Creating a Latin rectangle from a projective plane

Given a projective plane I'd like to form a latin rectangle from the lines. In particular, I'd like to take each line from the plane, order the elements in some way, and stick them into the matrix as a column.

I liked to know whether this is possible. Better yet, I'd love an algorithm for creating the matrix.

• For what it's worth, this can be done for the 7-point plane: $$\matrix{1&2&3&4&5&6&7\cr2&5&7&1&3&4&6\cr3&7&4&5&6&2&1\cr}$$ – Gerry Myerson Mar 17 '15 at 22:59

By a theorem of Singer, the automorphism group of $PG(2,q)$ contains a cyclic subgroup $\langle\sigma\rangle$ which acts regularly on points and regularly on lines. Fix a base block, $B$, and list its elements in the first column of your matrix in any order you choose. Then form the rest of the columns by applying $\sigma$ repeatedly to the first row. For the seven point plane, you can take $\sigma = (1, 2, 3, 4, 5, 6, 7)$ and $B = \{1, 2, 4\}$, giving the solution
$1 2 3 4 5 6 7 \\ 2 3 4 5 6 7 1 \\ 4 5 6 7 1 2 3$
• Regarding the theorem of Singer you mentioned, I haven't been able to find a reference for that. I have found a weaker version that only applies when $q$ is a power of a prime. Do you have a reference? Thanks again for all your help! – user43928 May 26 '15 at 21:40
• I don't have a reference to hand for Singer's result - it is simply about the existence of an element of order $q^{n}-1$ in $\GL(n,q)$. This seems to be what you are referring to - it applies only when $q$ is prime power. If you know of an example of a projective plane where $q$ is not a power of a prime, you should publish it. You'll be quite famous! – Padraig Ó Catháin May 31 '15 at 22:59