Feb 26: Groups plan; uniqueness and automorphism group of
Pi_2 (again) and Pi_3


The second part of this course focuses on (usually finite) groups that
can be studied using some of the designs or strongly regular graphs
we've encountered and/or can shed light on them.  The groups we will
focus on are:

@ The symmetric and alternating groups S_n and A_n;

@ The linear and groups GL_n(k), SL_n(k) and projective linear groups
PGL_n(k), PSL_n(k) (usually k is finite); and

@ Mathieu's sporadic highly transitive subgroups M_n of S_n
(n=11, 12, 22, 23, 24).

In particular we'll prove the simplicity of A_n (n>4) and
PSL_n(k) (n>2, or n=2 and #k>3), explaining how simplicity fails
in small cases like A_4 and PSL_2(F_3), and also of M_n (which will
lead us to the corresponding Steiner systems); and use some of our
designs and graphs to explain some other remarkable behavior like
coincidences among some of these groups (listed below) and the
automorphism group of S_6 (which, uniquely among all S_n, is
larger than its group of inner automorphisms).  We'll temporarily
leave our textbook, relying on my text and TeX notes and some other
sources, except for using some of Chapter 6 (on Aut(S_6)).

The isomorphisms we'll address are as follows (listed according to
group size):

   12  A_4 =~= PSL_2(F_3)
'  24  S_4 =~= PGL_2(F_3)
   60  A_5 =~= (P)SL_2(F_4) =~= PSL_2(F_5)
' 120  S_5 =~= \Sigma L_2(F_4) =~= PGL_2(F_5)
  168  (P)SL_3(F_2) =~= PSL_2(F_7)
' 336  Aut((P)SL_3(F_2)) =~= PGL_2(F_7)
  360  A_6 =~= PSL_2(F_9)
20160  A_8 =~= (P)SL_4(F_2)

Each ' denotes a group G' that is not simple but is Aut(G) where
G is the simple group in the preceding line [G':G] = 2.
For G=(P)SL_3(F_2), an outer automorphism is inverse transpose,
corresponding to the action of G on the dual 3-dimensional space over F_2.
As we'll see Aut(A_6) is more complicated, containing A_6 with index 4
rather than 2; in the F_9 picture it's "P\Gamma L_2(F_9)".
"(P)SL_n(k)" means the group is both PSL_n(k) and SL_n(k),
which as we'll see happens whenever the group of nth roots of unity
in k (i.e. the n-torsion group of k*) is trivial, which in turn
means n is relatively prime to |k*| = |k|-1.  The full list of
exceptional isomorphisms is about thrice as long, but includes
orthogonal, unitary, and symplectic groups, which we won't get to
in Math 155.  The ATLAS lists them all (though not, I think,
in a single place).


We'll introduce our approach by proving the uniqueness up to
automorphisms of the projective planes of order at most 5
and identifying their automorphism groups.  In each case we
do this by showing that all (hyper)ovals are equivalent, i.e.
by showing that a combinatorial projective plane Pi must have a
(hyper)oval O and that Pi can be reconstructed from O.

Today we review Pi_2 and prove the result for Pi_3.

For Pi_2 we have:

Theorem: Let Pi be a projective plane of order 2.  Then Pi has
7 hyperovals, namely the line complements.  If O is a hyperoval of Pi,
and O' is a hyperoval of some other projective plane Pi' of order 2,
then any bijection f: O --> O' extends uniquely to an isomorphism
Pi --> Pi'.

Proof: The complement of a line l is a hyperoval because it has the
correct size (7-3 = 4 = 2+2) and meets every line l' in either 0 or 2
points (the former iff l'=l).  Conversely any hyperoval O has Bin(4,2)=6
secants, so 7-6=1 passant line l, and since the complement of l has size
4 = |O| it must be the same as O.

Now given f: O --> O' we know the images in Pi of the four points of O
and the six secants; thus the remaining line of Pi, the unique passant
of O, must go to the unique passant of O'.  Since we know the images of
all lines, the images of all points are determined (since each point
is the unique intersection of the 3 lines containing it, or indeed of
any 2 of them).  This shows that the extension of f, if it exists,
is unique.

If O = {p1,p2,p3,p4} we then have the following roster of points and
lines in Pi: the lines are

  6 secants s_{ij} determined by p_i and p_j
  1 unique passant line l
  7 total

  4 points of O: p1, p2, p3, p4
  3 points off O: intersections s_{12} s_{34}, s_{13} s_{24}, s_{14} s_{23}
  7 total

Moreover we know the 3 points on each line: s_{ij} contains s_i, s_j,
and an "off O" point where s_{ij} intersects another secant; and l
contains the three "off O" points.  These satisfy the axioms of a
finite projective plane of order 2, as can be checked either directly
or by observing that we already have one example.  So we can reconstruct
O' from f(p1), f(p2), f(p3), and f(p4) to complete the isomorphism f,
QED.

Corollary: i) Pi_2 is the unique projective plane up to isomorphism.
ii) Aut(Pi_2) = (P)GL_3(F_2) = (P)SL_3(F_2), and this group acts
transitively on hyperovals of Pi_2.

Proof: (i) is clear: fix a hyperoval O in Pi_2; for any projective
plane Pi' of order 2 we can choose any of its 7 hyperovals O',
and any bijection f : O --> O', and extend f to an isomorphism
Pi --> Pi'.

(ii) Take Pi' = Pi_2.  Then there are 7 choices of O and 4! bijections,
whence |Aut(Pi_2)| = 168.  Since Aut(Pi_2) contains (P)GL_3(F_2),
a group of order 168, this must be the full group of automorphisms.
By our theorem these include automorphisms taking any oval O to
any other oval O'.  QED

Moreover: Fix a line l, let O be its complement, and consider the
stabilizer H, which is the subgroup of Aut(Pi_2) consisting of
automorphisms g such that g(O)=O (equivalently: such that g(l)=l).
Our Theorem gives an isomorphism from H to the group S_4 of permutations
of O.  On the other hand, H acts transitively on l; this recovers the
group homomorphism S_4 --> S_3.  Explicitly: if we label the points of O
p_1, p_2, p_3, p_4 then each point p of l is on two secants, and thus
determines a partition of O into two pairs -- and this partition
uniquely determines p as the intersection of the corresponding secants.
But there are only 3 partitions, so each one arises -- and this is
tantamount to the usual construction of the homomorphism S_4 --> S_3
by the action of S_4 on such partitions.


So what happens for Pi_3?

Theorem: Let Pi be a projective plane of order 3.  Then Pi has
234 ovals.  If O is a oval of Pi, and O' is a oval of some other
projective plane Pi' of order 3, then any bijection  f: O --> O'
extends uniquely to an isomorphism Pi --> Pi'.

Proof: We count ordered ovals (p1,p2,p3,p4).  There are:

  13 choices for p1;
  12 = 13-1 choices for p2 given p1;
  9 = 13-4 choices for p3 given p1 and p2
    (namely the complement of the line joining p1 and p2); and
  4 = 13 - 3 - 3*2 choices for p4 given p1, p2, p3
    (namely the complement of the three lines joining two of
    p1, p2, and p3, which are distinct because of how we chose p1,p2,p3)

for a total of 13*12*9*4 = 5616.  Since that's 4! times the number of
unordered ovals, the count is 5616/234 as claimed.

Now fix an ordered oval O = {p1,p2,p3,p4}.  It has four tangents,
call them t1,t2,t3,t4, and six secants, call them s_{i,j} for distinct
i,j in {1,2,3,4}.  Given f: O --> O' we know their images in Pi,
call them

  p1', p2', p3', p4';  t1', t2', t3', t4'; and s'_{i,j}).

There are six pairs of tangents, each determining an intersection point.
We claim that these are al distinct.  Indeed each point of Pi - O
lies on either 0 or 2 tangents -- not 1 or 3 by parity, nor 4 because
(aha) Pi has no hyperovals.  So once two tangents meet at a point
there are no others.  So we've located six more points of Pi
(the intersections of tangent pairs), and thus their images in Pi'.
The remaining 13-4-6 = 3 points of Pi each lie on two secants, so
determine a partition of O into two pairs -- and as in our analysis of
Pi_2 it follows that each of the three partitions arises once.
So we've determined the images of all 13 points, and this fixes
all the line images as well.  This proves that f, if it exists,
is unique.

The remaining 3 lines can be identified as follows.  Consider
the intersection of t1 and t2.  The four lines through it are
t1, t2, s(34), and a passant line l.  What other points are on l?
Not the intersection of t1 with t3 or t4 (because we already know 
where l meets t1), nor the intersection of t2 with t3 or t4
(for the same reason).  But there are only three l's to be shared among
six such points, so l must also contain the intersection of t3 and t4.
Each of the other two points of l lies on two secants.  There are 3 such,
and we cannot use the intersection of s(34) and s(12), because we
already know where each of these secants meets l.  So it must be
the other two.  This completes the description of Pi_3 starting from
a labeled oval; we could now check directly that it satisfies the
axioms of a projective plane but it is simpler to note that we already
have P^2(F_3) so we must have reconstructed it.  So we can complete the
reconstruction of f: Pi --> Pi' knowing O' and the images on O' of
p1, p2, p3, p4, QED.

Corollary: i) Pi_3 is the unique projective plane up to isomorphism.
ii) Aut(Pi_3) = PGL_3(F_3) = PSL_3(F_3), and this group acts
transitively on ovals of Pi_3.

Proof: (i) is clear: fix a hyperoval O in Pi_3; for any projective
plane Pi' of order 3 we can choose any of its 234 hyperovals O',
and any bijection f : O --> O', and extend f to an isomorphism
Pi --> Pi'.

(ii) Take Pi' = Pi_3.  Then there are 234 choices of O and 4! bijections,
whence |Aut(Pi_2)| = 13*234 = 5616.  But Aut(Pi_3) contains PGL_3(F_2),
a group of order 26*24*18/2 = 5616.  So this must be the full group of
automorphisms.  By our theorem these include automorphisms taking any
oval O to any other oval O'.  QED

Remark:  The fact that every point of Pi - O is on 0 or 2 tangents
(while of course every point of O is on exactly 1 tangent) means that
the tangents constitute an oval O* in the dual projective plane Pi*.
The tangents of O* in turn are the points of O, so O**=O as should be
the case for a duality.  This works for a conic over any field not of
characteristic 2 (we already saw that in characteristic 2 the tangents
to an oval all meet at a point).