# Difference between revisions of "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 $K$.

## 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.