# Equivalence of definitions of periodic nilpotent group

This article gives a proof/explanation of the equivalence of multiple definitions for the term periodic nilpotent group
View a complete list of pages giving proofs of equivalence of definitions

## Statement

The following are equivalent for a group $G$:

1. It is both locally finite (every finitely generated subgroup is finite) and nilpotent.
2. It is both periodic (every element has finite order) and nilpotent.
3. It is a nilpotent group and also a group generated by periodic elements: it has a generating set where all the elements have finite orders.
4. Its abelianization is a periodic abelian group.
5. It is a nilpotent group and all the quotient groups between successive members of its lower central series are periodic abelian groups.
6. It is a restricted direct product of nilpotent p-groups, with a common bound on their nilpotency class.

### Variant for a set of primes

If $\pi$ is a set of primes, the following are equivalent for a group $G$:

1. It is nilpotent and every finitely generated subgroup of $G$ is a finite $\pi$-group (i.e., all the prime factors of its order are in $\pi$.
2. It is nilpotent and every element has finite order and the order is a $\pi$-number.
3. It is nilpotent and generated by $\pi$-elements: elements whose orders are all $\pi$-numbers.
4. It is nilpotent and its abelianization is a $\pi$-group.
5. It is nilpotent and all the quotient groups between successive members of its lower central series are abelian $\pi$-groups.
6. It is a restricted direct product of nilpotent p-groups, $p \in \pi$, with a common bound on their nilpotency class.

## Facts used

1. All successive quotients between lower central series members of a group are homomomorphic images of tensor powers of the abelianization.
2. Equivalence of definitions of periodic abelian group: This states in particular that for an abelian group, being periodic is equivalent to being locally finite.
3. Local finiteness is extension-closed

## Proof

### (1) implies (2)

This is obvious from the general fact that locally finite implies periodic. We do not use that $G$ is nilpotent for this implication.

### (2) implies (3)

This is obvious from the general fact that any periodic group is a group generated by periodic elements -- in fact, we could take the generating set as the set of all elements. Note that this step again does not use anything about $G$ being nilpotent.

### (3) implies (4)

Suppose $S$ is a generating set for $G$ that comprises only elements of finite order. Under the quotient map from $G$ to its abelianization, the image of $S$ is a generating set for the abelianization of $G$. Since the abelianization is an abelian group, being generated by a set of elements all of finite order is equivalent to the whole group being a periodic group, and hence a periodic abelian group.

### (4) implies (5)

Denote by $\gamma_i(G)$ the $i^{th}$ member of the lower central series of $G$. Each of the successive quotients $\gamma_i(G)/\gamma_{i+1}(G)$ is a homomorphic image of a tensor power of the abelianization of $G$ by Fact (1). Since the abelianization of $G$ is periodic, each $\gamma_i(G)/\gamma_{i+1}(G)$ is periodic.

### (5) implies (1)

We use Fact (2) to note that all the quotients between successive members of the lower central series are in fact locally finite. Fact (3) now completes the proof.