# Characteristic not implies characteristic-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 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.

## 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 $(81,3)$).