# Self-normalizing not implies contranormal

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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)
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## 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.