Re: Simple group of order 60

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

Date: 4 Nov 2004 03:44:50 -0800

Jyrki Lahtonen wrote:
>
> Are we sure that there can't be 15 Sylow 2-subgroups?????
Good point. I would like to look at this, and the overlapping of the
Sylow 2
subgroups.
> Careful here! In theory you might be able to form quite a few 2-Sylow
> subgroups using only 15 non-identity elements, if you cannot exclude
> the possibility of two Sylow subgroups having non-trivial
intersections!!!
> E.g. (not relevant here, but anyway) the group C_2 x C_2 x C_2 x C_2
> (that has exactly 15 elements of order 2) has no less then 35
subgroups
> of order 4. Are we sure that something like that cannot happen here??
No. Does it? I will think about this and post if I find out about it.

> Did you miss the reason for the size of the conjugacy of an
involution
> (=element of order 2) being at least 15? Are you familiar with the
> formula

I didn't miss it, I forgot about it ;-)

> the size of the conjugacy class of x = |G|/|C_G(x)|?

Yes, I know about this.
> In my first post I showed that if x is an involution, then
> |C_G(x)| is at most 4. Consequently the size of its conjugacy class
> is at least 15.

Yes. I reread your 1st post, and its clear that |C_G(x)| = 4 (can't be
2--too many
element then), and thus [G:C_G(x)] = 15 gives all the elements in G.
All 15 elements
are conjugate, thus all are of order 2, and there are no elements of
order 4. OK.

Now for the Sylow 2-subgroups. They are all Z_2 x Z_2 (Klein)
subgroups.

I just reread your argument in your 1st post. They can't intersect for
the reason
you gave; |C_G(x)| = 4. So n_2 = 5.

> Cheers,
>
> Jyrki

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