Element that is in $\phi^{-1}(Z(F (G/F(G)))$ I'm studying an article but I'm not able to understand one of his statements. I have the following hypotheses:

*

*$G$ 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) $\mathrm{mdc}(|\phi(J/F(G))|,|F(G)|)=1$.


*$x_r\in G$ 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).
 A: 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
