March 22: extending Pi_4 leads us to simplicity of A_n
(see page 22 of the textbook and the Rotman handout)


We saw(*) that if a projective plane Pi of order n is extendable
then either n=2 or n=4, and in both cases Pi is algebraic.
For n=2, the extension is the Hadamard 3-(8,4,1) design.
For n=4 it must be a 3-(22,6,1) Steiner system.  We investigate
this case nex; it also leads us to the 4-(23,7,1) and 5-(24,8,1)
systems and the Mathieu groups, via a construction that the text
attributes to L\"{u}neburg (page 22).

(*) assuming the nonexistence of a projective plane of order 10,
or at any rate of an extendable one.


A 3-(22,6,1) design D has 77 blocks.  Choosing a point p and identifying
the derived design D_p with Pi_4, we see that the blocks are:

  21: p + line
  56: hyperoval

The intersection triangle is as follows (see Table 1.1 on page 21,
and ignore the leftward diagonals that start 759 and 253):

              77 
            56  21
          40  16   5
        28  12   4   1
      20   8   4   0   1
    16   4   4   0   0   1
  16   0   4   0   0   0   1

in particular any two of the 56 hyperovals intersect in 0 or 2 points.
It turns out that this is an equivalence relation on the 168 hyperovals,
with three equivalence classes.  We could check this combinatorially,
but what does it "mean"?


Recall that all hyperovals are equivalent under Aut(Pi_4), and
every element of Aut(Pi_4) is either in PGL_3(F_4) or is
the Galois conjugate of an element of PGL_3(F_4).  So say we have
two hyperovals O and O'.  We can get from O to O' by some g0 in Aut(Pi_4).
Then g in Aut(Pi_4) takes O to O' iff  g = g0 h  with h(O)=O, i.e. with
h in the stabilizer Stab_O.  We have already identified Stab_O with the
group S_6 of permutations of O; and we've also seen that a permutation
pi of O comes from PGL_3(F_4) [without a Galois conjugation] iff pi is even.
So, composing with an odd permutation if necessary, we can assume that
g0 is in PGL_3(F_4), and then g in PGL_3(F_4) takes O to O' iff
g = g0 h  for some h in the subgroup A_6 of Stab_O.


Now I claim:

1) There exists a well-defined surjective homomorphism
det: PGL_3(F_4) --> F_4^*

2) A_6 is in the kernel PSL_3(F_4) of det

whence it follows that

3) All g taking O to O' have the same determinant.

Thus we can define an equivalence relation on the 168 hyperovals by defining

O ~ O' <==> there exists g in PSL_3(F_4) taking O to O'

that partition the hyperovals into 3 equivalence classes each of size 168/3=56.

It turns out that this is the equivalence relation we need to extend Pi_4.

Proof of (1): we know from linear algebra that there is a surjective
homomorphism det: GL_3(F_4) --> F_4^*.  The point is that it takes
every scalar 3x3 matrix to 1, and thus "factors through the quotient":
for any nonzero scalar c and 3*3 matrix g we have det(cg)=det(g),
so the determinant can be defined even when we identify cg with g.

Proof of (2): [ad hoc] A_6 is generated by 3-cycles; indeed A_n is
generated by 3-cycles for all n (yes, even n<3...): we saw that
S_n is generated by simple transposition *that move a fixed element*,
so A_n is generated by products of pairs of such transpositions,
and any such product is either a 3-cycle or the identity.  So, it is
enough to show that all 3-cycles have determinant 1.  Indeed it is
enough to show that *one* 3-cycle has determinant 1 because any two
3-cycles are conjugate in A_6 (see below).  But we already constructed
such a 3-cycle, the coordinate permutation of our hyperoval consisting of
the three unit vectors plus (1:1:1), (1:a:b), (1:b:a), and this linear
transformation visibly has determinant 1.  [We could also have used
diag(1,a,b), which together with that coordinate permutation yields a
3-Sylow in A_6; more about this later.]

Better proof of (2): The kernel of det: A_6 --> F_4^* is normal
and of index 1 or 3.  But A_6 is simple, and of order >3.
Therefore ker(det) is all of A_6, as claimed.


But how do we know that A_6 is simple?  In fact we prove that A_n is
simple for n >= 5.  We follow Rotman... [including conjugacy of all
3-cycles in A_n for n >= 5, promised above for n=6.  NB for our
purposes it's enough that for any 3-cycles c and c' either c or c^{-1}
is conjugacte to c' in A_n, and that's easier and true even for
n=3 and n=4 (and vacuously for n<3...).  Further simplification:
at the top of page 51, instead of citing a "second isomorphism theorem"
show directly that if H is a normal subgroup of G and F is any subgroup
then the intersection of H with F is normal in F.]  Similar techniques
are used to prove the simplicity of the other families, as we'll see
for PSL_n(F_q).