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:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
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 $G/H$.			Step (1)
3	Any homomorphism from $V$ to $G$ is trivial.			Steps (1), (2)	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, any homomorphism from $V$ to $G$ is trivial.
4	$V$ is characteristic in $K$.			Steps (2), (3)	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 by step (3), so the image of $V$ in $K$ must be in $V$.
5	The centralizer of $V$ in $VH$ equals $H$.			Steps (1), (2)	By definition of the wreath product action, $H$ centralizes $V$. Since $S$ is centerless, $V$ is also centerless. Thus, $C_{VH}(V)$ contains $H$ but has trivial intersection with $V$, forcing $C_{VH} (V) = H$.
6	The centralizer of $V$ in $K$ equals $H$.			Steps (2), (5)	Step (5) already shows that $C_{VH}(V) = H$, so it suffices to show that $C_K(V) \le VH$. To see this, note that, by the construction in step (2), any element of $K$ outside $VH$ permutes the direct factors of $V$ as an element of $G$ outside $H$. This action is nontrivial, so the action is nontrivial, and hence elements outside $VH$ cannot centralize $V$. This forces $C_K(V) \le VH$, completing the proof.
7	$H$ is characteristic in $K$.	Fact (1)		Steps (4), (6)	Step-fact combination direct.

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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.