Subgroup property between normal and subnormal-to-normal is not transitive

Statement
Suppose $$p$$ is a subgroup property that is weaker than the property of being a fact about::normal subgroup, but is stronger than the property of being a fact about::subnormal-to-normal subgroup: in other words, every subnormal subgroup satisfying $$p$$ is normal. Then, $$p$$ is not a fact about::transitive subgroup property.

Particular cases

 * Pronormality is not transitive
 * Weak pronormality is not transitive
 * Paranormality is not transitive
 * Polynormality is not transitive
 * Weak normality is not transitive

Facts used

 * 1) uses::Normality is not transitive

Proof
Since normality is not transitive (fact (1)), we can construct groups $$H \le K \le G$$ such that $$H$$ is normal in $$K$$ and $$K$$ is normal in $$G$$, but $$H$$ is not normal in $$G$$. We then have:


 * $$H$$ satisfies property $$p$$ in $$K$$ and $$K$$ satisfies property $$p$$ in $$G$$: This follows because $$p$$ is weaker than normality.
 * $$H$$ does not satisfy property $$p$$ in $$G$$: By construction, $$H$$ is subnormal in $$G$$, so if $$H$$ satisfies property $$p$$ in $$G$$, $$H$$ is normal in $$G$$, contradicting our assumption.

Thus, $$p$$ is not transitive.

This shows that any example of normality not being transitive yields an example of $$p$$ not being transitive.