P-group not implies nilpotent

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., p-group) need not satisfy the second group property (i.e., nilpotent group)
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Statement

A p-group (i.e., a possibly infinite group in which the order of every element is the power of a fixed prime $p$) need not be nilpotent.

Related facts

Similar facts

• p-group not implies solvable: This is true, at least for large enough primes, where we can take the Tarski groups.

Opposite facts

• Prime power order implies nilpotent: Any finite $p$-group (which is the same as a group of prime power order) must be nilpotent. For full proof, refer: Prime power order implies nilpotent
• Locally finite Artinian p-group implies hypercentral

Proof

McLain's example

Further information: restricted regular wreath product of group of prime order and quasicyclic group

For the given prime $p$, let $H$ be the quasicyclic group for $p$; concretely, $H$ is the group of $(p^{n})^{th}$ roots of unity in $\mathbb {C}$ for all nonnegative integers $n$. Clearly, $H$ is a $p$-group.

Let $G$ be the restricted regular wreath product of the group of prime order with $H$. In other words, $G$ is the restricted external wreath product of the group of prime order with $H$ having the left regular action. Equivalently, $G$ is the semidirect product of the additive group of the group ring $\mathbb {F} _{p}[H]$ by $H$ acting via left multiplication. We claim the following:

1. $G$ is a $p$-group: $G$ is a semidirect product of two $p$-groups. In particular, it is the extension of one $p$-group (the additive group of the group ring) by another (the multiplicative group $H$ living as a subgroup of the group of units of the group ring); hence it is a $p$-group.
2. $G$ is a metabelian group: The derived length of $G$ is two. In fact, the additive group $\mathbb {F} _{p}[H]$ is an abelian normal subgroup of $G$ with Abelian quotient.
3. $G$ is centerless: This is clear by inspection.

Tarski's examples

For any prime $p$ for which a Tarski group exists, the Tarski group is an example of a p-group that is not nilpotent. In fact, it is not even solvable.

Tarski groups do not exist for all primes.