Timeline for On the Groups of Order $(p^2+1)/2$
Current License: CC BY-SA 3.0
19 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 22, 2016 at 17:42 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 22, 2016 at 15:47 | answer | added | yakov | timeline score: 1 | |
| Feb 26, 2014 at 11:14 | comment | added | BHZ | Since we need a normal abelian Sylow subgroup for $G$ for a prime divisor of the order of $G$ can we say that in the above statement the $r$ Sylow subgroup has the stated properties | |
| Feb 26, 2014 at 9:01 | comment | added | Geoff Robinson | To try to cook up examples it might help to consider the prime factors of $\frac{p+1}{2} + i\frac{p-1}{2}$ in the ring of Gaussian integers $\mathbb{Z}[i].$ Here, by prime, I mean an irreducible element of $\mathbb{Z}[i],$ not a rational prime. | |
| Feb 25, 2014 at 20:01 | comment | added | Derek Holt | I didn't say that the answer must be no. But I would be very surprised indeed if it could proved that the answer was yes! | |
| Feb 25, 2014 at 11:08 | comment | added | BHZ | Thanks for the comments. Theoritically it seems that the answer is No but as Verret checked the numbers it seems that the answer is Yes. But as Derek pointed out the answer must be NO. Many thanks for your helps | |
| Feb 24, 2014 at 14:20 | comment | added | Derek Holt | I think this is as much of a question in number theory as in group theory. I would guess that the conjecture is false, but it could be very hard to find a counterexample. For example, if we had $(p^2+1)/2 = rq^3$, with $q$ prime and $r|(q^2-1)$ then there would be a counterexample of that order. | |
| Feb 24, 2014 at 12:18 | comment | added | verret | ADDENDUM: there are three more candidates for $p$ between 3 and 4 million: $p=3319597,3456127,3636443$, and then none up to 10 million. | |
| Feb 24, 2014 at 12:11 | comment | added | verret | A small observation : since the group has odd order, it is soluble. In particular, it has SOME (minimal) abelian normal subgroup. Moreover, by "Groups of Cube-Free Odd Order", by Curran, we may assume that the group is not cube-free. Anyway, I checked the conjecture up to $p=3000000$. I was only checking that $n=(p^2+1)/2$ was not squarefree and that Sylow's theorem would not force a normal $q$-Sylow subgroup of order at most $q^2$ for some prime $q$. Up to $p=3000000$, the only exceptions are for $p=239$, when we get n=$13^4$ and $p=2905807$ when we get $n=5^4∗13∗61∗97∗137∗641$. | |
| Feb 24, 2014 at 6:35 | history | edited | BHZ | CC BY-SA 3.0 |
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| Feb 24, 2014 at 6:34 | comment | added | abx | So could you edit the question in grey? As it stands, the answer is obviously no. | |
| Feb 24, 2014 at 6:26 | comment | added | BHZ | Yes that's right. But we know that by Crescenzo work on the diophantine equation $p^2+1=2q^m$ there is not any other possible case if $p\ne 13$ | |
| Feb 24, 2014 at 6:21 | comment | added | Aaron Meyerowitz | You could check out $p=239$ when $\frac{p^2+1}{2}=13^4$ | |
| S Feb 24, 2014 at 6:01 | history | edited | Amit Kumar Gupta | CC BY-SA 3.0 |
little changes
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| S Feb 24, 2014 at 6:01 | history | suggested | gaoxinge | CC BY-SA 3.0 |
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| Feb 24, 2014 at 5:26 | review | Suggested edits | |||
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| Feb 24, 2014 at 4:51 | history | edited | BHZ | CC BY-SA 3.0 |
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| Feb 24, 2014 at 4:42 | history | asked | BHZ | CC BY-SA 3.0 |