Characteristic not implies characteristic-isomorph-free in finite

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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 subgroup) need not satisfy the second subgroup property (i.e., characteristic-isomorph-free subgroup)
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Statement

We can find a group G (in fact, we can choose G to be a finite group) and characteristic subgroups H,K of G that are not isomorphic to each other but distinct.

Related facts

Proof

Example of the infinite cyclic group

Let \mathbb{Z} be the infinite cyclic group: the group of integers under addition. Then, n\mathbb{Z} is a characteristic subgroup of \mathbb{Z} for any n, and all the n\mathbb{Z} are isomorphic for n a nonnegative integer. Thus, there are distinct characteristic subgroups that are isomorphic.

Example of a finite solvable group

Further information: symmetric group:S3

Let A be a nontrivial metabelian group that is also a centerless group, with derived subgroup B. Let C be a group isomorphic to B. Then, in the direct product G := A \times C, we have:

  • The subgroup H = \{ e \} \times C equals the center of A \times C, hence is characteristic.
  • The subgroup K = A  \times \{ e \} equals the derived subgroup of A \times C, hence is characteristic.

Thus, we have two distinct characteristic subgroups that are isomorphic.

A particular case of this is where A is the symmetric group on three letters. In this case, B is A3 in S3 and C is cyclic of order three.

Example of a finite nilpotent group

There are groups of order p^5 with characteristic maximal subgroups that are isomorphic. For instance, the group with GAP Group ID (243,5) has three pairwise isomorphic characteristic subgroups (each with group ID (81,3)).