Hall not implies automorph-conjugate

From Groupprops

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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Statement

A Hall subgroup of a group need not be conjugate to all its automorphs.

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Proof

We prove that if $r$ is an odd prime, $q$ is a power of a prime $p$ , and $g c d (r, q - 1) = 1$ , then any subgroup of index $(q r - 1) / (q - 1)$ in $S L (r, q)$ 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 $S L (r, q)$ 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 $S L (3,2)$ , where the two Hall subgroups are both isomorphic to $S 4$ .

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Finite group +, Hall subgroup +, and Automorph-conjugate subgroup +