Fully invariant of strictly characteristic implies strictly characteristic

Property-theoretic statement
Fully invariant * Strictly characteristic $$\le$$ Strictly characteristic

Verbal statement
Every fully invariant subgroup of a strictly characteristic subgroup is strictly characteristic.

Symbolic statement
Let $$H \le K \le G$$ such that $$H$$ is fully invariant in $$K$$ and $$K$$ is strictly characteristic in $$G$$, then $$H$$ is strictly characteristic in $$G$$.

Hands-on proof
Given groups $$H \le K \le G$$ such that $$H$$ is fully invariant in $$K$$ and $$K$$ is strictly characteristic in $$G$$. We need to show that for any surjective endomorphism $$\sigma$$ of $$G$$, $$\sigma$$ takes $$H$$ to within itself.

First, notice that since $$K$$ is strictly characteristic in $$G$$, $$\sigma(x) \in K$$ for every $$x \in K$$. Thus, $$\sigma$$ restricts to a function from $$K$$ to $$K$$. Since this function arises by restricting an endomorphism of $$G$$, it is an endomorphism of $$K$$.

Since $$H$$ is fully invariant in $$K$$, $$\sigma|_K$$ takes $$H$$ to within itself. But since $$\sigma|_K$$ is the restriction of $$\sigma$$ to $$K$$ in the first place, we conclude that $$\sigma$$ in fact takes $$H$$ to itself.

Using the function restriction formalism
In terms of the function restriction formalism:


 * The following is a function restriction expression for the subgroup property of normality:

Surjective endomorphism $$\to$$ Endomorphism


 * The following is a function restriction expression for the subgroup property of full invariance:

Endomorphism $$\to$$ Endomorphism

We now use the composition rule for function restriction to observe that the composition of fully invariant and strictly characteristic implies the property:

Surjective endomorphism $$\to$$ Endomorphism

Which is again the subgroup property of strict characteristicity.