Let pi be a set of prime numbers. A group H is a pi-group if all prime divisors of |H |lie in pi. The trivial group is a pi-group, vacuously.
Let G be a group. The set of normal pi-subgroups of G is nonempty, since e is a pi-group, and e is normal. Since G is finite, there exists at least one maximal normal pi-subgroup of G. Suppose N and M are both maximal normal pi-subgroups of G. Claim NM is a normal subgroup of G. It is a subgroup because N (or M) is normal. Moreover, for any g in G, (NM)g=NgMg=NM, since conjugation by g is an automorphism of G. Thus NM is a normal subgroup of G. By 2.8 (HW 3.1), |NM|=|N||M|/|N ^ M| (where ^ is html for intersection), so NM is also a pi-group. Since N is a subset of NM and N is a maximal normal pi-subgroup of G, we conclude NM=N, which implies M is a subset of N. Then, by maximality of M, M=N. Thus there is a unique maximal normal pi-subgroup of G.