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Definition: Let $G$ be a finite solvable group and $\Sigma \in \text{H}(G)$, the set of Hall systems of $G$. The normaliser of $\Sigma$ is defined as $$ N_G(\Sigma) = \{ g\in G \,|\, H=H^g \text{ for all} H \in \Sigma \}.$$ A system normaliser of $G$ is a subgroup of the form $N_G(\Sigma)$ for some $\Sigma \in \text{H}(G)$.

I have shown that a system normalizer of a finite solvable group covers the central chief factors and avoids the eccentric chief factors of $G$.

Lemma: Let $U$ be a subgroup of a finite group $G$. Let $W \leq V \leq G$. Then $V/W$ is covered by $U$ $\iff$ $[U \cap V: U \cap W]=[V:W]$

Theorem : The order of the system normalizer in a finite solvable group is the product of the order of the central chief factors in a chief series of $G$. [Reference: Finite Soluble Groups by Doerk and Hawkes]

I can't seem to reason why this is the case. The author says that it should follow easily from the lemma but I don't understand how.

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  • $\begingroup$ Presumably an eccentric chief factor is one that is not central? What is the definition of a subgroup avoiding a chief factor? $\endgroup$ – Derek Holt Apr 28 '17 at 16:57
  • $\begingroup$ @DerekHolt Yes. Precisely, If $H/K$ is an eccentric chief factor of $G$ if $H/K \nleq Z(G/K)$. We say that a subgroup $A$ of $G$ avoids a chief factor $H/K$ if $H\cap A \leq K$. $\endgroup$ – R Maharaj Apr 28 '17 at 17:04
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This is straightforward, using the definitions and the lemma. I am going to vote to transfer the question to MSE.

Let $1 =G_0 < G_1 < \cdots < G_n=G$ be a chief series of $G$, and $U$ a system normalizer in $G$. Then $|U| = \prod_{i=1}^n [U \cap G_i:U \cap G_{i-1}]$.

If $G_i/G_{i-1}$ is a central chief factor, then by the lemma $[U \cap G_i:U \cap G_{i-1}] = [G_i:G_{i-1}]$, which is the order of that chief factor.

Otherwise, $U$ avoids that chief factor, in which case $U \cap G_i \le G_{i-1}$, so $U \cap G_i = U \cap G_{i-1}$ and $[U \cap G_i:U \cap G_{i-1}] = 1$.

Hence $|U|$ is the product of the orders of the central chief factors.

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