If a finite projective plane $\pi\_1$ of order $m$ contains, as a sub plane, a
finite projective plane $\pi\_2$ of order $n$, then $m \geq n^2$ with equality holding only in the case of a Baer sub plane. Otherwise $m \geq n^2 + n$. (This is a theorem of Bruck that can be found in Hall's Group Theory book, I believe.) My
question is, how nearly can this bound be achieved?


For example, we have cases for $m=9$ and $n=2$, where $\pi\_1$ is one of the non-Desaurgian planes of order $9$, and $\pi\_2$ is the Fano plane. Is anything known about how close we can come to the bound in the general case? Has the bound been improved?



Subplanes of Finite Projective Planes
No, the bound has not been improved. Even a subplane of order 3 of a projective plane of order 12 has not yet been ruled out. (You can see this
in the arguments in the work on possible collneations of a projective plane of order 12, such as Janko-Van Trung (1980) [On projective planes of order 12 which have a subplane of order 3. I. J. Combin. Theory Ser. A 29, 254–256].) There's a result of Roth from 1964 that if the subplane is fixed by a non-trivial collineation then $n > m^2+m+1$ [Roth, Richard (1964). Collineation groups of finite projective planes. Mathematische Zeitschrift, 83(5), 409-421, Corollary 2 on page 418]. Of course, if you found a subplane with equality in the Bruck bound, you would have a projective plane of order not a prime power, and have solved a much bigger problem, by disproving the prime power conjecture for finite projective planes.


