- Show that every maximal subgroup of a \(p\dash\)group has index \(p\).
Solution
Let \(G\) be a finite \(p\)-group and let \(M<G\) be maximal.
First, every proper subgroup of a finite \(p\)-group is properly contained in its normalizer. Indeed, let \(M\) act by left multiplication on \(G/M\). The number of fixed cosets is congruent to \[ [G:M]\equiv0\pmod p. \] The fixed cosets are exactly the cosets \(gM\) with \(g\in N_G(M)\), so their number is \[ [N_G(M):M]. \] Since the coset \(M\) is fixed, this positive number is divisible by \(p\), hence is at least \(p\). Thus \[ M<N_G(M). \]
By maximality, \(N_G(M)=G\), so \(M\trianglelefteq G\). Then \(G/M\) is a nontrivial finite \(p\)-group with no proper nontrivial subgroup. Hence it has prime order, necessarily \(p\). Therefore \[ \boxed{[G:M]=p}. \]