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. $

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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!