Classifying Finite Nilpotent Groups

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.