From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: subgroups of SO(n) acting on S^{n-1}
Date: 3 Feb 2000 16:57:55 GMT
Newsgroups: sci.math
Summary: [missing]

In article <86u5fi$gk0$1@gannett.math.niu.edu>,  I wrote
> Keep in mind that there are several questions here. 
[... including]
>"classify the finite subgroups of  O(n)  and  SO(n)  up to conjugacy".

In article <871094$ev4$1@nnrp1.deja.com>,
David Bernier  <bernier@my-deja.com> wrote:
>If G and H are subgroups of SO(n), I consider them equivalent
>if there is a fixed element t in SO(n) such that
>H = t*G*(t^{-1}) .  Does this correspond to the idea of classifying
>up to conjugacy?

Yes, exactly. You've written the definition of what it means for two
subgroups of a group to be conjugate.

After I wrote,
> One may in fact
> classify the finite subgroups of  SO(n)  for any  n  given
> sufficient patience. As in the case n=3 there may be infinitely many
> such groups but there will be an abelian normal subgroup of bounded index
> and bounded rank.

Bernier wrote
>An illustration for n=3 would help to clarify the last point for me.

Finite, or indeed finitely-generated abelian groups may be written as
a product of cyclic group  Z/(n_i)  with each  n_{i+1}  dividing  n_i ;
the number of cyclic factors is what I meant by  "rank". Such a subgroup
of  SO(n)  is contained in one of the maximal abelian subgroups of
the matrix group, known as tori. (They're a little easier to visualize
over an algebraically closed field like the complex numbers  C: the
maximal tori in the unitary group are conjugate to the group of
diagonal matrices all of whose entries are of norm  1.) In particular,
the abelian subgroups must fit inside a product of  n  copies of the
circle group, and so must have rank at most  n.  Actually the situation
for SO(3)  is much more restrictive; the abelian subgroups are cyclic
except for copies of  Z/2 x Z/2.

So the finite subgroups of  SO(3)  fall into two classes: those with a
normal abelian subgroup of rank 1 (cyclic) and those with rank 0 (the 
trivial group) In each case, there is an upper bound on the index of this
subgroup: in the first case the index is  2  (the groups are either
themselves cyclic or else dihedral) and in the second case the index
(which is then actually the group order) is at most 60.

In some sense the "right" way to talk about all this is in the language
of  Lie groups; you pass from the tori to their normalizers (quotients
are the Weyl groups) and so on. I think this material in the context
you first asked it is due to Jordan, maybe 1880s?

>I think any finite group of order n is isomorphic to some
>subgroup of SO(2n).  The basic idea is to view G as acting on
>itself,

Yes, this is Cayley's theorem 

>[ M | 0 ]
>[---|---]
>[ 0 | M ]

SO(n+1)  is sufficient, as one may replace an orthogonal matrix  M  with
the matrix

[ M |   0    ]
[---|--------]
[ 0 | det(M) ]

of determinant  1.

dave