Let $H, K$ be incomparable subgroups of $G$. The following is **false**:


$
N\_G(H \cap K) = H \cap K \quad \Rightarrow \quad N\_G(H)=H \text{ and }
N\_G(K)=K
$


Here is a counter-example:


$
G = A\_6, \quad H = (C\_3 \times C\_3) : C\_2, \quad \quad K = S\_4.
$


(see [link text](http://www.math.hawaii.edu/~williamdemeo/groups/A6NormalizerCounterEx.pdf) for details)


Is it true that $N\_G(H \cap K) = H \cap K$ implies that at least one of $H$ or $K$ is self-normalizing? I doubt it, but I can't seem to find a counter-example. So, does anyone know of an example of the following?


A group $G$ with incomparable subgroups $H, K$ such that $H \lneq N\_G(H)$, $K \lneq N\_G(K)$,and $H\cap K = N\_G(H\cap K)$. 


Thank you!



A basic question about selfnormalizing subgroups
I think a counterexample is $G=S\_3\times S\_3\times S\_3$. 


Say $U$ is the "diagonal" $S\_3$ in $G$; so if $G\lt S\_9$ is generated by $(123)$, $(12)$, $(456)$, $(45)$ ,$(789)$ and $(78)$, then $U$ is generated by $(123)(456)(789)$ and $(12)(45)(78)$, and it is self-normalizing.


However $H=\langle U,(123)\rangle$ and $K=\langle U,(456)\rangle$ (both $\cong C\_3\rtimes S\_3$) are not self-normalizing and their intersection is $U$.


