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 be an odd prime. Consider the group of order obtained as a semidirect product of a cyclic group of order with a cyclic group of order . In other words, is given by:
Then, the cyclic group of order generated by is an automorph-conjugate subgroup that is not characteristic.
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 . Now, all cyclic groups of order lie inside the subgroup , which is elementary Abelian of order generated by and . There are exactly subgroups inside this of order , and one of them, the center (generated by ) is characteristic. Of the other subgroups, none is normal in , and hence, since the size of any conjugacy class of subgroups is a multiple of , they are all conjugate. Thus, is an automorph-conjugate subgroup.
Example of non-Abelian groups of prime-fourth order
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