Normal equals 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 potentially characteristic subgroup of $G$ in the following sense: there exists a group $K$ containing $G$ such that $H$ is a characteristic subgroup of $K$.

Related facts

Stronger facts

• NRPC theorem

Other related facts

• 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
2. Characteristic implies normal
3. Normality satisfies intermediate subgroup condition

Proof

Proof of (1) implies (2) (hard direction)

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

To prove: There exists a group $K$ containing $G$ 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).

Proof of (2) implies (1) (easy direction)

Given: A group $G$, a subgroup $H$ of $G$, a group $K$ containing $G$ such that $H$ is characteristic in $K$.

To prove: $H$ is normal in $G$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$H$ is normal in $K$.	Fact (2)	$H$ is characteristic in $K$	--	Given-fact-combination direct.
2	$H$ is normal in $G$.	Fact (3)	$H \le G \le K$	Step (1)	Given-step-fact combination direct.