# 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 $p=2$. Verrit'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. Verrit'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.