From: Nick Halloway <snowe@coyote.rain.org>
Subject: Re: structure of arbitrary abelian groups
Date: 2 Jan 2000 20:47:28 GMT
Newsgroups: sci.math
Summary: direct product versus direct sum

I wrote:
: On 30 Dec 1999, Dave Rusin wrote:

:> If  G  is an abelian group then the collection of its elements of finite
:> order forms a subgroup, its  torsion subgroup  T ; G/T  is then torsion-free,
:> But already the nicer result you would like to be true -- that  G  is
:> in fact the direct sum of  T  and a torsion-free subsgroup  H  of  G  -- is
:> false in general. An example of the failure would be the direct product
:>   (Z/2Z) x (Z/4Z) x (Z/8Z) x ...
:> whose torsion subgroup is the direct _sum_ of the cyclic factors.
:> These 'mixed' groups (for which T does not split) make it frustratingly
:> difficult to provide good structure theorems.

: I think your weird construction there actually splits into a direct
: product of infinite and torsion subgroups.  

I take it back.  That group can't be a direct product with its torsion 
subgroup.  Suppose you found an infinite subgroup H of G to be the other 
factor in the direct product.  H can't have any element that isn't a 
square but is not a square in only finitely many coordinates.  If it 
did have such an element h, you would have z^2 = h*s for s of finite 
order.  Then if G were a direct product, z = h'*s', so z^2 = h'^2 * s'^2,
so h'^2 would have to be a square.  

So, suppose z = (a, a, a^2, a^2, a^2^2, a^2^2, a^2^3, a^2^3, ...).  

If z = h*s for h in H, s of finite order, the h part has the same 
order as z after finitely many coordinates.  So, the h part has increasing
order.  If h isn't a square it would be a counterexample.  If h is a 
square, say h = y^2, then if y isn't in H, y*h' is finite order for 
some h' in H.  Since y^2*h'^2 is in H, y^2*h'^2 = 1.  So h has a 
square root in H.  So, you could keep on taking square roots of h 
until you got an element of H that isn't a square but isn't a square
in only finitely many coordinates.  

But you could complete G so it is a direct product, the product
A_2 x A_2 x ... is a direct product of its torsion subgroup with a 
maximal subgroup H with no nonidentity finite order elements.  G/H has 
only elements of finite order:  if x is not in H, x^m*h is of finite order
for some h in H.  Similarly G/H has no elements of odd order so all
elements have order a power of 2.  

If S_k is the subgroup of elements of G with order dividing 2^k, then 
the elements of order 2^k over H are in H*S_k:  by induction.  assume
true for k+1.

Suppose x is not in H, x^2^(k+1) is in H.  x^2 = h*s_k for some s_k in 
S_k by induction.  h has a square root in G, so h has a square root h' 
in H.  So, s_k is also a square, of s_(k+1) in S_(k+1).  So 
x = h'*s_(k+1).