Pronormality is not finite-intersection-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-intersection-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Intersection-closed subgroup property (?), .
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Statement

An intersection of two pronormal subgroups of a group need not be pronormal.

Definitions used

Pronormal subgroup

Further information: pronormal subgroup

A subgroup H of a group G is termed pronormal in G if, given any g \in G, H and g^{-1}Hg are conjugate in the subgroup they generate.

Facts used

  1. Join with any distinct conjugate is the whole group implies pronormal

Proof

The example of the symmetric group

Further information: symmetric group:S4

Let G = S_4 be the symmetric group on the set \{1,2,3,4\}. Let K = \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \} be the subgroup comprising the identity and double transpositions. Let H = \{ (), (1,2), (3,4), (1,2)(3,4) \} be the subgroup generated by two disjoint single transpositions. Then, H \cap K = \{ (), (1,2)(3,4) \} is a two-element subgroup.

  • K is pronormal: In fact, K is a normal subgroup of G.
  • H is pronormal: Any conjugate of H is either equal to H or intersects H trivially, in which case they generate the whole group (in other words, H is a subgroup whose join with any distinct conjugate is the whole group). Thus, H is pronormal in G (See fact (1)).
  • H \cap K is not pronormal: Indeed, this subgroup and a conjugate of it generate the subgroup K, within which they are not conjugate.