2-subnormal not implies hypernormalized: Difference between revisions

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A [[2-subnormal subgroup]] of a group need not be [[hypernormalized subgroup|hypernormalized]].
A [[2-subnormal subgroup]] of a group need not be [[hypernormalized subgroup|hypernormalized]].
==Related facts==
===Stronger facts===
* [[Abnormal normalizer and 2-subnormal not implies normal]]: In fact, the same example used here works for that as well.


==Proof==
==Proof==

Revision as of 22:08, 19 September 2008

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., 2-subnormal subgroup) need not satisfy the second subgroup property (i.e., hypernormalized subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about 2-subnormal subgroup|Get more facts about hypernormalized subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property 2-subnormal subgroup but not hypernormalized subgroup|View examples of subgroups satisfying property 2-subnormal subgroup and hypernormalized subgroup

Statement

Verbal statement

A 2-subnormal subgroup of a group need not be hypernormalized.

Related facts

Stronger facts

Proof

An example in the symmetric group on four letters

Let G be the symmetric group on four letters {1,2,3,4} and H be the two-element subgroup generated by (13)(24).

Then, H is normal in the subgroup K={(),(12)(34),(13)(24),(14)(23)}, which is normal in G. So H is 2-subnormal in G.

On the other hand, the normalizer NG(H) is a dihedral subgroup of order eight, which is a self-normalizing subgroup.