Re: paper claiming projective planes must have prime power order
 From: hagman <google@xxxxxxxxxxxxx>
 Date: Sat, 14 Feb 2009 08:28:22 0800 (PST)
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 n1 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(n1) rows of n1 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(n1) 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(n1)
unique ordered pairs (1,2),(1,3),…,(1,n),(2,1),(2,3),….,(n,1),(n,2),…,
(n,n1). (Otherwise, the n1 Latin squares would not be orthogonal.)
In other words, the sets of permutations must be sharply 2transitive.
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 n1
permutations in G that are fixedpointfree, since these permutations
form a Latin square. (One can use an elementary countingargument to
show that there are at most n1 permutations in G that are fixedpoint
free, since G is sharply 2transitive.) 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?
.
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