Pronormality is not commutator-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., commutator-closed subgroup property).
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Statement

It is possible to have a group G and pronormal subgroups H, K of G such that the commutator [H,K] is also a pronormal subgroup.

Related facts

Proof

Example of the symmetric group of degree four

Further information: symmetric group:S4

Let G be the symmetric group on the set \{1,2,3,4 \}. Let H = K be a 2-Sylow subgroup of G, say:

H = \{ (), (1,2,3,4), (1,3)(2,4), (1,4,3,2), (1,3), (2,4), (1,2)(3,4), (1,4)(2,3) \}.

Then, H is isomorphic to a dihedral group of order eight. [H,K] = [H,H] = \{ (), (1,3)(2,4) \}.

  • H is pronormal in G: In fact, H is a Sylow subgroup of G, and Sylow implies pronormal.
  • [H,H] is not pronormal in G: This subgroup is conjugate to \{ (), (1,2)(3,4) \} by the permutation (2,3), but these two subgroups are not conjugate in the subgroup they generate.