Let \(G\) be a finite group and \(H \trianglelefteq G\) a normal subgroup of order \(p^k\). Show that \(H\) is contained in every Sylow \(p\)-subgroup of \(G\).
Solution
Let \(P\) be an arbitrary Sylow \(p\)-subgroup of \(G\).
1. There exists \(g\in G\) such that \(H\le gPg^{-1}\).
The subgroup \(H\) is a \(p\)-subgroup. By Sylow’s containment theorem, every \(p\)-subgroup of \(G\) is contained in a conjugate of a Sylow \(p\)-subgroup. Hence \(H\le gPg^{-1}\) for some \(g\in G\).
2. One has \(H\le P\).
Conjugating the inclusion in 1 by \(g^{-1}\) gives \[ g^{-1}Hg\le P. \] Since \(H\trianglelefteq G\), normality gives \(g^{-1}Hg=H\). Therefore \(H\le P\).
3. Thus \(H\) is contained in every Sylow \(p\)-subgroup of \(G\).
The Sylow subgroup \(P\) was arbitrary.