Equivalence of definitions of transitively normal subgroup

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 $$K$$.

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 $$N$$.

(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.