The following are definition from Finite Soluble Groups by Doerk and Hawkes.

Defintion 1: A subgroup $H$ of a finite group $G$ is said to be normally embedded in $G$ if each Sylow subgroup of $H$ is a Sylow subgroup of some normal subgroup of $G$.

Defintion 2: A subgroup $H$ of $G$ is said to be pronormal in $G$ if for all $g\in G$, $H$ and $H^g$ are conjugate in $\langle H, H^g \rangle$

Definition 3: A subgroup $H$ of a finite group $G$ is said to be locally pronormal in $G$, if each Sylow subgroup of $H$ is pronormal in $G$.

It is clear from the definitions that a normally embedded subgroup of a finite group is locally pronormal, as Sylow subgroups of a normal subgroup of $G$ are pronormal in $G$.

Proposition: Let $G$ be a finite group and $H$ is normally embedded subgroup of $G$. If $H$ is subnormal in $G$, then $H \unlhd G$

Since $H$ is subnormal in $G$, we may assume that $H \unlhd K \leq G$. Let $P$ be a Sylow $p$-subgroup of $G$. Since $H$ is normally embedded in $G$, it it locally pronormal, and so $P$ is pronormal in $G$. We then get that $G = N_G(P)K$ (follows from properties of pronormal subgroups)

Let $g\in G$. Then $g = ab$ where $a\in N_G(P)$ and $b\in K$. Now $P^g = P^{ab} = P^b \leq H^b = H$. I'm not sure how to proceed to show that $g\in N_G(H)$.


Here is a proof of the proposition in the revised question.

Lemma. If $H$ is subnormal in $K$ and the prime $p$ does not divide $|K:H|$, then all Sylow $p$-subgroups of $K$ are contained in $H$.

I will leave the proof of the lemma to you. You can do it by induction on the length of the subnormal chain from $H$ to $K$.

Now suppose that $H$ is normally embedded and subnormal in $G$. Let $P$ be a Sylow $p$-subgroup of $H$ for some $p$. Then $P \in {\rm Syl}_p(K)$ for some $K$ with $H \le K \unlhd G$. Since $H$ is subnormal in$G$, it is subnormal in $K$ and, by the lemma, all Sylow $p$-subgroups of $K$ are contained in $H$.

Now, for $g \in G$, $P^g \in {\rm Syl}_p(K)$, so $P^g \in H$. Since this is true for all Sylow subgroups of $H$, and $H$ is generated by its Sylow subgroups, it follows that $H \unlhd G$.

  • $\begingroup$ Thank you for your assistance. The proof is elegant. $\endgroup$ – R Maharaj Aug 25 '17 at 14:15

Note: This is an answer to an earlier version of this question.

I don't think this claim is true. Or, some key assumption is missing.

For example, if we select a prime $p$ such that $p\nmid |H|$, then $P=\{1_G\}$. Hence $N_G(P)=G$, so the assumption $G=N_G(P)K$ holds trivially.

But you have surely seen occasions where $H$ is not a normal subgroup of $G$.

Simply assuming that $P$ is non-trivial is not going to fix this problem. For example, we could have $G=G_1\times P$ and similarly adjust $K,H$ from a counterexample with a trivial $P$.

  • 2
    $\begingroup$ I spent quite some time trying to figure out how to twist Frattini's argument to allow me to conclude. Considering the set of Sylow subgroups of $K$ lying above $P$. Nothing... :-) $\endgroup$ – Jyrki Lahtonen Aug 24 '17 at 22:21
  • 2
    $\begingroup$ Yes, I did the same, the only thing you can conclude is that $G=N_G(P)H$ and that does not help much. @R Maharaj - I wonder where this question is coming from. $\endgroup$ – Nicky Hekster Aug 25 '17 at 8:00
  • $\begingroup$ @JyrkiLahtonen, I have added another assumption to the hypothesis. I still can't seem to get the desired conclusion. $\endgroup$ – R Maharaj Aug 25 '17 at 9:24
  • 1
    $\begingroup$ @RMaharaj But a trivial Sylow subgroup is trivially a pronormal subgroup of $G$? The counterexamples we get from my recipe are unaffected by this extra assumption, no? Any assumption about a single Sylow subgroup of $H$ is not going to say much about $H$. $\endgroup$ – Jyrki Lahtonen Aug 25 '17 at 9:33
  • 1
    $\begingroup$ @RMaharaj I wonder whether the assumption might have been that $G=N_G(P)K$ should hold for all Sylow subgroups of $H$ for all $p$. The counterexamples I have in mind don't seem to work in that version. I'm afraid that variant may be outside my range. May be Derek Holt will show up and can comment? $\endgroup$ – Jyrki Lahtonen Aug 25 '17 at 11:04

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.