Let \(O_p(G)\) be the intersection of all Sylow \(p\)-subgroups of \(G\). Show that \(O_p(G) \normal G\), and that it is the unique maximal normal \(p\)-subgroup of \(G\) (containing all other normal \(p\)-subgroups).
Solution
Let \[ O_p(G)=\bigcap_{P\in\Syl_p(G)}P. \]
1. The subgroup \(O_p(G)\) is normal in \(G\) and is a \(p\)-group.
Conjugation by any \(g\in G\) permutes the Sylow \(p\)-subgroups, so \[ gO_p(G)g^{-1} =\bigcap_{P\in\Syl_p(G)}gPg^{-1} =O_p(G). \] Thus \(O_p(G)\trianglelefteq G\). Since it is contained in every Sylow \(p\)-subgroup, it is itself a \(p\)-group.
2. Every normal \(p\)-subgroup of \(G\) is contained in \(O_p(G)\).
Let \(N\trianglelefteq G\) be a \(p\)-subgroup. Choose a Sylow \(p\)-subgroup \(P\) with \(N\le P\). For every \(g\in G\), normality gives \[ N=gNg^{-1}\le gPg^{-1}. \] Every Sylow \(p\)-subgroup is conjugate to \(P\), hence \(N\) lies in every Sylow \(p\)-subgroup. Therefore \[ N\le\bigcap_{Q\in\Syl_p(G)}Q=O_p(G). \]
Thus \(O_p(G)\) is the unique largest normal \(p\)-subgroup of \(G\).