Nilpotent not implies ACIC

From Groupprops

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., nilpotent group) need not satisfy the second group property (i.e., ACIC-group)
View a complete list of group property non-implications | View a complete list of group property implications |Get help on looking up group property implications/non-implications
Get more facts about nilpotent group|Get more facts about ACIC-group

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a nilpotent group. That is, it states that in a nilpotent group, every subgroup satisfying the first subgroup property (i.e., automorph-conjugate subgroup) need not satisfy the second subgroup property (i.e., characteristic subgroup)
View all subgroup property non-implications in nilpotent groups $|$ View all subgroup property implications in nilpotent groups $|$ View all subgroup property non-implications $|$ View all subgroup property implications

Statement

A nilpotent group need not be ACIC: in a nilpotent group, every automorph-conjugate subgroup need not be characteristic.

Proof

Example of a non-Abelian group of prime-cubed order

Further information: prime-cube order group:p2byp

Let $p$ be an odd prime. Consider the group $P$ of order $p 3$ obtained as a semidirect product of a cyclic group of order $p 2$ with a cyclic group of order $p$ . In other words, $P$ is given by:

$\langle a,b \mid a^{p^2} = b^p = e, ab = ba^{p+1} \rangle$

Then, the cyclic group $C$ of order $p$ generated by $b$ is an automorph-conjugate subgroup that is not characteristic.

$C$ is clearly not characteristic because it is not normal, as it is the non-normal part in a semidirect product.

To see that it is automorph-conjugate, observe that any automorphism must send it to another cyclic group of order $p$ . Now, all cyclic groups of order $p$ lie inside the subgroup $Ω 1 (P)$ , which is elementary Abelian of order $p 2$ generated by $b$ and $a p$ . There are exactly $p + 1$ subgroups inside this of order $p$ , and one of them, the center (generated by $a p$ ) is characteristic. Of the other $p$ subgroups, none is normal in $P$ , and hence, since the size of any conjugacy class of subgroups is a multiple of $p$ , they are all conjugate. Thus, $C$ is an automorph-conjugate subgroup.

Example of non-Abelian groups of prime-fourth order

Fill this in later

Facts about Nilpotent not implies ACICRDF feed

Fact about	Nilpotent group +, ACIC-group +, Automorph-conjugate subgroup +, and Characteristic subgroup +
Particular example	Prime-cube order group:p2byp +

Nilpotent not implies ACIC

From Groupprops

Contents

Statement

Proof

Example of a non-Abelian group of prime-cubed order

Example of non-Abelian groups of prime-fourth order

Views

Personal tools

Search

Navigation

lookup

Credits

request/feedback

Toolbox