Transitive implies intermediate subgroup condition implies closed under former

Statement
Suppose $$p$$ is a transitive subgroup property and $$q$$ is a subgroup property satisfying the intermediate subgroup condition. Consider the group property:

$$\alpha = p \implies q$$

In other words, a group $$G$$ satisfies property $$\alpha$$ if every subgroup of it satisfying property $$p$$, also satisfies property $$q$$.

Then, if a group satisfies property $$\alpha$$, so does any subgroup of it with property $$p$$.

Applications

 * ACIC is characteristic subgroup-closed
 * T is normal subgroup-closed
 * Dedekind is subgroup-closed

Proof
Given: A group $$G$$ with property $$\alpha$$, a subgroup $$H$$ of $$G$$ satisfying property $$p$$ in $$G$$. A subgroup $$K$$ of $$H$$ satisfying property $$p$$ in $$K$$

To prove: $$K$$ satisfies property $$q$$ in $$H$$

Proof: It's a three-step proof:


 * Since $$p$$ is transitive, and $$K$$ satisfies $$p$$ in $$H$$, and $$H$$ satisfies $$p$$ in $$G$$, then $$K$$ satisfies $$p$$ in $$G$$
 * Since $$G$$ has property $$\alpha$$, $$K$$ must have property $$q$$ in $$G$$
 * Since $$q$$ satisfies the intermediate subgroup condition, and $$K \le H \le G$$, $$K$$ must satisfy $$q$$ in $$H$$