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I having problems understanding one of the parts in Proof 5.2.2 in Kurzweil & Stellmacher group regarding Centralizers of Fitting subgroup.

A question about this proof has been asked here before:

Question on Proof that $O_p(C/(C\cap F(G)) = 1$ for $C = C_G(F(G))$.

Lemma: Let $C := C_G(F(G))$. Then for each prime number $p$ we have $$ O_P(C/(C\cap F(G))) = 1. $$

Proof: Let $P$ be the inverse image of $O_p(C/(C\cap F(G))$ in $C$. Then $P$ is normal, and $P$ is nilpotent since $C \cap F(G) \le Z(C)$ (a Theorem is referred, which says that $G$ is nilpotent if $G/Z$ is nilpotent for each $Z\le Z(G)$). Hence $P \le F(G) \cap C$ and $O_p(C/(C\cap F(G)) = 1$.

The part I'm having trouble with is understanding why $P$ is a nilpotent subgroup. Supposing it is nilpotent, the rest of the proof is pretty clear to me. I also understand that If $P$ is a subgroup of $C$ then $P$ is nilpotent if $C$ is nilpotent. Now, $C$ would be nilpotent if $C/(C\cap F(G))$ is because $(C\cap F(G))$ is contained in $Z(C)$, and i also understand that $C\cap F(G)$ is as a subgroup of a nilpotent group, but how can we say anything about the nilpotency of $C/(C\cap F(G))$?

Thank you!

Martin Brandenburg
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mseren
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  • It doesn't say that $C/(C \cap Z(G))$ is nilpotent. You need $P/(Z \cap F(G))$ to be nilpotent, which is true because it is a $p$-group. – Derek Holt Dec 06 '15 at 16:36
  • @DerekHolt I thought I might have to prove that $C/(C\cap F(G))$ is nilpotent and $P/(C\cap F(G))$ nilpotency will follow. however, did you mean to say that $P/(C\cap F(G))$ is nilpotent because it's a $p$-group? why is it one then? is it because it's a pre-image of a $p$-group? and also, I know that $C\cap F(G)$ is contained in $Z(C)$ but $Z(C)$ might contain elements which are not in $Z(P)$ and therefore perhaps $C\cap F(G)$ will not be in $Z(C)$ ? – mseren Dec 06 '15 at 19:32
  • $P/(C \cap F(G)) = O_p(C/(C \cap F(G))$ is a $p$=group by definition. You have written "I know that $C \cap F(G)$ is contained in $Z(C)$ ... and therefore perhaps $C \cap F(G)$ will not be in $Z(C)$", which makes no sense at all. – Derek Holt Dec 06 '15 at 20:17
  • @DerekHolt yes, sorry, i meant in $Z(P)$. so $P/(C\cap F(G))$ is a $p$-group, and therefore i need $C\cap F(G)$ to be in $Z(P)$ in order to prove nilpotency of $P$, right? (editing the previous comment). Ok, i think i can't edit it... – mseren Dec 06 '15 at 20:25
  • Yes that's right and that follows from $Z \cap F(G) \le Z(C)$ because $P \le C$. – Derek Holt Dec 06 '15 at 21:32
  • @DerekHolt so can i say that $Z(C)\cap P \le Z(P)$, therefore $C\cap F(G) \cap P \le Z(P)$, and because $C\cap F(G) \le P$ so $C\cap F(G) \le Z(P)$. sorry about the amount of questions i just wasnt sure what you meant by $Z$. – mseren Dec 06 '15 at 22:13
  • Sorry, the $Z$ was a mistake, I meanty $C \cap Z(G)$. – Derek Holt Dec 06 '15 at 22:55
  • @DerekHolt To talk about $P/(C\cap F(G))$, don't we need that $C\cap F(G)\subseteq P$? – Guest Jun 13 '22 at 20:49
  • @Guest Yes, but that is true by the definition of $P$. – Derek Holt Jun 13 '22 at 21:08
  • @DerekHolt How? I can't see that. – Guest Jun 13 '22 at 21:25
  • Do you not know what "the inverse image of" means? – Derek Holt Jun 13 '22 at 21:46
  • Yes. $x \in P$ if and only if $x(C\cap F(G)) \in O_p(C/C\cap F(G))$. – Guest Jun 13 '22 at 21:54
  • Oh, so $C\cap F(G)$ is the identity element of $C/C\cap F(G)$. Thus $C\cap F(G) \in O_p(C/C\cap F(G))$. In particular, $x(C\cap F(G)) =C\cap F(G)$ for all $x\in H$, so $H\subseteq P$. Correct? – Guest Jun 13 '22 at 22:01

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