which gives us because but since we have thus now since and we have in either case
similarly so we've proved that G has exactly pq subgroups of order r, at least r subgroups of order q and at least q subgroups of order p. since every two distinct cyclic subgroups of the
same prime order have trivial intersection and also every two cyclic subgroups of distinct prime orders have trivial intersection, a simple counting of elements of G shows that we must have:
which is the contradiction we needed! Q.E.D.