2
$\begingroup$

I recently come across this question: Number of prime factors of the order of a finite non-abelian simple group. What caught my attention is the second question:

Does there exist a sequence $\{S_n\}$ of non-abelian simple groups such that $|S_n|$ goes to infinity and $\{\omega(|S_n|)\}$ is bounded? ($\omega(m)$ being the number of distinct prime factors of $m$, e.g., $\omega(12)=2$)

It seems that at the moment we are not able to answer it, but what happen if we add the additional requirement that the sequence $\{S_n\}_{n\in\mathbb{N}}$ is strictly ascending? (that is, $S_n$ is isomorphic to a proper subgroup of $S_{n+1}$? Is anything known in this case? I feel it should be easier to find an answer in this case, but I came with nothing at hand.

$\endgroup$
4
  • 1
    $\begingroup$ "Being a subgroup" is maybe not always natural, as opposed to "being a subquotient". For instance, the standard embedding $\mathrm{SL}_n\to\mathrm{SL}_{n+1}$ does not induce an embedding from $\mathrm{PSL}_n$ into $\mathrm{PSL}_{n+1}$. $\endgroup$
    – YCor
    Commented Feb 19, 2023 at 11:36
  • $\begingroup$ @YCor I know, in fact "subgroup" is a way stronger requirement in this case than "subquotient", but that's why I feel it should be possible to prove the result in this case (to give a negative answer in fact) $\endgroup$
    – W4cc0
    Commented Feb 19, 2023 at 11:44
  • $\begingroup$ This doesn't answer your question, but if $p$ is a prime such that $6p+1$ and $12p+1$ are also primes then $|PSL(2,12p+1)|=12p(6p+1)(12p+1)$ has only five distinct prime factors. It would follow from the Generalised Bunyakovsky Conjecture that there are infinitely many of these, and they do seem pretty dense. For example between $10^{15}$ and $10^{15}+10^6$ there are $137$ of them, the first few of which are 1000000000007651, 1000000000012933, 1000000000021171, 1000000000026271, ... $\endgroup$ Commented Aug 6, 2023 at 12:13
  • $\begingroup$ Between $10^{50}$ and $10^{50}+10^6$ there are still six of them. $\endgroup$ Commented Aug 6, 2023 at 12:20

0

You must log in to answer this question.

Browse other questions tagged .