0
$\begingroup$

Frattini property: A subgroup $H$ of $G$ is said to have to the Frattini property if for any two intermediate subgroups $K$ and $L$ for $H$ with $K \unlhd L$, the inclusion $L \leq N_G(H)K$ holds.

Lemma: If $H$ satisfies the Frattini property in $G$ then every intermediate subgroup for $N_G(H)$ coincides with its normalizer in $G$.

I want to use the above facts but I'm not if my following proposition is true

Let $G$ be a group with $H \leq G$ satisfying the Frattini property and $B \unlhd G$. Then every intermediate subgroup for $HB$ coincides with its normalizers in $G$.

Now let $K$ be an intermediate subgroup for $HB$. To show that $K = N_G(K)$. $H$ satisfies the Frattini property in $HB$. Using the lemma, we get $N_{HB}(K) =K$

Considering $H\leq K \unlhd N_G(K)$, using the Frattini property, $N_G(K) \leq N_G(H)K$. If $N_G(H) \leq K$, I would be done but I can't seem to proceed further with the given information

$\endgroup$
  • $\begingroup$ It is not immediately clear to me what you mean by "every intermediate subgroup for $N_G(H)$". Try and write mathematically precise statements. $\endgroup$ – Derek Holt Mar 5 '17 at 21:23
  • $\begingroup$ Sorry on my part. By that statement, I mean that a subgroup $K$ is an intermediate subgroup for $N_G(H)$ if $N_G(H) \leq K \leq G$. This is the definition provided in the paper researchgate.net/publication/… $\endgroup$ – R Maharaj Mar 6 '17 at 14:50
  • 1
    $\begingroup$ But if your claim was true, then we could take $B=1$, and conclude that $N_G(H)=H$, which is clearly not always true. $\endgroup$ – Derek Holt Mar 6 '17 at 21:17

Your Answer

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

Browse other questions tagged or ask your own question.