Let be a solvable group of order , where is a prime not dividing . Let . If , prove that has a normal subgroup of order . Where is solvability required in the proof? [Note: This result holds for nonsolvable finite groups as well, as a special case of Burnside’s N/C-Theorem.]
By hypothesis, we have , and the Sylow -subgroups of intersect trivially. Thus has elements of order . Denote the remaining elements by .
Let denote the set of all prime divisors of except . By this previous exercise, there is a Hall -subgroup , which (by definition) has order , and none of whose elements have order . Thus . Moreover, if is any other subgroup of order , then is a Hall -subgroup (again by definition) and is conjugate to . But , so that in fact is the unique subgroup of of order . Thus is characteristic, hence normal, in .
Solvability was needed in order to guarantee the existence and conjugateness of Hall subgroups.