Potentially characteristic not implies normal-potentially characteristic: Difference between revisions

Latest revision as of 21:30, 30 May 2009

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., potentially characteristic subgroup) need not satisfy the second subgroup property (i.e., normal-potentially characteristic subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about potentially characteristic subgroup|Get more facts about normal-potentially characteristic subgroup

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Statement

It is possible to have a potentially characteristic subgroup that is not a normal-potentially characteristic subgroup.

Related facts

Weaker facts

Facts used

Proof

The proof follows from facts (1) and (2).

Example of the dihedral group

Further information: dihedral group:D8

Let $G$ be the dihedral group of order eight, and $H$ be one of the Klein four-subgroups.

$H$ is not a normal-potentially characteristic subgroup of $G$ : Using the fact that every automorphism is center-fixing and inner automorphism group is maximal in automorphism group implies every automorphism is normal-extensible, every automorphism of $G$ can be extended to an automorphism of $K$ for any group $K$ containing $G$ as a normal subgroup. But since there is an automorphism of $G$ not sending $H$ to itself, $H$ cannot be characteristic in $K$ .
$H$ is potentially characteristic in $G$ : for instance, we can realize $G$ as the $2$ -Sylow subgroup of the symmetric group of degree four, in such a way that $H$ becomes characteristic.

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 ==Facts used==
-# [[uses::Normal-extensible not implies normal]]: Actually, we need a slightly ''stronger'' version of this statement. Namely, we need that there is a [[normal-extensible automorphism]] of a [[finite group]] that is not a [[normal automorphism]]. The example given in the proof is a finite group, so the proof actually shows the stronger version.
+# [[uses::Potentially characteristic not implies normal-extensible automorphism-invariant]], which in turn follows from [[uses::Normal not implies normal-extensible automorphism-invariant in finite]] and [[finite normal implies potentially characteristic]].
-# [[uses::Finite normal implies potentially characteristic]]
+# [[uses::Normal-potentially characteristic implies normal-extensible automorphism-invariant]]
 ==Proof==
-By fact (1) (or rather, the stronger version) there exists a finite group <math>K</math>, a normal-extensible automorphism <math>\sigma</math> of <math>K</math>, and a normal subgroup <math>H</math> of <math>K</math> such that <math>\sigma(H) \ne H</math>.
+The proof follows from facts (1) and (2).
-* <math>H</math> is potentially characteristic in <math>K</math>: This follows directly from fact (2).
+===Example of the dihedral group===
-* <math>H</math> is not semi-strongly potentially characteristic in <math>K</math>: Suppose there exists a group <math>G</math> containing <math>K</math> as a normal subgroup. Then, since <math>\sigma</math> is normal-extensible, <math>\sigma</math> extends to an automorphism <math>\sigma'</math> of <math>G</math>. But <math>\sigma'(H) = \sigma(H) \ne H</math>, so <math>H</math> is not characteristic in <math>G</math>. Thus, there is no group containing <math>K</math> as a normal subgroup and <math>H</math> as a characteristic subgroup.
+{{further|[[Particular example::dihedral group:D8]]}}
+Let <math>G</math> be the dihedral group of order eight, and <math>H</math> be one of the Klein four-subgroups.
+* <math>H</math> is not a normal-potentially characteristic subgroup of <math>G</math>: Using the fact that [[every automorphism is center-fixing and inner automorphism group is maximal in automorphism group implies every automorphism is normal-extensible]], every automorphism of <math>G</math> can be extended to an automorphism of <math>K</math> for any group <math>K</math> containing <math>G</math> as a normal subgroup. But since there is an automorphism of <math>G</math> not sending <math>H</math> to itself, <math>H</math> cannot be characteristic in <math>K</math>.
+* <math>H</math> is potentially characteristic in <math>G</math>: for instance, we can realize <math>G</math> as the <math>2</math>-Sylow subgroup of the symmetric group of degree four, in such a way that <math>H</math> becomes characteristic.