Self-normalizing not implies contranormal

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 fact about::normal closure in the group is a proper normal subgroup.

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.