# Characterisation of Pronormal subgroups of finite solvable groups by its system normalizers

Definition: A subgroup $H$ of $G$ is said to be pronormal if for all $g\in G$, there exists $x\in \langle H, H^g \rangle$ such that $H^g = H^x$

Definition: Let $\Sigma$ be a Hall system of $G$. Then is system normalizer of $G$ is of the form $N_G(\Sigma) = \{ g\in G \,|\, H=H^g \;\text{ for all } H \in \Sigma \}$

Lemma: Let $G$ be a finite solvable group. Suppose that $H \leq G$ is pronormal in every proper subgroup of $G$ in which it is contained. Then $H$ is pronormal in $G$ $\iff$ $N_G(H)$ contains some system normalizer of $G$

Theorem: [Wood] Let $G$ be finite solvable group and $H$ be a subgroup of $G$. Then the following are equivalent:

(a) $H$ is pronormal in $G$;

(b) Whenever $H \leq L \leq G$, then $N_L(H)$ contains some system normalizer of $L$.

I have already shown (a) $\implies$ (b)

For (b) $\implies$ (a) Proceed by induction on $|G|$. (This allows to assume that $H$ is pronormal in every proper subgroup in which it is contained to use the Lemma). I'm not sure what should I do next to use the induction hypothesis. Do I use an arbitrary subgroup $L$ of $G$ such that $H \leq L \leq G$? or should I use a specific subgroup, say $L = \langle H, D \rangle$, where $D$ is some system normalizer of $G$, and then use the hypothesis of (b)?

• I am confused. You seem to have assumed in (b) that $N_G(H)$ contains a system normalizer of $G$. – Derek Holt May 3 '17 at 9:04
• @DerekHolt, I want use induction on $|G|$ to show (b) $\implies (a)$. I think the Lemma will be useful, but I'm not sure how to argue this. – R Maharaj May 3 '17 at 16:28
• I suggest you use induction on $[G:H]$. It is certainly true for $[G:H]=1$, and the lemma will provide the inductive step. – Derek Holt May 3 '17 at 18:03
• @DerekHolt, For $[G : H] =1$, we have $G =H$ and the result is true. For $[G:H] > 1$, we have that $H < G$. Why can we assume for the induction hypothesis that $H$ is pronormal in every proper subgroup of $G$ in which it is contained? (I have not done induction on the index of a subgroup before) – R Maharaj May 4 '17 at 6:57
• So putting $L=G$ in your first sentence, we have that some system normalizer of $G$ is contained in $N_G(H)$. – Derek Holt May 4 '17 at 19:42