The order of the system normalizer in a finite solvable group

Question

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.

Answer 1

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.