P-group not implies nilpotent

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.

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.
 * Locally finite Artinian p-group implies hypercentral

McLain's example
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.