4
$\begingroup$

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!

$\endgroup$
6
$\begingroup$

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 $U$.

$\endgroup$
  • $\begingroup$ A full description of my question and your answer (including GAP code) now appears here: math.hawaii.edu/~williamdemeo/groups/… On page 2 is a picture of the interval above the diagonal in the subgroup lattice of G. Thanks again! $\endgroup$ – William DeMeo Dec 9 '10 at 8:15
  • $\begingroup$ You're welcome! (Incidentally, I checked and this appears to be the smallest example.) $\endgroup$ – Tim Dokchitser Dec 9 '10 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.