P-group not implies nilpotent: Difference between revisions

Revision as of 21:59, 22 January 2012

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

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 derived 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

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.

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 # <math>G</math> is a <math>p</math>-group: <math>G</math> is a semidirect product of two <math>p</math>-groups. In particular, it is the extension of one <math>p</math>-group (the additive group of the group ring) by another (the multiplicative group <math>H</math> living as a subgroup of the group of units of the group ring); hence it is a <math>p</math>-group.
-* <math>G</math> is a [[metabelian group]]: The solvable length of <math>G</math> is two. In fact, the additive group <math>\mathbb{F}_p[H]</math> is an Abelian normal subgroup of <math>G</math> with Abelian quotient.
+* <math>G</math> is a [[metabelian group]]: The derived length of <math>G</math> is two. In fact, the additive group <math>\mathbb{F}_p[H]</math> is an abelian normal subgroup of <math>G</math> with Abelian quotient.
 * <math>G</math> is [[centerless group|centerless]]: This is clear by inspection.