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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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


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.

Facts used


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.