# 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 $H \le K \le G$ 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.