Abstract Algebra

[Lofty]

We have an abelian group G. Every element of G is of finite order but the group is not necessarily finite. Let p be some prime, then Ap is the subgroup containing all the elements of order p^x for any x and Bp is the subgroup containing all the elements that have order that cannot be divided by p. I need to show that G is isomorphic to Ap x Bp.
Also Sylow's theorems are off limits.

 

[TheBB]

Well I guess the isomorphism in question is

R_p: A_p \times B_p \to G
R_p(a,b) = ab

To show that it's an isomorphism you need that the inverse image of the identity is only the identity in A_p\times B_p (easy), and that it's onto G (harder).

For the latter, consider that every g\in G with order N=p^cq can be written
g = g^{p^cq+1}.
Also note that g^{p^c}\in B_p and g^q\in A_p.

 

[Stefan]

Funny how much the internet changed the way people do their homework.

 

[Lofty]

Well all my classmates that I talked to where also having a hard time on this problem and I knew there were people here who are good at maths.
Thanks for the help. I turned this problem in already with not much deserving of even partial credit. But it was only one of about ten questions like it and I got all the others.