Define the following incidence structure of rank three. The points are the elements of $\mathbb{Z}_7=$ {$0,\ldots,6$}. The lines of type 1 are the triples $(x,x+1,x+3)$ modulo $7$. The lines of type 2 are the triples $(x,x+1,x+5)$ modulo 7. Define the incidence relation as follows. A point is incident to a line of type 1 (resp.2) if it is contained in the line. A line of type 1 is incident to a line of type 2 if they have two points in common.

It is not difficult to check that this incidence structure is a geometry (in the sense of Buekenhout). Somehow it looks like two superposed Fano planes.

Here is my question: what is the full automorphism group of this geometry, and what is the type preserving automorphism group of this geometry?



Your geometry has the property that each of its rank 2 restrictions is a Fano plane. In particular, the type-preserving automorphism group (let's call it $G$) is a subgroup of the automorphism group of the Fano plane, which is $PSL(3,2)$. The group $G$ has the property that the pointwise stabilizer of any two points is trivial (indeed, if $g \in G$ fixes for instance $0$ and $1$, then it has to fix $3$ and $5$, and from that you deduce that it has to fix everything).

On the other hand, $G$ contains the Singer cycle $x \mapsto x+1 \pmod{7}$, and it contains, for instance, the element $(1 2 4)(3 6 5)$, so it has order divisible by $21$. It follows that $G$ is isomorphic to the Frobenius group of order $21$, which is a maximal subgroup of $PSL(3,2)$.

Note that in the full automorphism group of the geometry, you can interchange all types, since you can for instance interchange lines of type 1 and type 2 by the permutation $(24)(35)$, and by symmetry you can interchange every two types and hence the induced action on the types is $\operatorname{Sym}(3)$. Hence the full automorphism group of the geometry is an extension of $G$ by $\operatorname{Sym}(3)$ (which is a group of order $126$).

  • $\begingroup$ $S_{7}$ has no subgroup of order $126$, since a Sylow $7$-subgroup normalizer has order $42$ (and any group of order $126$ has a normal Sylow $7$-subgroup). Then how is this automorphism group defined? $\endgroup$ Apr 11 '11 at 17:31
  • $\begingroup$ It is no longer a subgroup of $S_7$, because an automorphism interchanging points with lines (of type $1$ or type $2$) is not an element of $S_7$. $\endgroup$ Apr 11 '11 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.