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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 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!

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up vote 5 down vote accepted

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 $U$.

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Thank you very much! – William DeMeo Dec 9 '10 at 1:51
A full description of my question and your answer (including GAP code) now appears here:… On page 2 is a picture of the interval above the diagonal in the subgroup lattice of G. Thanks again! – William DeMeo Dec 9 '10 at 8:15
You're welcome! (Incidentally, I checked and this appears to be the smallest example.) – Tim Dokchitser Dec 9 '10 at 10:27

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