# 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 Normal subgroup (?), but is stronger than the property of being a Subnormal-to-normal subgroup (?): in other words, every subnormal subgroup satisfying $p$ is normal. Then, $p$ is not a Transitive subgroup property (?).

## Facts used

1. 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.