Contranormal not implies self-normalizing
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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There exist situations of a group with a contranormal subgroup that is not self-normalizing, i.e., it is properly contained in its normalizer.
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.
- Every finite group is a subgroup of a finite simple group
- Self-normalizing satisfies intermediate subgroup condition
A generic example
Given any finite group , we can embed inside a finite simple group . Thus, given a finite group and a nontrivial subgroup of , we see that is contranormal in .
On the other hand, if is not self-normalizing in , it cannot be self-normalizing in . That's because if were self-normalizing in , then would be self-normalizing in .
Thus, it suffices to find a finite group , with a nontrivial subgroup that is not self-normalizing.
Some specific realizations of this:
- Let be the alternating group on four letters, and be the normal Klein-four subgroup (comprising the identity element and double transpositions). Then, is the alternating group on five letters, with embedded as the permutations fixing the last letter. Clearly, is contranormal in , but is not self-normalizing because its normalizer contains .