Nilpotent not implies ACIC

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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., 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.

Example of non-Abelian groups of prime-fourth order

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