Normal equals retract-potentially characteristic

This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup
View a complete list of pages giving proofs of equivalence of definitions

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. 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).