Normal-potentially characteristic not implies characteristic

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., normal-potentially characteristic subgroup) need not satisfy the second subgroup property (i.e., characteristic subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal-potentially characteristic subgroup|Get more facts about characteristic subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal-potentially characteristic subgroup but not characteristic subgroup|View examples of subgroups satisfying property normal-potentially characteristic subgroup and characteristic subgroup

Statement

It is possible to have a group $G$ and a subgroup $H$ such that both the following hold:

• $H$ is not a characteristic subgroup of $G$.
• There exists a group $K$ containing $G$ as a normal subgroup and $H$ as a characteristic subgroup.

Facts used

1. subgroup of finite abelian group implies abelian-potentially characteristic

Proof

By fact (1), any subgroup of a finite abelian group can be realized as a characteristic subgroup in some bigger abelian group. Thus, if $G$ is a finite abelian group and $H$ is a non-characteristic subgroup of $G$, there exists some abelian group $K$ containing $G$ in which $H$ is characteristic. Note that $G$ is normal in $K$, because $K$ is abelian.

For instance, if we take $G$ to be the Klein four-group, and $H$ as one of the subgroups of order two, then we can take $K$ as a direct product of Z4 and Z2, containing $G$ as one of its cyclic subgroups.