Let $H, K$ be incomparable subgroups of $G$.  Then 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][1] 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!

  [1]: http://www.math.hawaii.edu/~williamdemeo/groups/A6NormalizerCounterEx.pdf