Famously from every finite projective plane $P$ we can create a dual projective plane $P'$ by taking the points of $P'$ to be the lines of $P$ and the lines of $P'$ to be the points of $P$.
It is also well known that the smallest projective plane, the Fano plane, is isomorphic to its own dual. I am interested in the set of all isomorphisms between the Fano plane and its dual.
Perhaps an easier way to think about these maps is to take the Heawood graph: the bipartite graph whose 14 vertices represent the 7 points and 7 lines of the Fano plane with an edge connecting a point-vertex and a line-vertex whenever in the Fanoplane the point lies on the line.
This graph has 336 graph-automorphisms: 168 mapping point-vertices to point-vertices and line-vertices to line-vertices – a normal subgroup of the automorphsim group isomorphic to the symmetry group of the fano plane. And it has 168 automorphisms mapping point-vertices to line-vertices and vice-versa: the unique non-group coset for this normal subgroup. It are these 168 symmetries in the non-trivial coset that I'm interested in. My question is:
Are these all of order 2?
Obviously they are all of even order, but all being of order 2 is a much stronger claim. I don't see a clear reason for this when merely staring at the Fano plane, but in the mean time I wasn't able to construct an example of higher order. I hope someone can clarify this!
Best Answer
Let $G$ be the full automorphism group of the Heawood graph and $H$ the subgroup induced from the automorphisms of the Fano plane. Let $\tau$ be a representative of a nontrivial coset.
Suppose all elements outside of $H$ have order two. Then $e=(\tau h)^2=\tau h\tau h=(\tau h\tau^{-1})h$ implies $\tau h\tau^{-1}=h^{-1}$ for all $h\in H$ (in analogy with the dihedral symmetries of a regular polygon, as you note in the comments). But this implies $h\mapsto h^{-1}$ is an automorphism of $H$, which is equivalent to $H$ being abelian. But $H\cong\mathrm{PGL}_3\mathbb{F}_2$ is not abelian, a contradiction.