Re: Simple group of order 60

From: Van Jacques (vanjac12_at_yahoo.com)
Date: 11/03/04 

Date: 3 Nov 2004 14:19:43 -0800

Jyrki Lahtonen wrote:
> Van Jacques wrote:
> > Jyrki Lahtonen wrote:
> >
> > The only thing I couldn't understand is why there are no elements
of
> > order 4
>
> There was no room! All the 60 elements had been accounted for, see
below.

But it seems to me that it was assumed that all the 15 elements had
order 2,
not shown.

>
> > I don't understand what you mean by "a non-trivial homomorphic
image of
> > order at most 24", or what you are arguing here.
> > Perhaps you are talking about the action of G by conjugation
> > on the 4 Sylow 3-subgps, which would map G to a subgp of S_4.
> > If so, let f: G --> S_4 , where f_g: H_i --> f_g(H_i) = g H_i g^-1
> > for each i, so f(G) < S_4, and f(G) =~ G/ker(f)
> > where ker(f) < G is a non-trivial normal subgroup of G ==> G not
> > simple.
> > Is this what you were talking about?
>
> Correct.
>
> > So n_3 = 10.
> >
> >
>
> [snipped]
>
> > You say "As there are no elements of order 4 in G"
> > (I put the 4 in, but that's what you meant to say),
> > I know this is true, but why is it true? I don't see how to exclude
> > this possibility.
>
> Yes, that's what I meant. Nice that you can fill in a crucial
parameter!
> There simply is no room for element of order 4: We know of 1 element
> of order 1, 24 of order 5, 20 of order 3. There must be at least 1
> of order 2, but then its entire conjugacy class of 15 must consist
> of elements of order 2. 1+15+20+24=60 => no elements of order 4, (and
> only a single conjugacy class of elements of order 2).

I fear I am missing something simple, but why must the conjugacy class
of the elements
of order 2 have 15 elements? Why not 5 elements? And then another class
with 10 elements
of order 4?
We have n_2 = 5 (can't be 3, and 15 would give too many elements).

All of Sylow subgps are either Z_4 or else all of them are Z_2 x Z_2.

If they are Z_4, there are 5 elements of order 2 and 10 of order 4.

If all elements with |x| = 4 are in the same class,
[G:C(x)] = 10 ; |C(x)| = 6 , where C(x) = centralizer of x.

If N_1 = <x>, then N_1 < C(x) accounts for 4 ef time elements of C(x).
I have a hard time seeing where the other 2 elements of C(x) would
come from, and perhaps this would make |C(x)| = 6 impossible,
but I fear I may be going astray here.

The elements of order 2 would be of form x^2, and C(x) < C(x^2),

[G:C(x^2)] = 5 ; |C(x^2)| = 12 . All these divide 60, so I have

failed to rule out elements of order 4.

I am probably not thinking in the right direction at all.
I do know that "There must be at least 1 element of order 2", and that
all elements in a conj. class have the same order (though not vice
versa).

But I don't see how this has anything to do with why there can't be
elements of order
4, or why the 15 remaining elements can't break up into 2 conj. classes
(or more)
with 5 elements of order 2 in one class, and 10 elements of order 4 in
the other,
for a total of 60 elements.

Van

> Cheers,
>
> Jyrki

Relevant Pages

  • Re: The group GL_3(2)
    ... Van Jacques wrote: ... _provided the eigenvalues are in the field k you are working in_, ... > We make a list of possible characteristic and minimal polynomials: ... > note that this conjugacy class contains all the remaining elements. ...
    (sci.math)
  • Re: Simple group of order 60
    ... Van Jacques wrote: ... Nice that you can fill in a crucial parameter! ... only a single conjugacy class of elements of order 2). ...
    (sci.math)