Group Chains, Classifying Finite Nilpotent Groups

Classifying Finite Nilpotent Groups

This is a rather surprising theorem. A finite group G is nilpotent iff all its sylow subgroups are normal, iff G is the direct product of its sylow subgroups.

Let G be nilpotent and let H be a proper subgroup of G. We will show that the normalizer of H is larger than H.

Remember that G has a normal series, namely the preimages of its central series. Write this series in ascending order, from e up to G.

Fix i so that Ni is the last subgroup wholly contained in H. Since H is a proper subgroup of G, there is another subroup in the normal series, namely Ni+1. And Ni+1 contains elements outside of H.

Let y be an element in Ni+1-H, and let x represent a coset of Ni in H. By construction, cosets of Ni lie in the center of G/Ni when they are also contained in Ni+1. Since y represents such a coset, we can write yxNi = xyNi. Since Ni is normal, yNi is the same as Niy. Thus yxNi = xNiy. This holds for every x in H, hence y is an element that normalizes H, although y is not in H. The normalizer is properly larger than H.

Now let G be a finite nilpotent group and let J be a sylow subgroup. If J = G we are done. So let J be a proper subgroup of G. Let H be the normalizer of J in G. Recall that the normalizer of H is the same as H, which is a contradiction, unless H = G. Therefore every sylow subgroup is normal in G.

Next assume sylow subgroups are normal and let x be an arbitrary element of G. If x belongs to two separate sylow subgroups, its order is a power of a prime p, and a power of some other prime q, which is impossible. The various sylow subgroups are pairwise disjoint.

Let H and K be two sylow subgroups and let D be the group generated by H and K. We know that D is the direct product of H and K. This remains normal in G, since conjugation can be applied per component, and H and K are normal in G.

If J is a third sylow subgroup, it is disjoint from D, since everything in D has order divisible by p and q, and everything in J has order divisible by some other prime r. Again, D and J are both normal in G, and they generate a normal subgroup that is isomorphic to their direct product. Thus H J and K span a subgroup that is isomorphic to their 3-way direct product. This continues until G becomes the direct product of all its sylow subgroups.

Finally, assume G is the direct product of its sylow subgroups. every finite p group is nilpotent, and the finite direct product of nilpotent groups is nilpotent, hence G is nilpotent.

Applications

For every m dividing |G|, there is a subgroup of order m. If a sylow subgroup is based on the prime p, there is a subgroup of order pk, where pk divides m. This holds for all primes, and their direct product produces the desired subgroup of order m.

Any subgroup or quotient group of a finite nilpotent group is nilpotent. This because the subgroup or quotient group of G is still a direct product of sylow subgroups.