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I'm trying to understand the proof of the following which is stated in Kurzweil and Stellmacher:

Let $G$ be a finite group and $C:= C_G(F(G))$. Then $$O_p(C/C\cap F(G))=1$$ for every prime $p$.

Here, $F(G)$ is the Fitting subgroup of $G$, i.e., the product of all nilpotent normal subgroups of $G$ and $O_p(H)$ denotes the intersection of all Sylow $p$-subgroups of a group $H$. Also, a group is defined to be nilpotent if all its subgroups are subnormal.

My attempt: Let $H=C\cap F(G)$ and let $\pi:C\to C/H$ be the quotient map. Let $P=\pi^{-1}(O_p(C/H))$. If $P\subseteq H$, then $\pi(P)= 1$ and so $O_p(C/H)=1$. Therefore it suffices to show that $P\subseteq H$. As $P\subseteq C$, we are only left to show that $P\subseteq F(G)$.

As the $F(G)$ is the largest normal nilpotent subgroup of $G$, it suffices to show that $P$ is normal in $G$ and $P$ is nilpotent.

Normal: As $C$ is a normal subgroup of $G$, so if $P$ is a characteristic subgroup of $C$, then $P$ is also a normal subgroup of $G$......?

Nilpotent: As subgroups of nilpotent groups are nilpotent, we only need to show that $C$ is nilpotent. However, as $H\subseteq Z(C)$, we know that $C$ is nilpotent if and only if $C/H$ is nilpotent.....

This is where I'm stuck. I don't know how to show that $P$ is a characteristic subgroup of $C$ and how to show that $C/H$ is nilpotent.

Edit: Okay, I made some attempt on the nilpotent part: $Z(C)$ is a characteristic subgroup of $C$ and $C$ is a normal subgroup of $G$, so $Z(C)$ is a normal subgroup of $G$. Also, $Z(C)$ is nilpotent. Therefore $Z(C)\subseteq F(G)$. Of course, $Z(C)\subseteq C$. It follows that $H=Z(C)$. So, we need to show that $C/Z(C)$ is nilpotent....

I don't know if this is any help.

Martin Brandenburg
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Using the answers and comments from here and here, I have been able to complete the proof:

Normal: Since normality is preserved by pre-images, we show that $O_p(C/H)$ is normal in $G/H$. This is actually true because $O_p(C/H)$ is a characteristic subgroup of $C/H$ and $C/H$ is normal in $G/H$.

Nilpotent: As $H\subseteq Z(C)\subseteq Z(P)$, it suffices to show that $P/H$ is nilpotent. Indeed, $P/H=\pi(P)=O_p(C/H)$ is nilpotent.

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