April 5: Introduction to subgroups of PSL_2(F) and Galois' theorem;
the identification of PSL_2(F_7) with Aut(Pi_2) = GL_3(F_2)

For the groups PSL_2(F) we can describe not only normal subgroups
(as we saw the only nontrivial examples are for |F|=2 or 3)
but all subgroups, and this is part of a beautiful story that
starts with Galois -- or arguably with Greek mathematics, via
the connection with finite subgroups of SO_3(R) -- and turns up
in a remarkable variety of mathematical contexts.

We introduce this topic via Galois' theorem on large subgroups H of
G = PSL_2(F_p), i.e. subgroups whose index [G:H] is small (without
being as small as 1).  For any group G there is a close connection
between transitive actions of G and small-index subgroups, and Galois'
theorem can be stated that way too.

The general picture is as follows.  Let G be a finite group acting
transitively on a set X.  Then the point stabilizer is a subgroup of G
with index |X|.  Conversely let H be a subgroup of G.  Then H has
[G:H] cosets gH, and those cosets are permuted transitively (from
the left) by G (and indeed if we started with a transitive action of
G on X and took H to be a point stabilizer then the cosets could be
identified bijectively with elements of X).  Let K be the kernel of this
permutation representation of G on the cosets.  Thus K consists of all
k in G such that kgH=gH for all g in G.  This condition is equivalent to
g^{-1}kg H = H, i.e. K consists of all k all of whose conjugates are in H.
So K is the intersection of all the G-conjugates gHg^{-1} of H.
This intersection is a normal subgroup of G (this can be checked
directly from the definition, and also follows from the fact that K is
the kernel of a group homomorphism); and K is strictly smaller than G
iff H is.  In particular, if G is simple then K is trivial and G acts
faithfully on the [G:H] cosets, i.e. is a subgroup of the symmetric
group of permutations of [G:H] objects.

When  G = PSL_2(F_q), the action on P^1(F_q) has a point stabilizer
of index q+1 (and we saw last time it's the "a^2 x + b  group").

Theorem (Galois): If q is prime then any proper subgroup of PSL_2(F_q)
has index at least q+1 except when q is 2, 3, 5, 7, or 11, when the
minimal index is q.

Remark: We shall see that the theorem holds even with q is a prime power
with the single further exception that if q=9 then the minimal index is 6.

Today we shall show for prime q that one cannot get below q (that's easy)
and describe the five exceptional cases, all of which have to do with
designs or groups that we've seen already.

Let G=PSL_2(F_q).  For q=2 and q=3 we know already that G is isomorphic
with S_3 and A_4 respectively.  For S_3, clearly the minimal index of
a proper subgroup is 2, attained uniquely by the normal subgroup A_3.
For A_4, there is no subgroup of index 2 (such a group would be normal
but the conjugacy class sizes 1,3,4,4 contain no subsum of 6) and we
know one of index 3 (the normal 4-group), which is unique (its complement
consists entirely of elements of order 3, so cannot be contained in
a group of order 4).

Assume q>3.  Then G is simple.  It follows that if G has a normal
subgroup of index n then G is a subgroup of the symmetric group S_n
of order n.  But then the order of G is a factor of |S_n| = n!.
But if q is an odd prime then |G| = (q^3-q)/2 is a multiple of q,
whence n is at least q.  Since we already know a subgroup of index q+1
the only remaining question is whether there is one of index q.

(Note that this argument fails when q is a nontrivial prime power.
For q=9 the formula (q^3-q)/2 still works, and gives 360=6!/2, so once
we find an index-6 subgroup it will follow immediately that PSL_2(F_9)
is isomorphic with A_6 !)

For q=5,7,11 the index-q subgroups are isomorphic with A_4, S_4, A_5
respectively -- check that their sizes 12, 24, 60 are the correct
(q^2-1)/2 in each case.  The fact that these are the rotation groups
of the regular solids is no accident, and the fact that there's no such
solid beyond the icosahedron and dodecahedron is closely related with
our approach to Galois' result that there are no examples past q=11.

As with q=2 and q=3 the maps G --> S_q for q=5,7,11 tell us something
special about each of these groups G.  Since G is simple, the image
must be in A_q.  For q=5, the size of A_q is barely enough to
accommodate G, and we recover the isomorphism between G and A_5.
For q=7 and q=11 we get G as proper subgroup of A_q, but one that
respects an interesting structure.  For q=7 this is a Pi_2 structure,
and gives us the identification of G with Aut(Pi_2) -- yes, we're
identifying the group structures of 3x3 matrices mod 2 with certain
2x2 matrices mod 7 (mod {1,-1})!  For q=11 it turns out to be the
Paley biplane -- you've already seen it has 660 symmetries with
point stabilizer A_5.  In each of the q=7 and q=11 cases there are
actually two conjugacy classes of index-q subgroups that are switched by
elements of PGL_2(F_q) not in PSL_2(F_q).  This gives rise to the
duality of the corresponding design.  We'll also see the q=11 case
(PSL_2(F_11) as a transitive subgroup of A_11), and also q=9
(PSL_2(F_9) = A_6), in our description of how M_12 fits in M_24.

For the rest of today's lecture we give a design-theoretic proof of
the identification of PSL_2(F_7) with GL_3(F_2), using Pi_2 and the
3-(8,4,1) design (and the simplicity of PSL_2(F_7), proved last week).