# What exactly does the definition of a nilpotent group mean?

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" ?

– MPW
Jul 18 '20 at 4:25

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., $$[G',G'] and so on, you want that the commutator works with one of the $$G'$$ replaced by the whole of $$G$$. So $$H=[G',G], and then $$[H,G] 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.

• It is not correct. Nilpotent infinite groups is one of the easiest classes of infinite groups. Jul 18 '20 at 11:38
• The structure of such groups being very tame is not the same as the definition being easy to understand. The definition of sofic group is not nearly as easy as conjecturally their structure is, being as they might be all groups. And since nilpotent groups contain all infinite abelian groups, which are a mess, I do not agree. You might be thinking of finitely generated nilpotent groups. Jul 18 '20 at 11:51
• The definitions are trivial to understand (as well as the definition of sofic groups). Of course it depends on who teaches it. Jul 18 '20 at 12:02
• The definition is trivial to recite. It is not trivial to understand, certainly not for undergraduates. Jul 18 '20 at 12:03
• Quite understandably (at least coming from a Lie-algebra perspective), you use language similar to that of linear maps, but I think that it may be confusing to refer to a map of groups as the $0$ map. Perhaps it is worth mentioning explicitly that (a) $\operatorname{ad}_x$ is usually not a homomorphism, and (b) saying that a power $\operatorname{ad}_x^n$ is the $0$ map means that it is the constant map at the identity of $G$? Jul 18 '20 at 15:28

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 $$1=Z_0 such that $$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.