Pronormal and subnormal implies normal

Verbal statement
A subgroup of a group that is both pronormal and subnormal must be normal.

Related facts

 * Pronormal implies intermediately subnormal-to-normal: This can further be generalized to polynormal implies intermediately subnormal-to-normal.
 * Normalizer of pronormal implies abnormal
 * Pronormal implies self-conjugate-permutable (finite groups)

Facts used

 * 1) Pronormality satisfies intermediate subgroup condition: A pronormal subgroup in the whole group is also pronormal in any intermediate subgroup (this is used in the first proof).

Proof
The proof involves two steps.

Pronormal and 2-subnormal implies normal
A 2-subnormal subgroup is a subgroup realized as a normal subgroup of a normal subgroup. We prove that a pronormal 2-subnormal subgroup is normal.

Given: $$H \triangleleft K \triangleleft G$$, and $$H$$ is pronormal in $$G$$.

To prove: $$H$$ is normal in $$G$$. Take $$g \in G$$. We must show that $$H^g = H$$.

Clearly $$H^g$$ must lie inside the normal closure of $$H$$ in $$G$$, and since $$K$$ is a normal subgroup containing $$H$$, $$H^g$$ must lie inside $$K$$. Now, since $$H$$ is pronormal in $$G$$, $$H$$ and $$H^g$$ are conjugate subgroups in $$<H,H^g$$, and thus are conjugate subgroups inside $$K$$.

This, combined with the fact that $$H$$ is normal inside $$K$$, tells us that $$H^g = H$$.

Overall proof
The overall proof proceeds by induction, using the above lemma. The induction statement for $$n$$ says:

Any pronormal subgroup of subnormal depth at most $$n$$ is normal.

The statement is proved by showing that any pronormal subgroup of subnormal depth $$n$$ is of subnormal depth at most $$n-1$$, and then appealing to induction. Here's how. Suppose $$H$$ is pronormal in $$G$$ and:

$$H = H_0 \triangleleft H_1 \triangleleft H_2 \triangleleft \ldots \triangleleft H_n = G$$

$$H$$, being pronormal in $$G$$ is also pronormal in every intermediate subgroup (property-theoretically, pronormality satisfies the intermediate subgroup condition). So $$H$$ is pronormal in $$H_2$$. But $$H$$ is also 2-subnormal in $$H_2$$, and the previous lemma tells us that any pronormal 2-subnormal subgroup is normal. So $$H \triangleleft H_2$$, and we get:

$$H = H_0 \triangleleft H_2 \triangleleft \ldots \triangleleft H_n = G$$

Thus $$H$$ has subnormal depth at most $$n-1$$, and induction applies.

Converse
The converse is clearly true: any normal subgroup is pronormal as well as subnormal.