Full invariance is transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., fully invariant subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
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Statement

Statement with symbols

Suppose $H\leq K\leq G$ 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 $G$.

Related facts

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. Balanced implies transitive

Proof

Using function restriction expressions

This subgroup property implication can be proved by using function restriction expressions for the subgroup properties
View other implications proved this way |read a survey article on the topic

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.