Re: paper claiming projective planes must have prime power order

On Feb 12, 10:44 pm, 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.

When I wrote this, I thought that a sharply 2-transitive set of
permutations must be a group. Now, I'm not so certain. I wrote the
author of the paper asking him for a proof of this.


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. Also, N is a normal subgroup in G: A conjugate
of any element in N must also be fixed-point-free. But since the only
permutations in G that are fixed-point-free are in N, the nonidentity
elements of N must all be conjugate to one another. Hence, the
nonidentity elements of N must have the same order, some prime p
divisible by n. If n is not a power of the prime p, then by Cauchy’s
theorem N must have an element of order not equal to p and not equal
to 1, which is impossible, since all elements of N have order either p
or 1. Hence, n must be a prime power.

Notice that this argument doesn’t require the theorem that Mehendale
quotes from Aschbacher’s book. You just need Cauchy’s theorem for
this.

There you have it - an elementary proof of a famous conjecture that
has been open for a very long time and that has been up on arxiv.org
for more than two years now, yet the conjecture is still considered by
an open problem by the mathematics community at large. Can anyone find
a flaw in this argument? I can’t.

.

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