Characteristic-isomorph-free not implies normal-isomorph-free in finite
This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., characteristic-isomorph-free subgroup) need not satisfy the second subgroup property (i.e., normal-isomorph-free subgroup)
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We may have a group with a characteristic subgroup such that there are no other characteristic subgroups isomorphic to it, but there are other normal subgroups isomorphic to it.
Example of an Abelian group
Let be any prime, and let . In other words, is a direct product of cyclic groups of order and respectively.
Consider the subgroup : the set of all elements of that can be written as powers. This set is .
- is a characteristic subgroup: It is defined as the set of powers, so it is clearly invariant under conjugation.
- There is no other characteristic subgroup of of order : All the subgroups of order in lie inside . Further, automorphisms of the form permute all the other subgroups.
- There are other normal subgroups of of order : In fact, there are of them: the other subgroups of .
Thus, is characteristic-isomorph-free in but is not normal-isomorph-free in .