Characteristic not implies potentially fully invariant

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., potentially fully invariant subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about characteristic subgroup|Get more facts about potentially fully invariant subgroup

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

Statement

It is possible to have a subgroup $H$ of a group $G$ such that $H$ is a characteristic subgroup of $G$ but is not a potentially fully invariant subgroup of $G$, i.e., there is no group $K$ containing $G$ such that $H$ is a fully invariant subgroup of $K$.

Facts used

1. Normal not implies potentially fully invariant: There exists an example of a group $J$ and a normal subgroup $H$ of $J$ such that $H$ is not fully invariant in any group $K$ containing $J$.
2. NPC theorem: This states that if $H$ is a normal subgroup of $J$, there exists a group $G$ containing $J$ such that $H$ is a characteristic subgroup of $G$.

Proof

Let $J$ and $H$ be a group-subgroup pair as given by fact (1). By fact (2), there exists a group $G$ containing $J$ such that $H$ is a characteristic subgroup of $G$.

We want to show that there is no group $K$ containing $G$ such that $H$ is fully invariant in $K$. The reason for this is: if such a group $K$ exists, it would also contain $J$, so we'd have a group containing $J$ in which $H$ is fully invariant. This contradicts the choice of $H$ and $J$ as being examples for fact (1).