Normal equals retract-potentially characteristic

Statement
The following are equivalent for a subgroup $$H$$ of a group $$G$$ :


 * 1) $$H$$ is a normal subgroup of $$G$$.
 * 2) $$H$$ is a retract-potentially characteristic subgroup of $$G$$ in the following sense: there exists a group $$K$$ containing $$G$$ as a retract such that $$H$$ is a characteristic subgroup of $$K$$.

Related facts

 * NPC theorem
 * Finite NPC theorem
 * Finite NIPC theorem
 * Fact about amalgam-characteristic subgroups: finite normal implies amalgam-characteristic, periodic normal implies amalgam-characteristic, central implies amalgam-characteristic

Facts used

 * 1) uses::Characteristicity is centralizer-closed

Proof
Given: A group $$G$$, a normal subgroup $$H$$ of $$G$$.

To prove: There exists a group $$K$$ containing $$G$$ as a retract such that $$H$$ is characteristic in $$K$$.

Proof:


 * 1) Let $$S$$ be a simple non-abelian group that is not isomorphic to any subgroup of $$G$$: Note that such a group exists. For instance, we can take the finitary alternating group on any set of cardinality strictly bigger than that of $$G$$.
 * 2) Let $$K$$ be the restricted wreath product of $$S$$ and $$G$$, where $$G$$ acts via the regular action of $$G/H$$ and let $$V$$ be the restricted direct power $$S^{G/H}$$. In other words, $$K$$ is the semidirect product of the restricted direct power $$V = S^{G/H}$$ and $$G$$, acting via the regular group action of $$G/H$$.
 * 3) Any homomorphism from $$V$$ to $$G$$ is trivial: By definition, $$V$$ is a restricted direct product of copies of $$S$$. Since $$S$$ is simple and not isomorphic to any subgroup of $$G$$, any homomorphism from $$S$$ to $$G$$ is trivial. Thus, for any homomorphism from $$V$$ to $$G$$ is trivial.
 * 4) $$V$$ is characteristic in $$K$$: Under any automorphism of $$K$$, the image of $$V$$ is a homomorphic image of $$V$$ in $$K$$. Its projection to $$K/V \cong G$$ is a homomorphic image of $$V$$ in $$G$$, which is trivial, so the image of $$V$$ in $$K$$ must be in $$V$$.
 * 5) The centralizer of $$V$$ in $$K$$ equals $$H$$: By definition, $$H$$ centralizes $$V$$. Using the fact that $$S$$ is centerless and that inner automorphisms of $$S$$ cannot be equal to conjugation by elements in $$G \setminus H$$, we can show that it is precisely the center.
 * 6) $$H$$ is characteristic in $$K$$: This follows from the previous two steps and fact (1).