Pronormality is normalizing join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) satisfying a subgroup metaproperty (i.e., normalizing join-closed subgroup property)
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Statement

Suppose $H,K\le G$ are pronormal subgroups such that $K\le N_G(H)$: in other words, $K$ normalizes $H$. Then the join of subgroups $⟨H,K⟩$ (also the same as $HK$) is also a pronormal subgroup.

Related facts

Similar facts: statements about normalizing join-closedness

Some facts with very similar proofs:

• Weak pronormality is normalizing join-closed
• Intermediate isomorph-conjugacy is normalizing join-closed
• Intermediate automorph-conjugacy is normalizing join-closed

Other facts about normalizing joins, but with a different kind of proof:

• Subnormality is normalizing join-closed

Related facts about pronormality

• Nilpotent join of pronormal subgroups is pronormal
• Pronormality is not join-closed
• Pronormality is not intersection-closed

Join-closedness of some related properties

• Paranormality is strongly join-closed
• Polynormality is strongly join-closed

Proof

(Note: This proof adopts the right-action convention. It can be adapted to the left-action convention).

Given: A group $G$, pronormal subgroups $H,K\le G$ such that $K\le N_G(H)$.

To prove: $HK$ is pronormal in $G$. Specifically, for any $g\in G$, our goal is to find a $z\in ⟨HK,(HK{)}^g⟩$ such that $(HK{)}^z=(HK{)}^g$.

Proof:

1. (Given data used: $H$ is pronormal in $G$): First, consider $g$ on $H$. By pronormality of $H$, there exists $x\in ⟨H,H^g⟩$ such that $H^x=H^g$. Thus, $H^{g{x}^{-1}}=H$.
2. (Given data used: $K$ normalizes $H$): Recall that $K\le N_G(H)$, so $K^{g{x}^{-1}}\le N_G(H^{g{x}^{-1}})=N_G(H)$. Thus, $⟨K,K^{g{x}^{-1}}⟩\le N_G(H)$.
3. (Given data used: $K$ is pronormal in $G$): By pronormality of $K$, there exists $y\in ⟨K,K^{g{x}^{-1}}⟩$ such that $K^y=K^{g{x}^{-1}}$. In particular, $y\in N_G(H)$, so $H^y=H$.
4. Now consider $z=yx$. We claim that $(HK{)}^z=(HK{)}^x$(No problem data used):. We have $H^z=H^{yx}=(H^y{)}^x=H^x$ (since $H^y=H$). We already know that $H^x=H^g$ (step (1)), so $H^z=H^g$. Similarly, $K^z=(K^y{)}^x=(K^{g{x}^{-1}}{)}^x=K^{g{x}^{-1}x}=K^g$. Thus, $z$ acts like $x$ on both $H$ and $K$, and so $(HK{)}^z=(HK{)}^x$.
5. We now check that $z\in ⟨(HK),(HK{)}^g⟩$ (No problem data used): For this, observe that $x\in ⟨H,H^g⟩\le ⟨HK,(HK{)}^g⟩$. Also, $y\in ⟨K,K^{g{x}^{-1}}⟩$. The subgroup $K$ is contained in $HK$. The subgroup $K^g$ is contained in $(HK{)}^g$, and the element $x$ is in $⟨HK,(HK{)}^g⟩$ as argued already, so the subgroup $K^{g{x}^{-1}}$ is also in $⟨HK,(HK{)}^g⟩$. Thus, $⟨K,K^{g{x}^{-1}}⟩\le ⟨HK,(HK{)}^g⟩$, and hence $y\in ⟨HK,(HK{)}^g⟩$. Thus, $z=yx$ is also in the subgroup $⟨HK,(HK{)}^g⟩$.

References

• Finite soluble groups with pronormal system normalizers by John S. Rose, Proceedings of the London Mathematical Society, ISSN 1460244X (online), ISSN 00246115 (print), Volume 17, Page 447 - 469(Year 1967): ^{Weblink on Oxford Journals pageMore info}, Proposition 1.7, Page 450