Group of order 720
rsv...@gmail.com
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.
Thanks.
Derek Holt
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,
we easily get a contradiction.
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 >.
rsv...@gmail.com
Kira Yamato
> rsv...@gmail.com 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.
I have some questions about the proof.
>
>
> 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)
Just filling so details here.
Ok. So, let P be a fixed element in Syl_3(G). You are considering the
action of P under conjugation on Syl_3(G). Let N_P(P') be the
stabilizer of P' in Syl_3(G). Then N_P(P') is a subgroup of G, so it
must be of order 1,3,or 9. The orbit of P' in Syl_3(G) has length
(1) |O_{P'}| = |P : N_G(P')|.
So, any orbit will be of size 1, 3 or 9.
If P fixes every points of Syl_3(G), then we can use Cayley's theorem
to pick a subgroup Q of P of order 3.
If P does not fix every points of Syl_3(G), then there must be some
orbit of length > 1.
If there is an orbit of length 3, then we pick Q to be the
corresponding stabilizer group, which by (1) will be of order 3.
If there is no orbit of length 3, we must have some orbits of length 9.
But the possibilities for 40 - 9k are
40-9 = 31
40-18 = 22
40-27 = 13
40-36 = 4
and these must correspond to number of orbits of length 1. But the
kernel of an action is a subgroup, so it must divide the order of P.
But none of the four numbers above divides 9. This is a contradiction.
So, there must be an orbit of length 3, which corresponding to a
stabilizer subgroup Q of P of order 3.
So far so good.
> such that N := N_G(Q) has more than 1 Sylow 3-subgroup.
I have a question here. It's probably due to my lack of knowledge
about p-groups.
Ok, so Q is a subgroup of P a p-group. I know that naturalizers of
subgroups of p-groups always grow, so
Q < N.
But how do you conclude that it must grow so large as to contain more
than 1 Syl_3 subgroup?
> So it has at least 4, and we get |N| = 36 or 72.
And how did you conclude that it must contain at least 4? And why
would the order then be 36 or 72? Can you show the arithmetics please?
I'll probably have more questions about the remaining proof. Sorry for
asking so much, but I really want to learn this.
Thanks.
--
-kira
Kira Yamato
> rsv...@gmail.com 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)
Just filling the details here.
Ok. So, let P be a fixed element in Syl_3(G). You are considering the
action of P under conjugation on Syl_3(G). N_P(P') is the stabilizer
of P' in Syl_3(G). N_P(P') is a subgroup of P; so it must be of order
1,3,or 9. The orbit of P' in Syl_3(G) has length
(1) |O_{P'}| = |P : N_P(P')|.
So, any orbit will be of size 1, 3 or 9.
If P fixes every points of Syl_3(G), then we can use Cayley's theorem
to pick a subgroup P of order 3 to be Q.
If P does not fix every points of Syl_3(G), then there must be some
orbit of length > 1.
If there is an orbit of length 3, then we pick Q to be the
corresponding stabilizer group, which by (1) will be of order 3.
If there is no orbit of length 3, we must have some orbits of length 9.
But the possibilities for 40 - 9k are
40-9 = 31
40-18 = 22
40-27 = 13
40-36 = 4
and these must correspond to number of orbits of length 1. But the
kernel of an action is a subgroup, so it must divide the order of P.
But none of the four numbers above divides 9. This is a contradiction.
So, there must be an orbit of length 3, which corresponding to a
stabilizer subgroup Q of P of order 3.
So far so good.
> such that N := N_G(Q) has more than 1 Sylow 3-subgroup.
I have a question here. It's probably due to my lack of knowledge
about p-groups.
Ok, so Q is a subgroup of P a p-group. I know that naturalizers of
subgroups of p-groups always grow, so
Q < N.
But how do you conclude that it must grow so large as to contain more
than 1 Syl_3 subgroup?
> So it has at least 4, and we get |N| = 36 or 72.
And how did you conclude that it must contain at least 4? And why
would the order then be 36 or 72? Can you show the arithmetics please?
>
--
-kira
Kira Yamato
confusing. Please see the other thread with the corrections.
Derek Holt
Sorry, these sorts of arguments arise frequently when analyzing simple
groups of various orders, but they can be difficult to follow in
isolation!
What we have shown so far is that the group Q of order 3 fixes more
than one point in the conjugation action on the Sylow 3-subgroups. The
stabilizer of each point is the normalizer of some Sylow 3-subgroup of
G. So we have shown that Q normalizes more than one Sylow 3-subgroup.
But if a p-subgroup of G normalizes a Sylow p-subgroup, then it must be
contained in that Sylow p-subgroup. So Q is contained in more than one
Sylow 3-subgroup of G. But the Sylow 3-subgroups have order 9 and so
are abelian, so each of these Sylow 3-subgroups normalizes Q. Hence the
normalizer N of Q contains more than one Sylow 3-subgroup.
So N has at least 4 Sylow 3-subgroups, and hence it is has order at
least 36. Also |N| is a divisor of |G| and is divisible by 9, so the
possibilities are 36, 45, 72, 90, 144, 180, 360.
Groups of order 45 and 90 have a unique Sylow 3-subgroup, so they are
ruled out. Also if the order was 144, the index would be at most 5,
which is clearly impossible. That leaves 36 and 72.
Derek Holt.
Jim Heckman
wrote in message <2006123019443275249-kirakun@earthlinknet>:
> On 2006-12-30 07:27:21 -0500, "Derek Holt" <ma...@warwick.ac.uk> said:
[...]
FYI, here's a much simpler way to see that there must be an orbit
of length 3.
The conjugation action of P on Syl_3(G) must contain exactly one
orbit of length 1, namely P itself, since otherwise P would
normalize another Sylow 3-subgroup P', and <P,P'> would be a
3-subgroup larger than a Sylow 3-subgroup, which is impossible.
And since (40-1) <> 0 (mod 9), not all of the other orbits can have
length 9, so ...
[...]
--
Jim Heckman
Giggling Volcanic Under-shorts
which bring it...
Wait, you're a Google-posting wool-wit with a gmail address.
Never mind.
Kira Yamato
<rot13(reply-to)@none.invalid> said:
Ah... Thank you. In fact, yours is the correct way of showing
existence of Q, since I actually made a mistake in my original attempt:
>> If there is no orbit of length 3, we must have some orbits of length 9.
>> But the possibilities for 40 - 9k are
>> 40-9 = 31
>> 40-18 = 22
>> 40-27 = 13
>> 40-36 = 4
>> and these must correspond to number of orbits of length 1. But the
>> kernel of an action is a subgroup, so it must divide the order of P.
First, it has nothing to do with the kernel of the action. In fact,
the assumption that there is an orbit of length 9 dictates that the
kernel must be trivial since the action group P has order 9 also.
Moreover, in the general case of a group G acting on a set X, the
number of orbits of length 1 does not need to divide the order of the
group at all, since I can always add a new element x into X and declare
that
xg = x for all g in G.
I got crosswired with the class equation. It works for the class
equation because X happens to be G itself and the set of orbits of
length 1 in that case happens to be Z(G), which is a subgroup.
So, yours is the correct way to show the existence of this subgroup Q.
>> But none of the four numbers above divides 9. This is a contradiction.
>> So, there must be an orbit of length 3, which corresponding to a
>> stabilizer subgroup Q of P of order 3.
Thanks.
--
-kira