# Proving a subgroup is normal from the given conditions

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 $H \unlhd G$.

• Thank you for your assistance. The proof is elegant. – 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 $P$.

• 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... :-) – Jyrki Lahtonen Aug 24 '17 at 22:21
• 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. – Nicky Hekster Aug 25 '17 at 8:00
• @JyrkiLahtonen, I have added another assumption to the hypothesis. I still can't seem to get the desired conclusion. – R Maharaj Aug 25 '17 at 9:24
• @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 $H$. – Jyrki Lahtonen Aug 25 '17 at 9:33
• @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? – Jyrki Lahtonen Aug 25 '17 at 11:04