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 $G$ :

$H$ is a normal subgroup of $G$ .
$H$ is a retract-potentially characteristic subgroup of $G$ in the following sense: there exists a group $K$ containing $G$ as a retract (i.e., $G$ has a normal complement in $K$ ) such that $H$ is a characteristic subgroup of $K$ .

Related facts

Facts used

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 $G$ as a retract 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 at least 5 and also 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 the quotient group $G / H$ (in particular, the restriction of the action to $H$ is trivial).		$H$ is normal in $G$	Step (1)	Normality of $H$ is what allows us to say that the action passes to the quotient group $G / H$ as a regular group action.
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 $K / V ≅ G$ 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 $V H$ equals $H$ .			Steps (1), (2)	By definition of the wreath product action, $H$ centralizes $V$ . Since $S$ is centerless, $V$ is also centerless. Thus, $C_{V H} (V)$ contains $H$ but has trivial intersection with $V$ , forcing $C_{V H} (V) = H$ .
6	The centralizer of $V$ in $K$ equals $H$ .			Steps (2), (5)	Step (5) already shows that $C_{V H} (V) = H$ , so it suffices to show that $C_{K} (V) \leq V H$ . To see this, note that, by the construction in Step (2), any element of $K$ outside $V H$ permutes the direct factors of $V$ as an element of $G$ outside $H$ . The permutation action is nontrivial, so the whole action is nontrivial, and hence elements outside $V H$ cannot centralize $V$ . This forces $C_{K} (V) \leq V H$ , completing the proof.
7	$H$ is characteristic in $K$ .	Fact (1)		Steps (4), (6)	Step-fact combination direct
8	$H$ is characteristic in $K$ and $G$ is a retract of $K$ .			Steps (2), (7)	Step-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)	$H \leq G \leq K$	Step (1)	Given-step-fact combination direct