3
$\begingroup$

A few days ago I asked a question (Groups of order $p(p^2+1)/2$) about a finite group of order $p(p^2+1)/2$ and I got a lot of useful information about it. Thanks for the nice and very helpful answers. Now I have a question:

Is it possible that we can conclude that any group of order $(p^2+1)/2$, where $p>13$ and $p\ne 239$ is a prime, has an abelian and normal Sylow subgroup?

For small $p$, i.e., $p<1000$, we can see that most of the time there exists an odd prime $p'$ which is large enough that the subgroup of that order is normal and abelian. Of course this is not always true.

$\endgroup$
10
  • 1
    $\begingroup$ You could check out $p=239$ when $\frac{p^2+1}{2}=13^4$ $\endgroup$ Feb 24, 2014 at 6:21
  • 2
    $\begingroup$ 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 $n=5^4∗13∗61∗97∗137∗641$. $\endgroup$
    – verret
    Feb 24, 2014 at 12:11
  • 1
    $\begingroup$ ADDENDUM: there are three more candidates for $p$ between 3 and 4 million: $p=3319597,3456127,3636443$, and then none up to 10 million. $\endgroup$
    – verret
    Feb 24, 2014 at 12:18
  • 2
    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Feb 24, 2014 at 14:20
  • 1
    $\begingroup$ 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! $\endgroup$
    – Derek Holt
    Feb 25, 2014 at 20:01

1 Answer 1

1
$\begingroup$

If for any prime $q$ dividing $\frac12(p^2+1)$ we have $q^3$ divides $\frac12(p^2+1)$, then the answer is `not'. This is purely number-theoretic question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.