# Nilpotent not implies ACIC

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)
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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)
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## 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 $\Omega_1(P)$, 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.