# Nilpotency of fixed class is subgroup-closed

This article gives the statement, and possibly proof, of a group property (i.e., nilpotent group) satisfying a group metaproperty (i.e., subgroup-closed group property)
View all group metaproperty satisfactions | View all group metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for group properties
Get more facts about nilpotent group |Get facts that use property satisfaction of nilpotent group | Get facts that use property satisfaction of nilpotent group|Get more facts about subgroup-closed group property

## Statement

Suppose $G$ is a nilpotent group and $H$ is a subgroup of $G$. Then, $H$ is also a nilpotent group and its nilpotency class is at most equal to the nilpotency class of $G$.

Note that we say that a group is nilpotent "of class $c$" if its nilpotency class is at most $c$. The statement can thus be reformulated as saying that the property of being nilpotent of class $c$ is closed under taking subgroups.

## Proof

### Proof using the upper central series definition

The idea here is to show that the upper central series ascends at least as fast as the series obtained by intersecting $H$ with the upper central series of $G$.

### Proof using the lower central series definition

The idea here is to show that the lower central series descends at least as fast as the series obtained by intersecting $H$ with the lower central series of $G$.

### Proof using the central series definition

The idea here is to show that intersecting with $H$ every member of any central series of $G$ gives a central series of $H$.

### Proof in terms of iterated commutators being trivial

The proof in terms of this definition is tautological.