- Show that every maximal subgroup of a \(p\)-group is normal.
Solution
Let \(G\) be a finite \(p\)-group and let \(M<G\) be maximal.
1. One has \(M<N_G(M)\).
By the normalizer condition for finite \(p\)-groups, every proper subgroup is properly contained in its normalizer. Since \(M<G\), \[ M<N_G(M). \]
2. Hence \(N_G(M)=G\).
We have \[ M<N_G(M)\le G. \] Because \(M\) is maximal, there is no subgroup strictly between \(M\) and \(G\). Therefore \(N_G(M)=G\).
3. Therefore \(M\trianglelefteq G\).
The equality \(N_G(M)=G\) says precisely that every element of \(G\) normalizes \(M\). Hence \(M\) is normal.