What exactly does the definition of a nilpotent group mean?

Question

I'm studying nilpotent and solvable group and find it pretty hard to tell what the definition of a nilpotent group is after.

For example, a group is solvable iff it has a solvable series (that is, a subnormal series such that each factor is abelian). This equivalent definition tells something clearly about the structure of the group for me.

Then what about a nilpotent group? Since it's a condition stronger than solvable, in which part does it strengthen the equivalent defination above? Is there a true proposition like "a group is nilpotent iff it has a subnormal series such that each factor is abelian and something else" ?

Answer 1

There are several equivalent definitions of nilpotent groups. The one most similar to the definition of solvable groups given in the OP is this.

A group is nilpotent iff there exists a normal series $$1=Z_0<Z_1<Z_2... <Z_n=G$$ such that $Z_i/Z_{i-1}$ is central in $G/Z_{i-1}$ for every $i=1,..., n$.

In particular the series is subnomal and all factors are abelian (because the center of any group is abelian), so nilpotent groups are solvable.

Answer 2

A nilpotent group is one of the concepts that is most difficult to grasp, particularly for infinite groups. If $G$ is a finite nilpotent group then it is just a direct product of $p$-groups, and that's normally enough to satisfy yourself.

A soluble group $G$ of length $n$ is one where you take the commutator subgroup $G'$ and this has length $n-1$. That, together with the fact that the trivial group is soluble of length $0$, is enough to understand the class. In particular, if $N$ is a normal subgroup and both $N$ and $G/N$ are soluble, then $G$ is soluble.

Nilpotent is similar, but you need that the commutator for the normal subgroup is 'compatible' with the whole group. So instead of checking that $G'$ is soluble, i.e., $[G',G']<G'$ and so on, you want that the commutator works with one of the $G'$ replaced by the whole of $G$. So $H=[G',G]<G'$, and then $[H,G]<H$ and so on until you hit the trivial group.

So if we make the map $\mathrm{ad}_x:G\to G$ given by $y\mapsto [x,y]$ then this map is nilpotent, i.e., some power is the 'zero' map (i.e., sends every element to the identity). This is not true for soluble groups, e.g., $G=S_3$ with $x=(1,2)$. If you know ring theory, the analogue would be the difference between a subring and an ideal, where we want a compatibility between the multiplication on the whole of the ring, not just the subset, to move from a subring to an ideal.

Whereas $G$ is soluble if and only if both $G/N$ and $N$ are soluble, the same statement cannot hold for nilpotent groups because it gives no information that connects the commutator map for $N$ to that of $G$. A group $G$ is nilpotent if and only if both $G/N'$ and $N$ are nilpotent. That drop from $N$ to $N'$ gives us enough information to connect the two commutator structures.