# On the Groups of Order $(p^2+1)/2$

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.

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

## 1 Answer

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.