# What is the automorphism group of this geometry?

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?

Thanks.

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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 $126$).
$S_{7}$ 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? – DavidLHarden Apr 11 '11 at 17:31
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 $S_7$. – Tom De Medts Apr 11 '11 at 18:00