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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Statement

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

Related facts

Similar facts

• Hall not implies isomorph-automorphic
• Hall not implies order-isomorphic

Other implications of the same or similar examples

• Hall not implies intermediately automorph-conjugate
• Hall not implies procharacteristic

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

Proof

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 $S_{4}$.