Self-normalizing not implies contranormal

From Groupprops
Jump to: navigation, search
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., self-normalizing subgroup) need not satisfy the second subgroup property (i.e., contranormal subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about self-normalizing subgroup|Get more facts about contranormal subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property self-normalizing subgroup but not contranormal subgroup|View examples of subgroups satisfying property self-normalizing subgroup and contranormal subgroup

Statement

Property-theoretic statement

The subgroup property of being a self-normalizing subgroup does not imply the subgroup property of being a contranormal subgroup.

Verbal statement

It is possible to have a self-normalizing subgroup in a group whose Normal closure (?) in the group is a proper normal subgroup.

Proof

An infinite example

The simplest examples of self-normalizing subgroups that are not contranormal can be obtained in the infinite case. Specifically, we have the following fact: any free factor is a self-normalizing subgroup, but its normal closure is a proper subgroup (namely the kernel of the natural projection to the other free factor). Hence, any free factor gives an example of a self-normalizing subgroup which is not contranormal.

A finite example

Finite examples are somewhat harder to construct. One finite example is as follows: let G be the automorphism group of the symmetric group on six letters. Then, the symmetric group on six letters sits as a normal subgroup of index two inside G. Call this subgroup N. Now, define H as the subgroup of N comprising those permutations that fix the first letters. Then, H is isomorphic to the symmetric group on five letters. Clearly:

  • H is not contranormal in G, because it is contained in the proper normal subgroup N of G.
  • H is self-normalizing in G. For this, first observe that conjugating by any element in N \setminus H sends H to a subgroup of N fixing a different letter. Second, conjugating by any element in G \setminus N cannot preserve H because it sends transpositions to triple transpositions, which do not fix any letter.