>>> So, in summary, G *definitely* has subgroups for every prime factor, 
>>> and *might* have subgroups for some or all of the composite factors 
>>> as well?
>>
>> I believe that is correct.  In addition I think that G must have a 
>> subgroup for every prime power which still divides the order of G. 
>> Weather or not the composite factors have associated subgroups will 
>> indeed depend on the particular group G represents.
>>
>> As an example consider the group of rotations of a tetrahedron.  It 
>> has 12 elements but no subgroup of order 6.
> 
> http://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)
> 
> It seems that for a *commutative* group, my original statements are 
> correct. I found many other interesting properties, but failed to find 
> any confirmation of your assertion about prime subgroup orders...

Wait, here we go:

http://en.wikipedia.org/wiki/Sylow%27s_theorem

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