# Re: paper claiming projective planes must have prime power order

On 13 Feb., 04:44, Craig Feinstein <cafei...@xxxxxxx> wrote:
For people who don’t have time to read Mehendale’s whole paper, I’ll
summarize his argument in the paper arXiv:math/0611492, which claims
that a finite projective plane must have prime power order:

The author proves that if there are n-1 orthogonal Latin squares of
order n, then n must be a prime power. (Bose’s Theorem tells us that
this is equivalent to proving that a finite projective plane must have
prime power order.)

Consider the n(n-1) rows of n-1 Latin squares of order n as a set of
permutations in the group of permutations S_n. For instance if n=5 and
a row is 3 2 4 1 5, then this means that 1 goes to 3, 2 goes to 2, 3
goes to 4, 4 goes to 1 and 5 goes to 5. Without loss of generality,
assume that one of the Latin squares has the identity permutation [1 2
… n] as a row. Call the set of rows in this Latin square N. And
without loss of generality, assume that the first column in each of
the Latin squares is [1 2 … n]. Call the set of n(n-1) rows G.

Then it is not difficult to see that any two columns of G (when all of
the rows are placed on top of one another) must contain the n(n-1)
unique ordered pairs (1,2),(1,3),…,(1,n),(2,1),(2,3),….,(n,1),(n,2),…,
(n,n-1). (Otherwise, the n-1 Latin squares would not be orthogonal.)
In other words, the sets of permutations must be sharply 2-transitive.
This implies that the set G of permutations is a group, since there is
an identity permutation in G.

Here ...

Furthermore, the set N consists of the identity permutation and n-1
permutations in G that are fixed-point-free, since these permutations
form a Latin square. (One can use an elementary counting-argument to
show that there are at most n-1 permutations in G that are fixed-point-
free, since G is sharply 2-transitive.) Since N is sharply transitive,
N is a subgroup of G.

.... and here the author seems to use a theorem(?) like

If G is a group operating on a set X and S is a subset
containing 1 (i.e. 1eScG) and ExeX:AyeX:E!seS:sx=y,
then S is a group.

However, this "theorem" is wrong, e.g. because the permutations in
12345
21453
34512
53124
45231
do not form a group.
So maybe there's something else used in thess places -- but what?
.