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 $K\leq H\leq G$, $K$ must satisfy $q$ in $H$