Let \(O_p(G)\) be the intersection of all Sylow \(p\)-subgroups of the finite group \(G\). Show that \(O_p(G)\trianglelefteq G\), that \(O_p(G)\) is a \(p\)-group, and that it contains every normal \(p\)-subgroup of \(G\).
Solution
Let \[ O_p(G)=\bigcap_{P\in\operatorname{Syl}_p(G)}P. \]
1. The subgroup \(O_p(G)\) is normal in \(G\).
For every \(g\in G\), conjugation sends Sylow \(p\)-subgroups to Sylow \(p\)-subgroups and therefore permutes the set \(\operatorname{Syl}_p(G)\). Hence \[ gO_p(G)g^{-1} =\bigcap_{P\in\operatorname{Syl}_p(G)}gPg^{-1} =\bigcap_{Q\in\operatorname{Syl}_p(G)}Q =O_p(G). \]
2. The subgroup \(O_p(G)\) is a \(p\)-group.
It is contained in every Sylow \(p\)-subgroup, hence in particular in one \(p\)-group. Every subgroup of a finite \(p\)-group is a \(p\)-group.
3. Every normal \(p\)-subgroup \(N\trianglelefteq G\) is contained in every Sylow \(p\)-subgroup of \(G\).
Fix \(P\in\operatorname{Syl}_p(G)\). Since \(N\trianglelefteq G\), the product \(NP\) is a subgroup of \(G\). Both \(N\) and \(P\) are \(p\)-groups, so \[ |NP|=\frac{|N|\,|P|}{|N\cap P|} \] is a power of \(p\). Thus \(NP\) is a \(p\)-subgroup containing the Sylow subgroup \(P\). Maximality of \(P\) among \(p\)-subgroups gives \(NP=P\), hence \(N\le P\).
4. Therefore \(O_p(G)\) is the unique maximal normal \(p\)-subgroup of \(G\).
By 1 and 2, \(O_p(G)\) itself is a normal \(p\)-subgroup. By 3, every normal \(p\)-subgroup \(N\) satisfies \[ N\le\bigcap_{P\in\operatorname{Syl}_p(G)}P=O_p(G). \] Hence \(O_p(G)\) contains every normal \(p\)-subgroup, which is exactly the asserted maximality and uniqueness.