Math Is Fun Forum

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Clarification needed: p-primary components of an Abelian group

Definition: an Abelian group G is p-primary for any prime p, if for every g in G there is n>0 such that g*p^n = 0

So then given any arbitrary Abelian group, not necessary p-primary, we can define it's p-primary components:

Gp = {g in G : there exist n>0 such that g*p^n = 0} for any prime p.

So that means that for any prime p, an Abelian group has a p-primary component Gp.

So then there is a theorem that says that G is the direct sum (finite!) of it's p-primary components. My confusion is, since for any prime p there is a p-primary component, that means that there are infinitely many p-primary components. In the decomposition of G into the direct sum of it's p-primary components, which of the infinitely many components do you choose for the decomposition into a finite direct sum of components?

I don't know if I'm explaining myself. A group G has infinite components, but there is a finite decomposition into direct sum of those components, how do you choose those finite components from the infinite list?

Thanks.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Clarification needed: p-primary components of an Abelian group

Are you sure that the theorem is not restricted to finite abelian groups G?
If this is the case, then for every prime p that does not divide the order of G, the p-primary component Gp will be trivial.
Hence, we need only take the direct sum over the finitely many p that divide the order of G.

If G is an infinite cyclic group, then Gp is trivial for all primes p, and so G is not a direct sum of p-primary components of G.

is also clearly not a finite direct sum of p-primary components.