Pronormality is not finite-intersection-closed: Difference between revisions

Latest revision as of 19:13, 22 February 2009

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 $g^{- 1} H g$ are conjugate in the subgroup they generate.

Facts used

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.

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 A subgroup <math>H</math> of a group <math>G</math> is termed '''pronormal''' in <math>G</math> if, given any <math>g \in G</math>, <math>H</math> and <math>g^{-1}Hg</math> are conjugate in the subgroup they generate.
+==Facts used==
+# [[uses::Join with any distinct conjugate is the whole group implies pronormal]]
 ==Proof==
@@ Line 27: / Line 30: @@
 * <math>K</math> is pronormal: In fact, <math>K</math> is a normal subgroup of <math>G</math>.
-* <math>H</math> is pronormal: Any conjugate of <math>H</math> is either equal to <math>H</math> or intersects <math>H</math> trivially, in which case they generate the whole group. In either case, <math>H</math> and that conjugate are conjugate in the subgroup generated.
+* <math>H</math> is pronormal: Any conjugate of <math>H</math> is either equal to <math>H</math> or intersects <math>H</math> trivially, in which case they generate the whole group (in other words, <math>H</math> is a [[subgroup whose join with any distinct conjugate is the whole group]]). Thus, <math>H</math> is pronormal in <math>G</math> (See fact (1)).
 * <math>H \cap K</math> is not pronormal: Indeed, this subgroup and a conjugate of it generate the subgroup <math>K</math>, within which they are not conjugate.