Timeline for What is the number of Sylow p subgroups in $S_p$?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 4, 2020 at 17:47 | comment | added | CA13 | @eraldcoil Each subgroup is cyclic. Suppose $P_1\cap P_2\ni a\neq e$. Then $\langle a\rangle =P_1, P_2$. So the subgroups would be equal. | |
| Jul 29, 2020 at 5:20 | comment | added | roly | Why the intersection of two subgroup must be trivial? | |
| Apr 14, 2020 at 16:05 | comment | added | Aryaman Jal | @Geoffrey Critzer I'm not sure that can be said either. Suppose $p=2$. Then the number of subgroups of order $2$ is the same as the number of involutions on $[n]$ which is equal to $\displaystyle \sum_{i=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2i}(2i-1)!!$ which is nowhere close to $\binom{n}{2}$. | |
| Nov 15, 2015 at 17:27 | comment | added | Geoffrey Critzer | Yes, Thank you Derek Holt. I agree. What I think we can say is this: The number of subgroups in S_n of order p (where p is a prime dividing n) is binomial(n,p)*(p-2)!. | |
| Nov 14, 2015 at 22:29 | comment | added | Derek Holt | Unfortunately it is not as simple as that. For example, when $n=m=6$, in addition to elements like $(1,2,3,4,5,6)$, you also have those like $(1,2)(3,4,5)$. | |
| Nov 14, 2015 at 21:35 | comment | added | Geoffrey Critzer | OK, Thanks. I see now that generally the number of cyclic subgroups of order m in S_n is binomial(n,m)*(m-1)!/phi(m) where phi(m) is the number of positive integers less than m that are relatively prime to m. | |
| Nov 14, 2015 at 21:15 | vote | accept | Geoffrey Critzer | ||
| Nov 14, 2015 at 20:14 | comment | added | Derek Holt | And it saves one click! | |
| Nov 14, 2015 at 20:06 | comment | added | Dietrich Burde | This is one line more than the solution of "see page number $2$" (which is good, of course). | |
| Nov 14, 2015 at 20:02 | history | answered | Derek Holt | CC BY-SA 3.0 |