**Definition:** A subgroup $H$ of $G$ is said to be *pronormal* in $G$ if for all $g\in G$, there exists $x \in \langle H, H^g \rangle$ such that $H^x =H^g$.


**Definition:** A subgroup $H$ of $G$ is said to be *weakly pronormal* in $G$ if for $g \in G$, there exists $x \in H^{\langle g \rangle} := \langle g^nHg^{-n} \, |\, n\in \mathbb{Z} \rangle$ such that $H^x =H^g$.


By the inclusion $\langle H, H^g \rangle \leq H^{\langle g \rangle}$, we have that pronormality implies weak pronormality. It is also known that they coincide for finite solvable groups. However it does not hold in general according to the [paper](http://mat.polsl.pl/groups/docs/Subbotin.pdf)



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> I need to construct an example to show that weak pronormality does not imply pronormality for finite groups in general. This would have to be a non-solvable finite group to start with.
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Weak Pronormality of a finite group
Here is a suggestion on a possible way to look for an example, though I have not pursued it myself: suppose that $H$ is a non-maximal subgroup of maximal order of a finite simple group $G$ and that $H$ is self-normalizing( given the first assumption, the second one is equivalent to requiring that $H$ is not a (proper non-trivial) normal subgroup of maximal order of any maximal subgroup of $G$). Suppose further that there is some $x \in G$ such that $x \not \in \langle H,H^{x} \rangle.$ Then $H$ is weakly pronormal in $G$, but not pronormal.


If $H$ were pronormal, then for $x$ as above we would have $H^{x} = H^{y}$ for some $y \in \langle H,H^{x} \rangle$, and then $xy^{-1} \in N\_{G}(H) =H.$ But then $x \in \langle H,H^{x} \rangle,$ contrary to hypothesis. Hence $H$ is not pronormal.


To prove that $H$ is weakly pronormal, we note that $H^{\langle y \rangle }$ is normalized by $y$ for any $y \in G.$


If there were some $y \in G$ with $H$ not conjugate to $H^{y}$ in $H^{\langle y \rangle},$ then $H < H^{\langle y \rangle} < N\_{G}(H^{\langle y \rangle})$ is a strictly increasing chain of proper subgoups of $G$. But this is a contradiction, since the choice of $H$ forces $H^{\langle y \rangle}$ to be maximal in $G$, and then $G = N\_{G}(H^{\langle y \rangle}),$ contrary to the simplicity of $G$.


