Pronormality is not finite-upper join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) not satisfying a subgroup metaproperty (i.e., finite-upper join-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Upper join-closed subgroup property (?), .
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We can have a subgroup H \le G such that H is a pronormal subgroup in two intermediate subgroups K and L, but H is not pronormal in the join \langle K, L \rangle.

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Further information: symmetric group:S3, symmetric group:S4

Let G be the symmetric group on the set \{ 1,2,3,4 \}, K be the subgroup comprising permutations on \{ 1,2,3 \}, L be the subgroup comprising permutations on \{ 1,2,4 \}, and H = K \cap L be the two-element subgroup generated by (1,2). Observe that:

  • H is a pronormal subgroup in both K and L. In fact, it is a maximal subgroup in both, and maximal implies pronormal.
  • \langle K, L \rangle = G.
  • H is not pronormal in G. That's because the subgroup generated by (3,4) is a conjugate of H, and these two subgroups are not conjugate in the subgroup they generate.