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

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

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

Let $G$ be the wreath product of the cyclic 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 $F_{p} [H]$ by $H$ acting via left multiplication. We claim the following:

$G$ 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.

$G$ is a metabelian group: The solvable length of $G$ is two. In fact, the additive group $F_{p} [H]$ is an Abelian normal subgroup of $G$ with Abelian quotient.
$G$ is centerless: This is clear by inspection.

Tarski's examples

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