Is there any group with self-normalizing Sylow $p$-subgroup? Let $G$ be a finite group and assume that $G$ is not a $p$-group. I am looking for an example that for every Sylow subgroup $P$ of $G$,
$$N_G(P)=P$$.
I have doubt whether such groups exist.
 A: There are no such groups. Let us a call a group strange if all of its Sylow subgroups are self-normalizing.
We first show that if $G$ is strange and $N \unlhd G$ then $G/N$ is strange. If not, then there exists $P/N \in {\rm Syl}_p(G/N)$ with $P/N \lhd K/N$ and $P \ne K$. Let $Q \in {\rm Syl}_p(G)$ with $QN=P$. Then, by the Frattini argument, $K = N_K(Q)N$, so $Q \ne N_K(Q)$, contradicting $G$ strange.
Now let $G$ be a group of minimal order that is strange and not a $p$-group, and let $N$ be a minimal normal subgroup of $G$. Then, since $G/N$ is strange, it must be a $p$-group for some prime $p$. Since $G$ is not a $p$-group, $N$ is not a $p$-group, and if $Q \in {\rm Syl}_q(N)$ for some $q \ne p$, then $Q \in {\rm Syl}_q(G)$ and, by the Frattini Argument, $G=N_G(Q)N$. Hence, since $G$ is strange, we must have $G=N$ and $G$ is simple.
In fact $G$ is nonabelian simple and it is known from the classification that $G$ has a cyclic Sylow $p$-subgroup $P$ for some prime $p$. (See, for example, the discussion here.) But now, by Burnside's Transfer Theorem, $P$ cannot be self-normalizing in $G$, contradiction.
A: What about $G=S_3$ and $P=\{(1),(12)\}$? Or do you mean for every prime $p$ dividing $|G|$? If so then see here. It is proved (Corollary 1.3)) that if $p$ and $q$ are different primes dividing the order of $G$, and $P \in Syl_p(G)$ and $Q \in Syl_q(G)$, then one of $P$ and $Q$ cannot be self-normalizing.
