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

CONVENTION WARNING: This article/section uses the right-action convention. The left and right action conventions are equally powerful and statements/reasoning here can be converted to the alternative convention (the main reason being that every group is naturally isomorphic to its opposite group via the inverse map). For more on the action conventions and switching between them, refer to switching between the left and right action conventions.

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