1. The problem statement, all variables and given/known data LET G BE A FINITE GROUP WHOSE ORDER IS DIVISIBLE BY THE PRIME P. SUPPOSE P^M IS THE HIGHEST POWER OF P WHICH IS A FACTOR OF |G|AND SET K=(|G|/P^M), THEN THE GROUP G CONTAINS AT LEAST ONE SUBGROUP OF |P^M|. I have the proof but can someone explain it in simpler terms? Maybe even a few examples please. 2. Relevant equations 3. The attempt at a solution Let X denote the collect of all subsets of G which have p^m elements and let G act on X by left translation so that the group element g is in G sends the subset A in X to gA. The size of X is the binomial coefficient kp^m choose p^m which is not divisible by p. Hence, there must be an orbit G(A) whose size is not a multiple of p. We have |G|=|G(A)|*|G(subA)|, consequently |G(subA)| is divisible by p^m. Now G(subA) is the stabilizer of A, so if a is in A and g is in G(subA), the ga is in A. This means that the whole right coset of G(subA)a is contained in A whenever a is in A and |G(subA)| cannot exceed p^m. Therefore, G(subA) is a subgroup of G which has order p^m.