Sylow Maximal Normal 89D276

1. **Problem Statement:** Prove that if $P$ is a Sylow $p$-subgroup of a finite group $G$, then $P$ is a maximal normal $p$-subgroup of its normalizer $N_G(P)$. 2. **Recall definitions and facts:** - A Sylow $p$-subgroup $P$ of $G$ is a maximal $p$-subgroup of $G$, i.e., a subgroup of order $p^n$ where $p^n$ divides $|G|$ and $p^{n+1}$ does not. - The normalizer $N_G(P) = \{g \in G : gPg^{-1} = P\}$ is the largest subgroup of $G$ in which $P$ is normal. - A $p$-subgroup is a subgroup whose order is a power of $p$. 3. **Goal:** Show that $P$ is a maximal normal $p$-subgroup of $N_G(P)$, meaning if $Q \triangleleft N_G(P)$ is a normal $p$-subgroup with $P \subseteq Q$, then $Q = P$. 4. **Proof:** - Since $P$ is a Sylow $p$-subgroup of $G$, it is maximal among $p$-subgroups of $G$. - Consider any normal $p$-subgroup $Q$ of $N_G(P)$ such that $P \subseteq Q$. - Because $Q$ is a $p$-subgroup of $N_G(P)$, and $N_G(P) \leq G$, $Q$ is also a $p$-subgroup of $G$. - Since $P$ is a Sylow $p$-subgroup of $G$, it is maximal among $p$-subgroups of $G$, so $Q = P$ or $Q$ is not larger than $P$. - But $P \subseteq Q$ implies $Q = P$. - Hence, $P$ is maximal among normal $p$-subgroups of $N_G(P)$. 5. **Summary:** - $P$ is normal in $N_G(P)$ by definition of normalizer. - Any normal $p$-subgroup of $N_G(P)$ containing $P$ must be equal to $P$ because $P$ is maximal as a $p$-subgroup in $G$. Therefore, every Sylow $p$-subgroup of $G$ is a maximal normal $p$-subgroup of its normalizer in $G$.