Full invariance is transitive

Statement with symbols
Suppose $$H \le K \le G$$ are groups (in words, $$H$$ is a subgroup of $$K$$ and $$K$$ is a subgroup of $$G$$). Then, if $$K$$ is a fully invariant subgroup of $$G$$ and $$H$$ is a fully invariant subgroup of $$K$$, we have that $$H$$ is a fully invariant subgroup of $$G$$.

Transitivity for related properties

 * Characteristicity is transitive
 * Injective endomorphism-invariance is transitive
 * Strict characteristicity is not transitive
 * Normality is not transitive
 * Homomorph-containment is not transitive

Related facts about full invariance

 * Fully invariant of strictly characteristic implies strictly characteristic
 * Full invariance is quotient-transitive
 * Full invariance does not satisfy intermediate subgroup condition
 * Full invariance is strongly intersection-closed
 * Full invariance is strongly join-closed

Facts used

 * 1) uses::Balanced implies transitive

Proof
Full invariance is the balanced subgroup property for endomorphisms. In other words, it can be expressed as:

endomorphism $$\to$$ endomorphism

This says that $$K$$ is fully invaraint in $$G$$ if every endomorphism of $$G$$ restricts to an endomorphism of $$K$$. By fact (1), any balanced subgroup property is transitive, hence full invariance is transitive.