What is the intersection of all Sylow $p$-subgroup's normalizer?
Intersection of all Sylow $p$-subgroups is generally denoted by $O_p(G)$ and it is one of the well studied topics in group theory as there are many theorems related to this.
Let $R$ be intersection of all Sylow $p$-subgroup's normalizer in $G$. It is easy to observe that $R$ is a characteristic subgroup of $G$ containing $O_p(G)$. I wonder the properties of $R$ and its relation with $O_p(G)$.
If anyone can find or observe something about $R$, I would be thankful.
For reference, here is what we've got so far:
Let $G$ be a finite group with Sylow $p$-subgroup $P$ and set $R= \bigcap N_G(P^g) = \bigcap N_G(P)^g$ to be the intersection of the Sylow $p$-normalizers.
$R$ is an important subgroup, it is the normal core of $N_G(P)$ and the kernel of the permutation action of $G$ on its Sylow $p$-subgroups. In particular, if one wanted to organize groups by how they acted on their Sylow $p$-subgroups, we'd need two invariants: (1) a transitive permutation group whose point stabilizer is the normalizer of a Sylow $p$-subgroup, and (2) $R$.
Consider $[P,R]$. Since $R$ normalizes $P$, we get $[R,P] \leq P$. However $R$ is itself a normal (characteristic) subgroup of $G$, so $[R,P] \leq R$ as well. In other words $[R,P] \leq R \cap P$.
Consider $R \cap P$, a $p$-subgroup normalizing each Sylow subgroup $P^g$. Since $(R \cap P) P^g$ is a subgroup, it is a $p$-subgroup, and so is actually equal to $P^g$, since $P^g$ is maximal amongst $p$-subgroups. Hence $R \cap P \leq P^g$ for all $g$. Taking the intersection we get $R \cap P \leq O_p(G)$. Since $O_p(G)$ is a $p$-subgroup of $R$ contained in $P$, we also get $O_p(G) \leq R \cap P$. Hence $R \cap P = O_p(G)$.
For any $G$-normal subgroup $X$, $X \cap P$ is a Sylow $p$-subgroup of $X$. Hence $O_p(G)$ is a normal $p$-subgroup of $R$. By Schur-Zassenhaus $R=Q \ltimes O_p(G)$ for some $p'$-subgroup $Q$. Since $[Q,P] \leq R\cap P = O_p(G)$, we get that $Q$ centralizes $P/O_p(G)$, but $Q$ need not centralize $O_p(G)$, lest it centralize all of $P$.
Indeed, I think one of the first things to decide is how much different $R$ is from $Z=\bigcap C_G(P^g) = \bigcap C_G(P)^g$. When $O_p(G)=1$, we get $R=Z$, so that $Q \leq Z$.
A survey of small groups (in progress) reveals a variety of structures of $R/Z$:
Amongst the isomorphism classes of groups $G$ with $|G|\leq 1000$ and the conjugation action of $G$ on its Sylow 3-subgroups isomorphic to $A_4$'s action, the quotients $R/Z$ occur with the following frequencies:
- $R=Z$, 1705 times
- $[R:Z] = 2$, 199 times
- $[R:Z] = 3$, 115 times
- $R/Z \cong C_4$, 5 times
- $R/Z \cong C_2 \times C_2$, 13 times
- $R/Z \cong S_3$, 49 times
- $R/Z \cong C_6$, 3 times
- $R/Z \cong C_8$, 1 times
- $R/Z \cong D_8$, 1 times
- $R/Z \cong Q_8$, 1 times
- $R/Z \cong C_3 \times C_3$, 151 times
- $R/Z \cong C_3 \times S_3$, 2 times
- $R/Z \cong \operatorname{Dih}(C_3 \times C_3)$, 17 times
- $R/Z \cong C_3 \ltimes C_9$, 9 times
- $R/Z \cong C_3 \ltimes (C_3 \times C_3)$, 26 times
- $R/Z \cong C_3 \times C_3 \times C_3$, 21 times
- $R/Z \cong \operatorname{Dih}(C_3 \times C_3 \times C_3)$, 1 times
$R_p(G)$ is $p$-solvable, but for every odd prime $p$ there is a finite group $G$ such that $R_p(G)$ is not solvable.
We consider solvability properties of $R_p(G) = \bigcap\{ N_G(P^g) : g \in G \}$ where $P$ is a Sylow $p$-subgroup of $G$. By the previous answer, $R_p(G) = Q \ltimes O_p(G)$ is $p$-closed, so definitely $p$-solvable ($p$-length 1, even).
If $p=2$, then clearly $R_2(G)$ is solvable by Feit–Thompson's odd order theorem. In cases where $|G|\leq 1000$, $R_p(G)$ is always solvable. However, in general this need not be true, since we can take $R_p(G) = G$ by taking any $G$ with a normal Sylow $p$-subgroup. Any such group is $p$-solvable, but need not be solvable. For example $G=A_5 \times C_7$ and $p=7$ works.
A slightly less trivial example (the smallest pefect example in fact) is $G=A_5 \times \operatorname{GL}(3,2)$ with $p=5$ or $p=7$ and $R_p(G)$ is the coprime direct factor. The next smallest perfect example is $G=\operatorname{SL}(2,5) \ltimes \operatorname{GF}(11)^2$ with $p=11$, and then $O_p(G)=P=Z:=\bigcap\{ C_G(P^g) : g \in G\}$ is a Sylow $p$-subgroup, and $R_p(G)/O_p(G)=\operatorname{SL}(2,5)$ is $11$-solvable (being of order coprime to 11) but not solvable.