Definition: A chief series of a finite group $G$ is a maximal normal series of $G$. In other words, $H_i \unlhd G$ for $i=0, \ldots, n$. $$ 1_G = H_0 \leqslant H_1 \leqslant H_2 \leqslant \dots \leqslant H_n = G$$ and there exists no normal subgroups strictly contained between $H_i$ and $H_{i+1}$.

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 $\Sigma \in \text{H}(G)$.

Lemma: Let $D= N_G(\Sigma)$. If $H/K$ is a chief factor of $G$ such that $H/K \leq Z(G/K)$ i.e. it is central then $D$ covers $H/K$. That is to say, $H \leq DK$.

Let $G$ be a finite solvable group with $N$ a minimal normal subgroup of $G$, $HN \unlhd G$ and $y \in G$ such that $G = HN \langle y \rangle$. Let $D= N_G(\Sigma)$ for some Hall system $\Sigma$ of $G$. Then $G = DHN$

I know that $G/HN \cong \langle y \rangle$, so that $G/HN$ is cylic and so abelian. Thus $G/HN = Z(G/HN)$. Thus for any chief series of $G$ passing through $HN$, the chief factors above $HN$ are all central, and $D$ covers them by the Lemma.

I'm not sure how $G = DHN$ follows. Is it possible that $G/HN$ is a chief factor of some chief series of $G$?. If this is the case then $D$ covers $G/HN$ so that $G \leq DHN$. Thus $G=DHN$.

  • $\begingroup$ $G/HN$ is a chief factor if and only if it has prime order. From the point of view of your question, the normal subgroup $HN$ is irrelevant because it is being factored out. The question reduces to let $G$ be a cyclic group and $D$ a subgroup that covers each of its composition factors. Then prove that $D=G$. This is straightforward. $\endgroup$ – Derek Holt May 1 '17 at 9:48
  • $\begingroup$ @DerekHolt, I just realised that I did not specify my question. I need to show that $G = DHN$ from the conditions highlighted. Would the same procedure you have pointed out in your comment still hold? $\endgroup$ – R Maharaj May 1 '17 at 11:46
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    $\begingroup$ That was the question I was addressing in my comment. Since the question is about $G/HN$, you may as well assume that $HN=1$. $\endgroup$ – Derek Holt May 1 '17 at 13:09
  • $\begingroup$ Okay a final question, you mention ''composition factors''. Are we considering a composition series of $G$ or a chief series of $G$? and why can we assume that $D$ covers all the these factors? $\endgroup$ – R Maharaj May 1 '17 at 15:42

Consider the chief series of $G$ given by $$\{1_G \} = B_0 \leqslant B_1 \leqslant \dots \leqslant B_n = G.$$ Since we are factoring out $HN$ from $G$ in $G/HN$, we may assume that $HN = \{1_G \}$. In this case, we have that $G/\{1_G \} \cong G$, which is cyclic, and hence abelian. Moreover, $B_{i+1}/B_{i} \leq G/B_i = Z(G/B_i)$ for $i=0, \ldots, n-1$, since $G/B_i$ is abelian. By the Lemma, $D$ covers $B_{i+1}/B_{i}$ for $i=0, \ldots, n-1$. Now $|G| = \prod \limits_{i=0}^{n-1} [B_{i+1} : B_i]$. Intersecting the chief series of $G$ with $D$ yields $|D| = \prod \limits_{i=0}^{n-1} [B_{i+1} \cap D : B_i \cap D]$. Since $D$ covers $B_{i+1}/B_{i}$ for $i=0,.., n-1$, we have that $[B_{i+1} \cap D : B_i \cap D] = [B_{i+1} : B_i ]$ for $i=0,.., n-1$. Therefore $|D| =\prod \limits_{i=0}^{n-1} [B_{i+1} : B_i] = |G|$. This implies that $G = D$, and so $G = DHN$


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