Centralizer of the radical is a subgroup of the radical?

Question

Let $G$ be a finite soluble group. We denote by $\mathfrak{N}$ the class of the finite nilpotent groups. The nilpotent radical, $G_\mathfrak{N}$, is the maximal subnormal nilpotent subgroup of $G$. I want to prove that $C_G(G_\mathfrak{N})\leq G_{\mathfrak{N}}$. How can I prove this?

Answer 1

Let $C=C_G(G_\mathfrak{R})$ and $K=C_G(G_\mathfrak{R})\cap G_\mathfrak{R}$. Then $[C,K]=1$ so $K\leqslant Z(C)$. In particular, $K\unlhd C$.

Since $G$ is solvable, so is the quotient $\overline{C}:=C/K$. Let $r=d_\ell(\overline{C})$ and $H$ be the complete inverse image of $\overline{C}^{(r-1)}$ in $G$. Then $H'\leqslant K$, but since $K$ centralizes $C$, $K$ also centralizes $H$, so $[H,H,H]=1$. Thus $H$ is nilpotent.

Now note that since $G_\mathfrak{R}$ is characteristic, so is $C$, and thus so is $K$. $\overline{H}$ is characteristic in $\overline{C}$, so $H$ is characteristic in $C$. Thus $H$ is characteristic in $G$. In particular, $H$ is a nilpotent normal subgroup, but $H\leqslant C$ and $H\not\leqslant K$, so $H\not \leqslant G_\mathfrak{R}$, which contradicts that $G_\mathfrak{R}$ is maximal.