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# Is the Monster a congruence subgroup?

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### Gerard Westendorp

ongelezen,
8 nov 2008, 21:00:0108-11-2008
aan
To check if the Monster group might be a congruence subgroup of the
modular group, I tried to find a congruence subgroup Gamma(N), that has
the same order as the monster group.

If N has prime decomposition (p1^a1)* (p2^a2) * ... = (p_i^a_i)
(Multiplying over repeated index as in Einstein convention)
then the order of Gamma(N) is

(1/2)*(p_i -1)*(p_i +1)*(p_i^[3a_i -2]).

Using this formula, if
N = 2*3*5*7*11*13*17*19*31*41*47*59*71

then the order of Gamma(N) turns out:
(2^46)*(3^16)*(5^7)*(7^4)*(11^1)*(13^1)*17*19*23*29*31*41*47*59*71

Compare to the order of the Monster:
(2^46)*(3^20)*(5^9)*(7^6)*(11^2)*(13^3)*17*19*23*29*31*41*47*59*71

So it does not quite work, Gamma(N) is slightly too small.
Adding extra powers of p_i appears to make Gamma(N) too big.

However, there are 2 interesting cases with more success.
One is for the Mathieu group M24. If
N = 5*7*23

Then the order of Gamma(N) is
(2^10)*(3^3)*5*7*11*23

This is the correct order for M24.

Also, if
N = 2*3*11

Then the order of Gamma(N) is
(2^6)*(3^3)*5*11

Which fits for M12.

It would be cool if M24 and M12 were congruence subgroups.
Having the correct order is not sufficient for the groups to be equal,
but it seems at least to be an interesting hint.

Now for M23, there is no N that fits. But M23 is a subgroup of M24.

So similarly, maybe there is some N, for which Gamma(N) has the Monster
as a subgroup in some interesting way.

to shed some light here.

Gerard

### jmilne

ongelezen,
10 nov 2008, 07:26:4910-11-2008
aan

A congruence subgroup of Gamma(1)=SL(2,Z) is usually defined to be a
subgroup containing Gamma(N) for some N. Here Gamma(N) is the subgroup
of matrices congruent to the identity matrix modulo N. In particular,
it is infinite.

However, you seem to be actually asking whether Gamma(1)/Gamma(N) can
be an interesting simple group. Obviously, it is not simple if N is
composite. On the other hand Gamma(1)/Gamma(p) is SL(2,F_p), which is
simple modulo its centre {\pm I} (except may be for some small p).

J.S. Milne

### Jacko

ongelezen,
14 nov 2008, 05:16:4414-11-2008
aan

On 9 Nov, 02:00, Gerard Westendorp <west...@xs4all.nl> wrote:
> To check if the Monster group might be a congruence subgroup of the
> modular group, I tried to find a congruence subgroup Gamma(N), that has
> the same order as the monster group.
>
> If N has prime decomposition (p1^a1)* (p2^a2) * ... = (p_i^a_i)
> (Multiplying over repeated index as in Einstein convention)
> then the order of Gamma(N) is
>

> ت ت(1/2)*(p_i -1)*(p_i +1)*(p_i^[3a_i -2]).

the exponent differences are all even except for 11

cheers
jacko

### Jacko

ongelezen,
14 nov 2008, 05:16:4514-11-2008
aan

the other factor is

(Gamma(11*13))^2/(11*2^12)

cheers
jacko

### Jacko

ongelezen,
14 nov 2008, 05:16:4614-11-2008
aan

M = Gamma(N)*Gamma(5)*Gamma(13)^2*Gamma(11)/2^8

It needs checking...

This implies 8 possible divisions of the Gamma(N) to reduce the 2^8

I maybe oing this all wrong but please explain..

cheers
jacko

### Gerard Westendorp

ongelezen,
1 dec 2008, 04:00:0101-12-2008
aan
jmilne wrote:
[..]

> However, you seem to be actually asking whether Gamma(1)/Gamma(N) can
> be an interesting simple group. Obviously, it is not simple if N is
> composite. On the other hand Gamma(1)/Gamma(p) is SL(2,F_p), which is
> simple modulo its centre {\pm I} (except may be for some small p).

Thanks, I missed some elementary lessons of group theory there.

It looks like the Monster is not Gamma(1)/Gamma(N) for some N.
But might it be another quotient of the Modular group?
I read that the Monster is a Hurwitz group, ie it is a quotient
of the (7,3) triangle group. I think the (7,3) triangle group is
in turn a quotient of the modular group; (the (inf,3) tiling).
That would mean the Monster is indeed a quotient of the
Modular group?

A related question:

The group generated by

(1 p) ( 1 0)
(0 1) , (-p 1)

seems to generate a subgroup of the congruence subgroup for p.

Gerard

### J.S. Milne

ongelezen,
2 dec 2008, 03:10:4702-12-2008
aan
On Dec 1, 10:00 pm, Gerard Westendorp <west...@xs4all.nl> wrote:
> It looks like the Monster is not Gamma(1)/Gamma(N) for some N.
> But might it be another quotient of the Modular group?

The quotients of Gamma(1)/{I,-I} are exactly the groups that can be
generated by two elements, one of order 2 and one of order 3.
Apparently, every finite simple group can be generated by two elements
(see Baez's recent post to this forum -- week 272). Perhaps the
experts on finite simple groups know which groups satisfy the
condition.

J.S. Milne

### Peter Mueller

ongelezen,
3 dec 2008, 05:43:0003-12-2008
aan
The monster group is generated by two elements of order 2 and 3,
respectively. Such a pair (with product of order 29) was used by John
Thompson in the 80s to realize the monster as a Galois group over the
rationals.

Most finite simple groups are generated by a pair of elements of order 2
and 3, see Tamburini, M. Chiara; Wilson, John S. On the $(2,3)$-generation
of some classical groups. II. J. Algebra 176 (1995), no. 2, 667--680 and
the references given there.

Peter M"uller (W"urzburg)

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