Contranormal not implies self-normalizing

Property-theoretic statement
The subgroup property of being contranormal is not stronger than, or does not imply, the subgroup property of being self-normalizing.

Verbal statement
There exist situations of a group with a contranormal subgroup that is not self-normalizing, i.e., it is properly contained in its normalizer.

Converse
The property of being self-normalizing does not imply the property of being contranormal, either.

However, there are properties that imply both the property of being contranormal and the property of being self-normalizing. These include:


 * Abnormal subgroup
 * Weakly abnormal subgroup

For instance, the normalizer of any Sylow subgroup is abnormal, and is hence both contranormal and self-normalizing.

Facts used

 * Every finite group is a subgroup of a finite simple group
 * Self-normalizing satisfies intermediate subgroup condition

A generic example
Given any finite group $$G$$, we can embed $$G$$ inside a finite simple group $$K$$. Thus, given a finite group $$G$$ and a nontrivial subgroup $$H$$ of $$G$$, we see that $$H$$ is contranormal in $$K$$.

On the other hand, if $$H$$ is not self-normalizing in $$G$$, it cannot be self-normalizing in $$K$$. That's because if $$H$$ were self-normalizing in $$K$$, then $$H$$ would be self-normalizing in $$G$$.

Thus, it suffices to find a finite group $$G$$, with a nontrivial subgroup $$H$$ that is not self-normalizing.

Some specific realizations of this:


 * Let $$G$$ be the alternating group on four letters, and $$H$$ be the normal Klein-four subgroup (comprising the identity element and double transpositions). Then, $$K$$ is the alternating group on five letters, with $$G$$ embedded as the permutations fixing the last letter. Clearly, $$H$$ is contranormal in $$K$$, but is not self-normalizing because its normalizer contains $$G$$.