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The following are subgroup embedding properties introduced by Bah and Borevich.

Definition 1: A subgroup $H$ of $G$ is said to be paranormal if for each $g \in G$, we have that $H^{\langle H, H^g \rangle}= \langle H, H^g \rangle$.

Proposition 1: A subgroup $H$ of $G$ is paranormal if and only if, for any $K \leq G$ containing $H$, all conjugates of $H$ lying in $K$ also lie in $H^K$ i.e., the normal closure of $H$ in $K$.

Definition 2: A subgroup $H$ of $G$ is said to be polynormal if for each $g\in G$, we have $H^{\langle g \rangle} = H^{H^{ \displaystyle \langle g \rangle}}$

Proposition 2: A subgroup $H$ of $G$ is polynormal if and only if $H^K = H^L$ for any subgroups $K$ and $L$ such that $H \leq K \unlhd L \leq G$.


It is clear that paranormality implies polynormality. However, the converse does not hold in general. An example showing this is given by V. I. Mysovskikh in here. But he uses an algorithm based on the table of marks (Burnside matrices). I'm looking for an intuitive counterexample to show that polynormality does not imply paranormality. Any help will be greatly appreciated.

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One way to ensure that a subgroup $H$ of $G$ is polynormal is to find an example in which there do not exist unequal $K$ and $L$ with $H \le K \lhd L \le G$. We could could look for an example of a simple group $G$ with subgroups $H <K < G$, where $H$ is maximal in $K$ and $K$ is maximal in $G$, so there are unlikely to be many intermediate subgroups between $H$ and $G$. And for $H$ to fail to be paranormal, we would like there to be $g \in G$ with $H^g < K$ and $H$ and $H^g$ not conjugate in $K$.

One example I found was $G = M_{12}$, $K = {\rm PSL}(2,11)$, and $H=A_5$. Then $K$ has two conjugacy classes of subgroups isomorphic to $A_5$, but they are fused in $G$. There are are only two intermediate subgroups between $H$ and $G$, both isomorphic to ${\rm PSL}(2,11)$.

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