## In a finite solvable group with no nontrivial normal subgroups relatively prime to p, the p-core contains its centralizer

Let $p$ be a prime dividing the order of the finite solvable group $G$. Assume $G$ has no nontrivial normal subgroups of order prime to $p$. Let $P = O_p(G)$ be the largest normal $p$-subgroup of $G$ (cf. this previous exercise). Prove that $C_G(P) \leq P$; i.e., that $C_G(P) = Z(P)$.

If $G$ is simple, then $G \cong Z_p$ for some prime $p$. In this case, $P = G$ (by definition), and we have $C_G(P) = Z(P) = G$.

Suppose $G$ is not simple. Then nontrivial normal subgroups of $G$ exist. Let $K \leq G$ be a minimal nontrivial normal subgroup; by this previous exercise, $K$ is an elementary abelian $q$-group for some prime $q$. By our hypothesis, in fact $q = p$. In particular, $P = O_p(G)$ is nontrivial.

Let $N = C_G(P)$, and let $\pi$ denote the set of primes dividing $|N|$ except $p$. ($p$ indeed does divide $N$ since, as a $p$-group, $Z(P)$ is nontrivial.) Note that $Z(P) \leq C_G(P)$ is normal and that $C_G(P) \leq N_G(P) = G$ is normal. Letting $H$ be a Hall $\pi$-subgroup of $N$ and $Q$ a Sylow $p$-subgroup, we have $N = QH$ by Lagrange. Note that $Z(P) \leq Q$. Suppose this inclusion is strict; that is, there are elements in $Q$ which are not also in $Z(P)$. Then these elements are not in $O_p(G)$. However, note that if $xO_p(G) \in G/O_p(G)$. (@@@)

We now consider $Z(P) H$. Since $Z(P) \leq P$, $H \leq C_G(Z(P))$. Now if $zh \in Z(P) H$, note that $(zh)H(zh)^{-1} = zhHh^{-1}z^{-1}$ $= zHz^{-1} = Hzz^{-1} = H$. Thus $H \leq Z(P) H$ is normal; since $H \leq Z(P) H \leq N$ is a Hall subgroup, it is in fact the unique subgroup of $Z(P) H$ of order $|H|$. Thus $H$ is characteristic in $Z(P) H$.

Recall that $C_G(P) \leq N_G(P) = G$ is normal, so that $H$ is normal in $G$. Since $|H|$ is prime to $p$ by construction, $H = 1$. Thus $C_G(P) = Z(P)$.