# Self-normalizing not implies contranormal

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) neednotsatisfy the second subgroup property (i.e., contranormal subgroup)

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## Contents

## 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 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 . Call this subgroup . Now, define as the subgroup of comprising those permutations that fix the first letters. Then, is isomorphic to the symmetric group on five letters. Clearly:

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