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 $G$?

5$\begingroup$ 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. $\endgroup$ – Derek Holt Jan 10 at 8:33

2$\begingroup$ For each nonabelian 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 $S$. $\endgroup$ – YCor Jan 10 at 9:10

3$\begingroup$ J. Wiegold proved some precise results about the minimum number of generators of a direct product of nonAbelian simple groups. $\endgroup$ – Geoff Robinson Jan 10 at 10:26

3$\begingroup$ This post is related, and this one. $\endgroup$ – Derek Holt Jan 10 at 12:07

2$\begingroup$ As a complement to my previous comment: if the $T_i$ are pairwise nonisomorphic, then the product is always 2generated. In particular, for every $d$ there exists such a product that is 2generated. Also for $d>19$, the group $\mathrm{Alt}_5^d$ is not 2generated. Does this answer your question? As Derek Holt says, giving more context would be helpful. $\endgroup$ – YCor Jan 10 at 14:47
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 JordanHolder 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.

$\begingroup$ By a result of Attila Maróti and M. Chiara Tamburini (Communications in Algebra, Vol. 41, No. 1 (2013), 3449), if $G$ is a nonabelian 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 $1 \leq n \leq 19$. $\endgroup$ – Mikko Korhonen Jan 13 at 5:04

$\begingroup$ Thanks a lot for all the experts who sent the answer to my question, and I have got a complete anwser. J. M. Pan $\endgroup$ – Jiangmin Pan Jan 13 at 9:19

$\begingroup$ @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. $\endgroup$ – Will Sawin Jan 13 at 18:24