Polynormal implies intermediately subnormal-to-normal

Property-theoretic statement
The subgroup property of being a polynormal subgroup is stronger than the subgroup property of being an intermediately subnormal-to-normal subgroup.

Symbolic statement
If $$H$$ is a polynormal subgroup of $$G$$, and $$K$$ is any intermediate subgroup containing $$H$$, such that $$H$$ is subnormal inside $$K$$, then $$H$$ is normal inside $$K$$.

Simplifying the statement
Note that polynormality satisfies the intermediate subgroup condition. In other words, if $$H \le K \le G$$, and $$H$$ is polynormal in $$G$$, then $$H$$ is polynormal in $$K$$. Thus, the statement reduces to the statement that any polynormal subnormal subgroup is normal.

In fact, we can further reduce the statement to saying that any polynormal 2-subnormal subgroup is normal (we need to again use that polynormality satisfies the intermediate subgroup condition).

Proving the simplified version
Suppose $$H$$ is a polynormal 2-subnormal subgroup of $$G$$. Then $$H$$ is normal inside its normal closure, and in particular inside any subgroup generated by $$H$$ with its conjugates. In particular, for any $$g \in G$$, $$H$$ is a normal subgroup of $$H^{}$$. But by the hypothesis of polynormality, there exists $$x \in H^{}$$ such that $$H^{} = H^{}$$. This implies that $$H$$ is contranormal inside $$H^{}$$, forcing that $$H = H^{}$$. This proof works for every $$g \in G$$, so $$H$$ is a normal subgroup of $$G$$.