# For odd primes $p$, are finite groups with self-normalizing Sylow $p$-subgroups solvable?

Is it the case that for odd primes $$p\geq5$$, all finite groups with self-normalizing Sylow $$p$$-subgroups are solvable? The simple group of order 168 shows that this conjecture does not hold for $$p=2$$. Verret's answer provides a counterexample when $$p=3$$.

Does this conjecture hold for any odd primes $$p\geq5$$? If so, is there a proof that does not rely on the CFSG?

Edit: Initially, I thought that I had proved that a minimal counterexample to this conjecture was necessarily simple. Verret's answer shows that this is not the case.

Fix a prime $$p$$ and let $$G$$ be a minimal counterexample to the conjecture. Suppose that $$G$$ is not simple.

1) Let $$H$$ be a nontrivial normal subgroup of $$G$$ and let $$P$$ be a Sylow $$p$$-subgroup of $$G$$. Then $$PH$$ is a subgroup of $$G$$ that contains $$N_G(P)$$ so $$N_G(PH)=PH$$. Also, $$PH/H$$ is a Sylow $$p$$-subgroup of $$G/H$$ with $$N_{G/H}(PH/H)=N_G(PH)/H=PH/H.$$ Then $$G/H$$ is solvable by the minimality of $$G$$. However, $$G$$ is not solvable so $$H$$ is not solvable.

2) Let $$H$$ be a proper normal subgroup of $$G$$ with $$G/H$$ simple. Then $$G/H$$ is cyclic of prime order. If $$G/H$$ is not cyclic of order $$p$$ then $$H$$ contains a Sylow $$p$$-subgroup of $$G$$ and thus Sylow $$p$$-subgroups of $$H$$ are self-normalizing. The minimality of $$G$$ provides a contradiction. This shows that $$G/H$$ is cyclic of order $$p$$. If $$p$$ does not divide the order of $$H$$ then $$G\cong H\rtimes_\varphi C_p$$ for some homomorphism $$C_p\to Aut(H)$$. However, Sylow $$p$$-subgroups of $$G$$ are self-normalizing so this homomorphism must be fixed-point-free. Then $$H$$ admits a fixed-point-free automorphism of prime order which contradicts the non-solvability of $$H$$. In summary, $$p$$ divides the order of $$H$$ and $$G/H$$ is cyclic of order $$p$$.

• I did a quick google search and, in the introduction of ac.els-cdn.com/S0021869302001205/…, it is claimed that $\mathrm{P\Sigma L}(2,27)$ is an example. – verret Aug 29 '18 at 19:39
• I checked and $\mathrm{PSL}(2,27)$ doesn't have the property. Doesn't that contradict some of the arguments above? – verret Aug 29 '18 at 19:44
• @verret Are you saying that the Sylow 3-subgroup of $PSL(2,27)$ is self-normalizing? – Thomas Browning Aug 29 '18 at 19:54
• I think the problem is in part 2). Why does $N_G(P)$ contain every Sylow $p$-subgroup of $G$? – verret Aug 29 '18 at 20:11
• I think you might be right. If $Q$ is a Sylow $p$-subgroup of $G$ containing $P$ then $P=Q\cap H$ and so $N_G(P)$ contains $Q$. Then $N_G(P)$ contains the subgroup of $G$ generated by the Sylow $p$-subgroups of $G$ that contain $P$. However, $N_G(P)$ might not contain Sylow $p$-subgroups of $G$ that do not contain $P$. – Thomas Browning Aug 29 '18 at 20:27

$\mathrm{P\Sigma L}(2,27)$ has a self-normalising Sylow $3$-subgroup.
(On the other hand, $\mathrm{PSL}(2,27)$ does not. There is a mistake in the proof that a minimal example must be simple, in part 2): there is no reason for $N_G(P)$ to contain every Sylow $p$-subgroup of $G$.)
According to http://www.ams.org/journals/proc/2004-132-04/S0002-9939-03-07161-2/S0002-9939-03-07161-2.pdf the only counterexamples for odd $p$ occur for $p=3$ and involve $\mathrm{PSL}(2,3^f)$ as a composition factor.