Hall not implies automorph-conjugate

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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.

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We prove that if r is an odd prime, q is a power of a prime p, and gcd(r,q-1) = 1, then any subgroup of index (q^r - 1)/(q-1) in SL(r,q) is a Hall subgroup.

This follows from order computation.

Now observe that the parabolic subgroup P_{r-1,1} has the required index, and hence is a Hall subgroup. By P_{r-1,1} we mean the subgroup of SL(r,q) comprising those elements where the bottom row has only one nonzero entry, namely the last.

Now consider P_{r-1,1} and its image under the transpose-inverse automorphism. For r > 2 (which is true if r 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 SL(3,2), where the two Hall subgroups are both isomorphic to S_4.