Timeline for A question on the number of subgroups of symmetric groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 28, 2013 at 7:31 | comment | added | Derek Holt | $H$ has a subgroup of index 2 contained in $A_n$ so the same argument will work for all sufficiently large $n$, and then it is just a matter of checking small $n$. | |
| Aug 28, 2013 at 2:52 | vote | accept | Marius Tarnauceanu | ||
| Aug 28, 2013 at 2:49 | comment | added | Marius Tarnauceanu | Sorry. My definition is again not clear: $G$ is a PSOS-group if $|\{K\leq G \,|\, |K|=|H|\}|$ divides |G| for any $H\leq G$. So the above answer is excellent. Thank you very much! A natural additional question: is the alternating group $A_n$ a PSOS-group? (the above argument does not work here because $H$ is not contained in $A_n$) | |
| Aug 27, 2013 at 11:24 | comment | added | Gerry Myerson | I agree that Derek's answer is more interesting, and probably more what OP wanted (and that my answer is easier). | |
| Aug 27, 2013 at 10:28 | comment | added | Stefan Kohl♦ | @GerryMyerson: Well -- I'm not sure that an order for which there is no subgroup qualifies as "possible order" in the sense of the question ... . | |
| Aug 27, 2013 at 9:42 | comment | added | Derek Holt | Yes, that makes it a lot easier! I thought the question would be more interesting (and more suitable for MO) if we looked for non-divisors other than 0. | |
| Aug 27, 2013 at 0:02 | comment | added | Gerry Myerson | $S_5$ has 120 elements, 30 is a divisor of 120 and so a "possible order" of a subgroup of $S_5$, the number of subgroups of order 30 in $S_5$ is zero, and zero is not a divisor of 120, so $S_5$ is not PSOS. Is it true that for $n\ge5$, $S_n$ has no subgroup of index 4? If so, that would be another way to show $S_n$ is not PSOS for $n\ge5$. And it's proved at crazyproject.wordpress.com/2010/06/08/… that for $n\ge5$, $S_n$ has no proper subgroup of index less than $n$, except for $A_n$. | |
| Aug 26, 2013 at 14:56 | history | answered | Derek Holt | CC BY-SA 3.0 |