Equivalence of definitions of transitively normal subgroup

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

Statement

The following are equivalent for a normal subgroup $K$ of a group $G$:

1. For any normal subgroup $H$ of $K$, $H$ is a normal subgroup of $G$.
2. For any normal automorphism $\sigma$ of $G$, the restriction of $\sigma$ to $K$ is also a normal automorphism of $K$.

Definitions used

Normal automorphism

Further information: Normal automorphism (?)

An automorphism $\sigma$ of a group $G$ is termed a normal automorphism if, for every normal subgroup $N$ of $G$, $\sigma$ restricts to an automorphism of $N$.

Proof

(1) implies (2)

Given: A group $G$, a subgroup $K$ of $G$ such that any normal subgroup $H$ of $K$ is also a normal subgroup of $G$. $\sigma$ is a normal automorphism of $G$.

To prove: $\sigma$ restricts to an automorphism $\sigma '$ of $K$ that is a normal automorphism of $K$.

Proof:

1. $K$ is normal in $G$: Since $K$ is a normal subgroup of itself, setting $H=K$ in the condition on $K$ yields that $K$ is normal in $G$.
2. $\sigma$ restricts to an automorphism $\sigma '$ of $K$: This follows from the previous step and the definition of normal automorphism.
3. For any normal subgroup $H$ of $K$, $H$ is a normal subgroup of $G$: This is by assumption.
4. For any normal subgroup $H$ of $K$, $\sigma '$ restricts to an automorphism of $H$: By the previous step, $H$ is normal in $G$, so, by assumption, $\sigma$ restricts to an automorphism of $H$. But since $H\leq K\leq G$ and the restriction of $\sigma$ to $K$ is $\sigma '$, the restriction of $\sigma$ to $H$ equals the restriction of $\sigma '$ to $H$.

From the last step, $\sigma '$ is a normal automorphism of $K$, completing the proof.

(2) implies (1)

Given: A group $G$, a subgroup $K$ of $G$ such that every normal automorphism of $G$ restricts to a normal automorphism of $K$. $H$ is a normal subgroup of $K$.

To prove: $H$ is a normal subgroup of $G$, i.e., for any $g\in G$ and $h\in H$, $ghg^{-1}\in H$.

Proof: Let $\sigma =c_{g}=x\mapsto gxg^{-1}$ be conjugation by $g$ (i.e., an inner automorphism).

1. $\sigma =c_{g}$ is a normal automorphism of $G$: By the definition of normal subgroup, $c_{g}$ restricts to an automorphism of every normal subgroup of $G$. Thus, $c_{g}$ is a normal automorphism of $G$.
2. The restriction of $c_{g}$ to $K$, which we call $\sigma '$, is a normal automorphism of $K$: This follows from the given data for $K$.
3. $\sigma '$ and hence, $\sigma$, restricts to an automorphism of $H$: Since $H$ is a normal subgroup of $K$, and $\sigma '$ is a normal automorphism of $K$, $\sigma '$ restricts to an automorphism of $H$. Hence, $\sigma$ restricts to an automorphism of $H$.
4. $\sigma (h)\in H$, i.e., $c_{g}(h)=ghg^{-1}\in H$: This follows immediately from the previous step.