Hall not implies automorph-conjugate

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

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 $$S_4$$.