I'm studying an article but I'm not able to understand one of his statements. I have the following hypotheses:


1. $G$ is a solvable group with trivial center, $J=\phi^{-1}(F(G/F(G)))$ and $J\_2=\phi^{-1}(Z(F (G/F(G)))$ where $\phi$ is the canonical homomorphism of $G$ to $G/F(G)$.  

i) $\mathrm{mdc}(|\phi(J/F(G))|,|F(G)|)=1$.
2. $x\_r\in G$ is a prime element such that:  

i) $x\_r\in J\_2$;  

ii) $C\_G(x\_r)F(G)$ is a Frobenius group with complement $C\_G(x\_r)$ and kernel $F(G)$;  

iii) $C\_G(x\_r)\cap F(G)=\{1\_G\}$;  

iv) $C\_G(x\_r)$ has odd order;  

v) $G=N\_G(\langle x\_r\rangle)F(G)$;  

vi) $N\_G(\langle x\_r\rangle)\cap F(G)=\{1\_G\}$;  

vii) $G/C\_G(x\_r)F(G)$ is cyclic.


In the article he considers a prime order element $c\in C\_G(x\_r)$ and shows that $c\in Z(F(C\_G(x\_r))$ (this I managed to show) and, because of that, he says that $c\in J\_2$ But I can't understand why $c\in J\_2$ Could someone help me?


The name of the article is: (The commuting graph of a soluble group) and it was written by: (Christopher Parker). My question is on page 845 within the proof of Theorem 1.1. and link is: (arxiv.org/abs/1209.2279).



Element that is in $\phi^{-1}(Z(F (G/F(G)))$
For simplicity let $\bar G=G/F(G)$ and for any subset or element $x$ of $G$, let $\bar x$ be the image of $x$ in $\bar G$ (the "bar convention"). By (v), $\langle \bar x\_r\rangle\triangleleft \bar G$. In particular, $\bar x\_r\in F(\bar G)$.


It suffices to show that for any $\bar y\in F(\bar G)$ of prime power order $p^a$, $[\bar c,\bar y]=1$. And for this it suffices to show that $\bar y\in C\_{\bar G}(\bar x\_r)$, since then $\bar y\in F(\bar G)\cap C\_{\bar G}(\bar x\_r)\le F(C\_{\bar G}(\bar x\_r))$, in which you have shown $\bar c$ is central. (Note that reduction modulo $F(G)$ induces an isomorphism $\alpha:C\_G(x\_r)\to C\_{\bar G}(\bar x\_r)$. By (iii), $\alpha$ is injective. Since $|x\_r|$ is coprime to $|F(G)|$, a Frattini argument shows that $\alpha$ is surjective.)


Finally, say $\bar x\_r$ has (prime) order $r$. If $p\ne r$, then $\bar y$ and $\bar x\_r$ are of coprime orders in $F(\bar G)$, so $\bar y\in C\_{\bar G}(\bar x\_r)$. If $p=r$, then $\bar y\in C\_{\bar G}(\bar x\_r)$ as $\langle \bar x\_r\rangle\triangleleft \bar G$. QED


