5

Let $G$ be some finite group. A component of a finite groups is a quasisimple subnormal subgroup. The layer $E(G)$ is the subgroup generated by all components of $G$, let $F(G)$ denote the Fitting subgroup, then $$ F^*(G) = F(G) E(G) $$ is called the generalized Fitting subgroup of $G$.

Let $L$ be some subnormal subgroup of $G$ in $F^*(G)$, then $L = (L\cap F(G))(L \cap E(G))$ and $L \cap E(G)$ is a product of components of $G$.

This statement is taken from Kurzweil/Stellmacher, Theorie der endlichen Gruppen (page 130). But why is it true?

I tried to prove it, or to understand a very concise sentence given after the statement. But it confuses me. Every component of $L$ is a component of $G$, hence $E(L) \le E(G)$ and also $F(L) \le F(G)$ as $L$ is subnormal. Now $E/Z(E)$ is a direct product of simple non-abelian groups, and as $F^*(G)$ is a central product we have $$ F^*(G) / Z(F^*(G)) \cong F(G) / Z(F(G)) \times E(G) / Z(E). $$ So I tried to deduce the statement by looking at the quotient $LZ(F^*(G)) / Z(F^*(G))$ and write it as a product using that it embedds into a direct product where the first part is nilpoent, and the second a product of simple non-abelian groups, from which the subnormal subgroups have an easy form as direct product of a subset of the factors. But I do not get it.

So could someone please supply a proof of the claim?

Martin Brandenburg
  • 181,922
StefanH
  • 18,586

1 Answers1

2

Here is a sketch proof. Any $g \in L$ can be written as $hg_1g_2 \cdots g_k$, where $h \in F(G)$ and the $g_i$ are in distinct components $E_i$ of $G$ with $g_i \not\in F(G)$.

Then, for each $i$, $E_i \unlhd F^*(G) \Rightarrow [E_i,g] \le E_i$, and $E_i$ quasisimple implies $E_i \le [E_i,g]$, so $E_i = [E_i,g]$ and then $L$ subnormal in $G$ implies $E_i \le L$. The required equality now follows.

Derek Holt
  • 96,726
  • I do not understand your last step. Don't we have $[E_i, g] = [E_i, g_i] \le E_i' = E_i$ as $E_i$ commutes with all $E_j$ for $j \ne i$ and with $F(G)$, so how could the last inequality be implied?? – StefanH Nov 27 '19 at 17:54
  • Which last inequality? I don't see any inequalities. – Derek Holt Nov 27 '19 at 19:29
  • I meant $E_i \le [E_i, g]$. – StefanH Nov 27 '19 at 19:30
  • That follows from $E_i$ quasisimple. $[E_i,G]$ is a normal subgroup. – Derek Holt Nov 27 '19 at 20:37
  • That would give $E_i = [E_i, G]$, which, as I see, would give $[E_i, g] \le E_i$ again, but you are talking about inclusion in the other direction? What am I missing... – StefanH Nov 27 '19 at 20:47
  • Sorry I meant $[E_i,g]$ is a normal subgroup of $E_i$. In geenral $[A,B]$ is normalized by $A$ and by $B$. – Derek Holt Nov 27 '19 at 22:11
  • Okay, but then quasisimplicity yields $[E_i, g] \le Z(E_i)$ or $[E_i, g] = E_i$. In the first case the Three subgroups lemma gives $[E_i, g] = 1$. But how to exclude that case? Also the phrasing of your conclusion suggest a different path, as you write "$E_i$ perfect implies $E_i \le [E_i, g]$", I do not see that perfect implies this inclusion... – StefanH Nov 27 '19 at 22:33
  • We have $g_i \not\in F(G)$ by definition. I replaced perfect by quasisimple. My answer was intended to be a rough sketch, not a detailed proof. – Derek Holt Nov 28 '19 at 08:18
  • Okay, got it. If $[E_i, g] = [E_i, g_i] = 1$ that would imply $g_i \in Z(E_i) \le F(G)$. Thanks for your effort! – StefanH Nov 28 '19 at 13:46