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I am interested in the following question:

Given a projective plane of order $n=2^a$, is its incidence matrix must contain the incidence matrix of the Fano plane? If not, is it true that for any $n$ of the form $n=2^a$ there exists a projective plane of this order whose incidence matrix contains the incidence matrix of the Fano plane?

The only thing that I managed to find is the following from "Fano configurations in subregular planes‏" (Fisher, 2010) "It is obvious that any translation plane of order $p^r$, for $p$ a prime, has a subplane of order $p$".

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It is a famous conjecture of Hanna Neumann that every non-Desarguesian plane contains a Fano subplane. All the small planes that we can compute with contain huge numbers of Fano subplanes, and it would be a great surprise to find a plane with no Fano subplanes. –  Gordon Royle Jul 26 '13 at 23:34
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1 Answer

up vote 3 down vote accepted

A projective plane over a field, i.e. one in which Pappos' theorem holds, does contain the Fano configuration if and only if the characteristic of the field is two. These are exactly the cases where the order is a power of two. So the latter statement is certainly true: for every such $n$ there exists the plane over $GF(2^a)$ which contains the Fano configuration. One way to see this is consider the homogenous coordinates of the Fano plane: you will find points with these coordinates in any of these larger planes as well, and the collinearities hold due to the characteristic two.

https://en.wikipedia.org/wiki/Projective_plane#Fano_subplanes states as much. It also states:

The situation in non-desarguesian planes is unsettled.

A plane in which Desargues' theorem holds but Pappos' does not would be a plane over a skew field that is not a field. Since a finite skew field will be real fields, the two theorems can be considered the same for finite planes. Therefore unless Wikipedia is out of date, your main question remains open.

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