Show that every maximal subgroup \(M\) of a finite \(p\)-group \(G\) is normal and has index \(p\).
Solution
We use the normalizer condition: if \(H<G\) is a proper subgroup of a finite \(p\)-group, then \[ H<N_G(H). \] Indeed, let \(H\) act on the left cosets \(G/H\) by left multiplication. Since \([G:H]\) is a positive power of \(p\), the number of fixed points is divisible by \(p\). A coset \(gH\) is fixed exactly when \(g^{-1}Hg=H\), so the fixed-point set is \(N_G(H)/H\). It contains \(H\), hence has at least one element; divisibility by \(p\) therefore gives \[ [N_G(H):H]\ge p, \] so \(H<N_G(H)\).
Apply this to a maximal subgroup \(M<G\). Then \[ M<N_G(M)\le G. \] Maximality forces \(N_G(M)=G\), hence \(M\trianglelefteq G\).
Now \(G/M\) is a nontrivial finite \(p\)-group. Since \(M\) is maximal, \(G/M\) has no nontrivial proper subgroup. By Cauchy’s theorem, a nontrivial finite \(p\)-group has an element of order \(p\); therefore \(G/M\) itself must have order \(p\). Hence \[ [G:M]=p. \]