March 31: Simplicity of M_12 and M_24;
start on simplicity of PSL_n(F_q) [n>2 or n=2 and q>3]

Theorem: M_12 and M_24 are simple.

Proof: In fact we'll show that if G is a 5-transitive permutation group
of any finite set X (of size at least 5), and the point stabilizer G_x
is simple, then so is G.  Note that by transitivity all G_x are conjugate
in G and thus isomorphic.  For M_12 and M_24 the point stabilizers are
M_11 and M_23, whose simplicity we proved last time.  Alas it is now known
(by CT = the classification theorem of finite simple groups) that the only
further examples are A_n with n >= 7, which we've already proved simple
in a different (though related) way.

Let H be a normal subgroup of G other than {1} and G itself.
We claim H is transitive.  Let h be any non-identity element,
so h(x_0)=y_0 for some distinct x_0,y_0.  We show that for any
distinct x,y there is a conjugate h' of h in G such that h(x)=y.
Indeed since G is 2-transitive we can find g such that g(x)=x_0
and g(y) = y_0.  Then g^{-1} h g(x) = g^{-1} h(x_0) = g^{-1} y_0 = y.

Now the intersection of H with G_x is normal (because the
intersection of H with _any_ subgroup G' is normal in G').
So by hypothesis this intersection is either {1} or all of G_x.
Since H is transitive, the latter case implies |H| = |X| |G_x| = |G|
so |H|=|G|.  So we're in the former case and |H|=|X|, whence H acts
simply transitively.  (Note that so far we've used only 2-transitivity
of G_x, and there are still examples such as the 4-group in S_4.)

But (as we did with A_n for n>6) we can use high-order transitivity
to find a nontrivial commutator in H that has a fixed point,
and that will give our desired contradiction.  Let h be a
non-identity element of H, so h(x_0)=y_0 as before, and choose
x_1 distinct from x_0 and h^{-1}(x_0) (which is possible since |X|>2).
Then y_1 := h(x_1) is distinct from x_0, x_1, and y_0 (from x_0 by
assumption, from y_0 because h is injective, and from x_1 because
H is simply transitive).  But by 4-transitivity we can then find,
for all distinct x,y,x',y' in X, a G-conjugate of h that takes 
x to y and x' to y'.  Once |X|>4 we can obtain a contradiction by
setting (x,y,x')=(x0,y0,x1) but y' not equal y1: then multiplying our
conjugate by h^{-1} yields a no-identity element of H (a commutator,
as promised) that has a fixed point.  This completes the proof.

Corollary: All the permutations in M_12 and M_24 are even.

Proof: Else the even permutations would form a normal subgroup
of index 2 (think about it -- it's Exercise 3.22 in Rotman),
contradicting the fact that M_12 and M_24 are simple
(and not of order 2).

Remark: The same is thus true also of the subgroups M_11 and M_23,
and also M_22 and M_21 = PSL_3(F_4).  It is not true of Aut(D_22)
and Aut(Pi_4), because those groups contain automorphisms that
come from elements of M_24 that switch \infty_1 and \infty_2.



So how do we prove M_22 simple?  That contains the 2-transitive simple
subgroup M_21 = PSL_3(F_4).  As Rotman shows (Theorem 9.24), 2-transitivity
suffices to prove that a nontrivial normal subgroup H must be an
"elementary abelian" group, i.e. (Z/lZ)^r for some prime l and
integer r>0 -- and this will be enough because 21 is not a prime power.
However we must first prove that PSL_3(F_4) is simple -- and that's
the first case to which Chapman's criterion does not apply, again
because 21 is not a prime.  We next turn to the simplicity of PSL_n(F_q)
(except PSL_2(F_2) and PSL_2(F_3) which are the non-simple groups
S_3 and A_4).