought, I decided to make a CW answer myself. Hope I'm not stepping on anyone's toes. – David E Speyer Jan 12 '21 at 15:14
  • @DavidESpeyer It's fine; hopefully OP will accept the answer. – YCor Jan 12 '21 at 15:36

1 Answers1

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This answer mostly summarizes what was written in the comments:

(1) If the $T_i$ are pairwise nonisomorphic, this will happen. Each of the $T_i$ will individually be $2$-generated (see here; this uses the classification of finite simple groups). Now, let $(g_i, h_i)$ be a pair of generators for $T_i$ and let $g = (g_1, \ldots, g_d)$ and $h = (h_1, h_2, \ldots, h_d)$; I claim that $g$ and $h$ generate $\prod T_i$. Indeed, let $G = \langle g,h \rangle$. Then $G$ surjects onto each $T_i$, so it has each $T_i$ as a Jordan-Holder factor. Using that the $T_i$ are pairwise nonisomorphic, $G = \prod T_i$.

(2) At the other extreme, for a fixed finite simple group $T$, the number of generators needed to generate $T^d$ goes to $\infty$ as $d \to \infty$; see here.

David E Speyer
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  • By a result of Attila Maróti and M. Chiara Tamburini (Communications in Algebra, Vol. 41, No. 1 (2013), 34-49), if $G$ is a non-abelian finite simple group and $G^n$ is not $2$-generated, then $n > 2 \sqrt{|G|}$. It follows that $G^n$ is $2$-generated for all $1 \leq n \leq 19$. – Mikko Korhonen Jan 13 '21 at 05:04
  • Thanks a lot for all the experts who sent the answer to my question, and I have got a complete anwser. J. M. Pan – Jiangmin Pan Jan 13 '21 at 09:19
  • @JiangminPan When you are satisfied with an answer, you can accept it by clicking the check mark next to the answer, which will let others know the problem is solved and give you and the answerer some points. – Will Sawin Jan 13 '21 at 18:24