Timeline for Groups of order $p(p^2+1)/2$
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 24, 2023 at 23:04 | comment | added | Jianing Song | The Sylow $p$-subgroup is always normal. By Schur-Zassenhaus theorem, a group of order $p(p^2+1)/2$ is the semidirect product of $C_p$ and a group of order $(p^2+1)/2$. Since $\gcd((p^2+1)/2,p-1)=1$, the only homomorphism from a group of order $(p^2+1)/2$ to $\operatorname{Aut}(C_p)$ is trivial, so the semidirect product is actually direct. As a result, the question is equivalent to asking about the groups of order $(p^2+1)/2$. | |
| Aug 8, 2019 at 3:46 | comment | added | Yemon Choi | @verret fair enough but the question is 5 years old and attracted some informative answers, so I don't see why one should seek to close it now | |
| Aug 8, 2019 at 2:26 | comment | added | verret | @YemonChoi I voted that this was a better fit at MSE. (The characterisation of orders with all groups abelian is well-known, and easy to google, and it's an easy exercise to find an example after that.) | |
| Aug 7, 2019 at 1:05 | comment | added | Yemon Choi | Someone has just voted to close this question as "off topic" - would they care to justify or explain this vote? | |
| Aug 5, 2019 at 4:30 | review | Close votes | |||
| Aug 7, 2019 at 1:05 | |||||
| Feb 23, 2014 at 23:40 | answer | added | Geoff Robinson | timeline score: 3 | |
| Feb 22, 2014 at 19:48 | answer | added | Lucia | timeline score: 12 | |
| Feb 22, 2014 at 19:46 | history | edited | BHZ | CC BY-SA 3.0 |
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| Feb 22, 2014 at 18:01 | answer | added | Stefan Kohl♦ | timeline score: 5 | |
| Feb 22, 2014 at 9:53 | comment | added | მამუკა ჯიბლაძე | For convenience, let me add a link to the previous comment, it's here | |
| Feb 22, 2014 at 0:48 | answer | added | Gerry Myerson | timeline score: 14 | |
| Feb 22, 2014 at 0:20 | comment | added | Richard Stanley | mathoverflow.net/questions/31553 (answer by R. Chapman) gives a characterization of those integers $n$ for which every finite group of order $n$ is abelian. | |
| Feb 21, 2014 at 23:57 | comment | added | Lucia | @GerryMyerson: Note that $5\times 13\times 17$ is coprime to $\phi(5\times 13\times 17)=4\times 12\times 16$. So by your(!) nice answer to mathoverflow.net/questions/148731/… there is only one group in this situation. One could also use your answer there to check this -- find $p$ with $n=p(p^2+1)/2$ squarefree but with $n$ not coprime to $\phi(n)$. The answer with $p=53$ is such an example. | |
| Feb 21, 2014 at 23:26 | vote | accept | BHZ | ||
| Feb 21, 2014 at 23:08 | comment | added | Johannes Hahn | @GerryMyerson: No. The sylow theorems tell us, that there are unique normal subgroups A,B of order 13 and 17. AB is a normal {13,17}-Hall subgroup and any 5-sylow C is a complement. Now any morphism $C\to Aut(AB) = Aut(A)\times Aut(B)\cong C_{12}\times C_{16}$ must be trivial since 5 divides neither 12 nor 16. Therefore the semidirect product $G=(AB)\rtimes C$ is actually a direct product, i.e. $G=A\times B\times C$ is cyclic. | |
| Feb 21, 2014 at 21:32 | answer | added | მამუკა ჯიბლაძე | timeline score: 25 | |
| Feb 21, 2014 at 21:32 | comment | added | Gerry Myerson | When $p=13$, the order is $5\times13\times17$. Now $13\times17\equiv1\pmod5$, which (I think) enables us to conclude that there's a nonabelian group of that order. | |
| Feb 21, 2014 at 21:22 | history | asked | BHZ | CC BY-SA 3.0 |