Timeline for recognition of symmetric groups in GAP
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Mar 13, 2021 at 19:29 | history | suggested | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |
Added a citation (that was stated to not work in the original answer)
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| Mar 13, 2021 at 18:52 | review | Suggested edits | |||
| S Mar 13, 2021 at 19:29 | |||||
| Nov 17, 2017 at 22:17 | comment | added | John Shareshian | If $p=2^r-1$ is a Mersenne prime, then the image of the embedding of $PGL_2({\mathbf F}_{2^r})$ in $S_{p+2}$ determined by the natural action on $1$-spaces contains a $p$-cycle. So, it seems that you need a cycle of prime order at most $n-3$. | |
| Nov 17, 2017 at 21:04 | comment | added | Igor Rivin | @NoamD.Elkies I fixed it, at last. | |
| Nov 17, 2017 at 21:03 | history | edited | Igor Rivin | CC BY-SA 3.0 |
fixed typos
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| Nov 17, 2017 at 18:26 | comment | added | Igor Rivin | @NoamD.Elkies Yes, of course. I was thinking of a slightly different problem (finding the Galois group of a polynomial), so did not reboot properly :) | |
| Nov 17, 2017 at 18:00 | comment | added | Noam D. Elkies | @Igor Rivin you wrote that "You rule out $A_n$ by randomly generating an odd permutation". But for a subgroup generated by a given subset $S$ of $S_n$, the subgroup is contained in $A_n$ iff $S$ is, which can be tested without "randomly generating" anything. | |
| Nov 17, 2017 at 17:27 | vote | accept | Vladimir Dotsenko | ||
| Nov 17, 2017 at 17:20 | comment | added | Vladimir Dotsenko | Thank you, Igor. This makes much more sense. I was very disappointed once the program told me that my group was $S_{50}$ right away but then ran out of memory once I asked to represent $(12)$ as a product of generators... But with the method you are describing it is not surprising at all! | |
| Nov 17, 2017 at 15:48 | comment | added | Igor Rivin | @NoamD.Elkies I don't think I said it was. You have to generate random elements (which is exactly what the answer says). | |
| Nov 17, 2017 at 15:47 | comment | added | Noam D. Elkies | Deciding between $A_n$ and $S_n$ is not a matter of luck: just check whether any of the given generators is odd. | |
| Nov 17, 2017 at 15:37 | history | answered | Igor Rivin | CC BY-SA 3.0 |