Given that an $n$-simple group is a group isomorphic to the direct product of $n$ simple groups, can arbitrarily many nonisomorphic finite 2-simple groups share the same order?
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$\begingroup$ There are only finitely many isomorphism classes of groups of any given finite order anyways, so I'm not sure I understand the question correctly. $\endgroup$– Achim KrauseMar 12, 2021 at 5:46
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$\begingroup$ The question is unclear, but there do exist infinitely many pairs $(G,H)$ of finite simple groups such that $G \not\cong H$ and $|G| = |H|$. $\endgroup$– spinMar 12, 2021 at 6:07
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$\begingroup$ Given an integer $k$, let $f(k)$ be the number of non-isomorphic $2$-simple groups of order $k$. I think what is being asked is whether $f(k)$ is bounded? For example, using $A$ and $B$ non-isomorphic simple groups of the same order, we can take $A\times A$, $A\times B$ and $B\times B$, which shows that $f(|A\times A|)\geq 3$... (In principle, we could also have $|C\times D|=|A\times A|$, with $|C|\neq |A|$...) $\endgroup$– verretMar 12, 2021 at 9:32
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$\begingroup$ Edit: fixed typo $\endgroup$– Daniel SebaldMar 12, 2021 at 14:36
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$\begingroup$ My previous comment (now deleted) was inaccurate. It is known that $|S|^m = |T|^n$ for non-isomorphic finite simple groups can only occur when $m=n$ and $|S|=|T|$, but that does not answer the question. There are many instances of $|S \times T| = |U \times V|$ with four different simple groups $S,T,U,V$, such as $|L_2(7) \times U_3(4)| = |A_6 \times {\rm Sz}(8)| = 10483200$ and $|L_2(8) \times M_{11}| = |L_2(11) \times U_3(3)| = 3991680$, so I do not know the answer to the question. $\endgroup$– Derek HoltMar 12, 2021 at 15:03
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1 Answer
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At least four such groups can share the same order: $$ |{\rm A}_7 \times {\rm PSp}(6,2)| = |{\rm PSU}(3,3) \times {\rm J}_2| = |{\rm A}_8 \times {\rm A}_9| = |{\rm PSL}(3,4) \times {\rm A}_9|. $$
answered Feb 9, 2023 at 16:19