Timeline for Number of prime factors of the order of an increasing sequence of finite non-abelian simple group
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 6, 2023 at 12:20 | comment | added | Dave Benson | Between $10^{50}$ and $10^{50}+10^6$ there are still six of them. | |
| Aug 6, 2023 at 12:13 | comment | added | Dave Benson | 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, ... | |
| Feb 19, 2023 at 11:44 | comment | added | W4cc0 | @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) | |
| Feb 19, 2023 at 11:36 | comment | added | YCor | "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 $\mathrm{PSL}_{n+1}$. | |
| Feb 19, 2023 at 11:33 | history | edited | YCor | CC BY-SA 4.0 |
added quotation, added tag, formatting
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| Feb 19, 2023 at 11:15 | history | asked | W4cc0 | CC BY-SA 4.0 |