Hall not implies automorph-conjugate
This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a finite group, every subgroup satisfying the first subgroup property (i.e., Hall subgroup) need not satisfy the second subgroup property (i.e., automorph-conjugate subgroup)
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A Hall subgroup of a group need not be conjugate to all its automorphs.
Other implications of the same or similar examples
Results of the opposite kind
- Nilpotent Hall implies isomorph-conjugate, or more generally, Nilpotent Hall subgroups of same order are conjugate
- Hall implies order-dominating in finite solvable
We prove that if is an odd prime, is a power of a prime , and , then any subgroup of index in is a Hall subgroup.
This follows from order computation.
Now observe that the parabolic subgroup has the required index, and hence is a Hall subgroup. By we mean the subgroup of comprising those elements where the bottom row has only one nonzero entry, namely the last.
Now consider and its image under the transpose-inverse automorphism. For (which is true if is an odd prime), the transpose-inverse has an invariant one-dimensional subspace while the original subgroup doesn't. Hence, the two subgroups cannot be conjugate. However, they are certainly automorphs (by the transpose-inverse automorphism). We thus have a Hall subgroup that is not automorph-conjugate.
A specific example is , where the two Hall subgroups are both isomorphic to .