# Re: Group of order 720

rsvirk@xxxxxxxxx wrote:
Does anyone know of a reference for the proof that there is no simple
group of order 720? Or can anyone provide a sketch of a proof? I am
down to the cases:

n_2 = 9, 15 or 45
n_3 = 10 or 40
n_5 = 36

where n_p denotes the number of p-sylow subgroups.

I wrote a sketch proof of this some time ago. Let me know if you need
any clarifications. There may be parts of the argument that can be
simplified, but all of the proofs that I have ever seen have proceeded
by showing that G would have to be equal to the Mathieu group M_10,
which is not simple, but of the form A_6.2.

Derek Holt.

Let G be simple of order 720 = 16 x 9 x 5.

By Sylow, |Syl_3(G)| = 1, 4, 16, 10 or 40.
Clearly not 1 or 4.
By Sylow, all groups of order 45 are abelian, so |Syl_3(G)| cannot be
16 by BTT (Burnside's Transfer Theorem).

We need to eliminate |Syl_3(G)| = 40.
If |Syl_3(G)| = 40, P in Syl_3(G) has an orbit of length 3 on Syl_3(G),
so there is a subgroup Q of order 3 (the pointwise stabilizer of this
orbit in P) such that N := N_G(Q) has more than 1 Sylow 3-subgroup.
So it has at least 4, and we get |N| = 36 or 72.

If |N| = 36 then N/Q has order 12 and has 4 Sylow 3-subgroups, so
N/Q = A_4, and since A_4 cannot act non-trivially on Q, Q is central
in N. Hence N has a normal subgroup T of order 4, and |N_G(T)| is
divisible by 8, so strictly contains N. Then the only possibility is
|N_G(T)| = 72, but then Q = O_3(N) is characteristic in N and hence
normal in N_G(T), contradiction, since N = N_G(Q).

So |N| = 72. Since |Aut(Q)| = 2, C(Q) has order at least 36, and a
subgroup R of order 12 in C(Q) must be abelian.
Consider the action of G on the 10 cosets of N. Let Q = <t>.
If t is a single 3-cycle, then by considering conjugates of t,
If t is 2 3-cycles, then an element u of order 2 in R must interchange
those cycles forming a 6-cycle tu. Since the 6-cycle is
self-centralizing
in S_6, an element in R outside of <tu> must fix all 6 points of the
6-cycle, so it must be a single transposition, which is impossible.
If t is 3 3-cycles, then an element of order 2 in R must interchange
2 of the 3-cycles and fix the other pointwise, so consists of 3
2-cycles,
and is an odd permutation, which is impossible in a simple group.

So |Syl_3(G)| = 10. Let P in Syl_3(G), N = N_G(P), so |N| = 72 and
G acts transitively by conjugation on Syl_3(G), which we denote by
{1,2,...,10} with P = 1, and N = G_1 the stabilizer of 1 in G.

If P is cyclic then it must act as a 9-cycle on {2,..,10}. Since
|Aut(P)| = 6, there is an element of order 2 in N which centralizes P,
and there is no way for such an element to act on {2,...,10}.

So P is elementary abelian. If a subgroup Q of P of order 3 fixes more
than one point, then N_G(Q) has more than 1 Sylow 3-subgroup, giving
a contradiction as before. So P acts fixed-point-freely on {2,..,10}.
In fact we can assume that P = < a,b > with
a = (2,3,4)(5,6,7)(8,9,10),
b = (2,5,8)(3,6,9)(4,7,10).
The stabilizer S = N_2 of 2 in N has order 8 and is a Sylow 2-subgroup
of N. Now S is contained in X_2, where X is the normalizer of N in the
symmetric group on {2,..,10}, and X can be identified with Aut(P) =
GL(2,3).
Note that the element (5,8)(6,9)(7,10) of X_2 is an odd permutation and
corresponds to an element of determinant -1 in GL(2,3).
Since SL(2,3) is the unique subgro p of index 2 in GL(2,3), it follows
that
the elements of SL(2,3) correspond to the even permutations in X_2.
So S corresponds to a Sylow 2-subgroup of SL(2,3), which is isomorphic
to Q_8. Since all such Sylow 2-subgroups are conjugate, we may assume
that S = < c,d > with
c = (3,5,4,8)(6,7,10,9),
d = (3,6,4,10)(5,9,8,7).

Note that G is 3-transitive, with no elements fixing more than 2
points.

Now N_G(S) must have order 16 and contain an element e outside of S
containing the cycle (1,2). e must also normalize a subgroup of order
4 in S, which we will take to be <c>. (The argument in the other two
cases, <d> and <cd> is similar.)
By multiplying e by an element of S, we may assume that e fixes the
point 3. Since e fixes at most 2 points, it must invert <c>, and hence
contains the cycle (5,8).
So there are just two possibilities, for e:
(1,2)(5,8)(6,7)(9,10) and (1,2)(5,8)(6,9)(7,10).
For the second of these, b*e fixes 3 points, which is impossible, so
e = (1,2)(5,8)(6,7)(9,10), and we hve

G = < a,b,c,d,e >.

In fact, this is a group of order 720, but it is the group M_10, which
is not simple. The subgroup < a,b,c,e > has order 360.

A proof of the uniqueness of the simple group of order 360 follows
similar lines to this one, and ends up proving that G = < a,b,c,e >.

.