# Contranormal not implies self-normalizing

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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., contranormal subgroup) need not satisfy the second subgroup property (i.e., self-normalizing subgroup)
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## Statement

### 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.

## Related facts

### Converse

The property of being self-normalizing does not imply the property of being contranormal, either. For full proof, refer: Self-normalizing not implies contranormal

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

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

## Proof

### 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 $G$.