# Intersection of $p$-subgroup normalizer

Let $Q \leq S$ with $S$ a Sylow $p$-subgroup of $G$. I am interested in conditions that guarantee $$R_Q = \bigcap\left\{ N_{S^g}(Q) : g \in N_G(Q) \right\}$$ is equal to $Q$.

For instance $Q=S$ suffices. If $Q<S$ then $Q<N_{S^g}(Q)$ for all $g \in N_G(Q)$, so it doesn't seem unreasonable for $Q<R_Q$.

I would be happy with a condition like “$N_S(Q)$ is a Sylow $p$-subgroup of $N_G(Q)$” or “all $p$-subgroups $Q$ of $G$.”

For the former Sylow condition, I have a very indirect proof of $R_Q=Q$, but I worry the proof might have a mistake, as I don't understand why $R_Q=Q$ is reasonable.

This is a relative version of the question “What is the intersection of all Sylow $p$-subgroup's normalizer?

• Kind of equivalent requierment. $N_{S^g}(Q)=N_G(Q)\cap S^g$ where $g\in N_G(Q)$ Thus we can say $N_{S^g}(Q)=(N_G(Q)\cap S)^g$ let $L=N_G(Q)\cap S$ and $H=N_G(Q)$ then $R_Q=Core_H(L)$ which is the largest normal sugroup of $H$ contained in $L$. Thus,$R_Q=Q$ means $Q$ is the largest normal subgroup of $N_G(Q)$ contained in $S$.($Q$ is obviously contained in $N_G(Q)$) Does it really have to be?? – mesel Feb 20 '14 at 20:56
• you are right $Q\leq R_Q$ since by above contruction $R_Q$ is laregest normal subgroup of $N_G(Q)$ contained in $S$ and $Q$ is normal in $N_G(Q)$ and contained in $S$... – mesel Feb 20 '14 at 21:08
• Thanks, that is called $p$-radical and is an excellent condition. – Jack Schmidt Feb 20 '14 at 21:19
• You are welcome. – mesel Feb 20 '14 at 21:22

## 1 Answer

Partial answer, mostly due to mesel:

If $N_S(Q)$ is a Sylow $p$-subgroup of $N_G(Q)$, then we calculate $R_Q = O_p( N_G(Q))$:

Suppose $Q < S \leq G$ is a chain of arbitrary subgroups in a group. If $g \in N_G(Q)$, then $N_{S^g}(Q) = N_S(Q)^g$, so

$$R_Q = \bigcap\left\{ N_{S^g}(Q) : g \in N_G(Q) \right\} = \bigcap\left\{ N_{S}(Q)^g : g \in N_G(Q) \right\} = \operatorname{Core}_{N_G(Q)}( N_S(Q) )$$

If $N_S(Q)$ is a Sylow $p$-subgroup of $N_G(Q)$, then $\operatorname{Core}_{N_G(Q)}( N_S(Q) ) = O_p( N_G(Q))$ by definition of $p$-core.

In general (when $Q<S$ and $S$ is a $p$-subgroup of $G$), we get $Q \leq R_Q \leq O_p(N_G(Q))$. Hence if $Q$ is “$p$-radical”, that is, if $Q=O_p(N_G(Q))$, then we get $Q=R_Q$.

I don't see why we couldn't have $Q=R_Q < O_p(N_G(Q))$ if $N_S(Q)$ is not a Sylow $p$-subgroup of $N_G(Q)$. Can anyone resolve that case?

• I am not sure if I have completely understood everything, but try this example. Let $G={\rm PSL}(2,17)$ and $Q$ a subgroup of order $2$. Then $N_G(Q)$ is a Sylow $2$-subgroup of $G$ and has order $16$, so $|O_p(N_G(Q))|=16$. Now $Q$ is contained in $9$ Sylow $2$-subgroups of $G$, but is only central in one of them. Its normalizer in each of the others has order $4$. Let $S$ be one of these. Then I calculate that $Q = \cap_{g \in N_G(Q)} N_S(Q)^g$. – Derek Holt Feb 21 '14 at 9:20
• Thanks, that is very helpful. I rechecked my proof that RQ=Q and found the hidden hypothesis, so this counterexample settles it for me. It looks like RQ=Q might just happen due to size constraints more than anything else (choosing the "wrong" S can shrink RQ, but Q is the smallest it could be). – Jack Schmidt Feb 21 '14 at 11:44