Group Chains, Central Series and Nilpotent Groups

Central Series and Nilpotent Groups

We've talked about the normal series of a group - now it's time for the central series.

Let C₀ = G, and let C₁ = G mod the center of G. If G is abelian we are done, but if not, continue in this manner. Let C₂ = C₁ mod the center of C₁, and so on. Thus the central series is actually a series of factor groups.

If the central series is finite, G is nilpotent. An abelian group is obviously nilpotent. Since a p group has a nontrivial center, every finite p group is nilpotent.

Direct Product

The finite direct product of nilpotent groups is nilpotent. The composite central series is the direct product of the individual central series. An infinite direct product may be nilpotent as well, if we can bound the lengths of the individual central series.

Quotient is Nilpotent

The quotient of a nilpotent group is nilpotent. Apply a homomorphism f(G) = H. By induction, the theorem holds for G/C, where C is the center. Let K be the kernel of f. The cosets of K that have representatives in C become elements in the center of H. In other words, C maps into D, where D is the center of H. Let J be the join of C and K, whence G/J maps onto H/D. Since G/C maps onto G/J, G/C maps onto H/D. Therefore, H mod its center is the homomorphic image of G mod its center, and is nilpotent by induction. This makes H nilpotent.

Subgroup is Nilpotent

If H is a subgroup of G, H is nilpotent. By induction, assume every subgroup of G/C (C is the center) is nilpotent. Let H intersect C in D. Each coset of C in G is now cut down to a coset of D. If there were two such cosets in one coset of C, that would imply two cosets of D in C, which is impossible. Furthermore, the cosets of D obey the rules of G/C, just like the cosets of C. In other words, we have a representation of G/C, using cosets of D. However, some of these cosets may be missing, i.e. the cosets that are not in H. The quotient H/D is a subgroup of G/C, and is nilpotent by induction. Now, the center of H could be larger than D, creating a quotient of H/D, which is also nilpotent. Since H mod its center is nilpotent, H is nilpotent.

Nilpotent is Solvable

Given a central series for G, build a corresponding normal series, in reverse order, as follows.

A trivial homomorphism maps G onto G, with kernel e. Thus C₀ = G, and N₀ = e.

Moving to C₁, a homomorphism maps G onto C₁, and its kernel is the center of G. Let N₁ be this kernel, the center of G.

A second homomorphism maps C₁ onto C₂. Combine this with the first homomorphism to get a map from G onto C₂. Let N₂ be the kernel of this map. Continue all the way to C_k = e, whence N_k = G.

Since each N_i is the kernel of a homomorphism, each N_i is normal in G. conjugate N_i by the elements in N_i+1, which all belong to G, and the result is always N_i. Thus N_i is normal in N_i+1, and we have a normal series for G, running in reverse order.

The factor group N_i+1/N_i is the center of C_i, which is abelian. All factor groups are abelian, the series is solvable, and the group is solvable. Every nilpotent group is solvable.

However, there are plenty of solvable groups that are not nilpotent. S₃ has no center, and no central series, even though it has an abelian subgroup Z₃ with an abelian factor group Z₂.