P-group not implies nilpotent

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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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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.

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Opposite facts


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.