0
$\begingroup$

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 $G=DHN$.

$\endgroup$
  • $\begingroup$ $G/HN$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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 $G = DHN$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.