Finitely Generated Abelian Groups

[Combinatorics]

Homework Statement

Let p be prime and let b_1 ,...,b_k be non-negative integers. Show that if :
G \simeq (\mathbb{Z} / p )^{b_1} \oplus ... \oplus (\mathbb{Z} / p^k ) ^ {b_k}
then the integers b_i are uniquely determined by G . (Hint: consider the kernel of the homomorphism f_i :G \to G that is multiplication by p^i . Show that f_1 , f_2 determine b_1.
Proceed similarly )

Homework Equations

The question, together with the theorem it should help me prove is attached (this question should help me prove the uniqueness part of the theorem)

The Attempt at a Solution

I've tried using the hint, and considering the homomorphism f_1 : G \to G that is defined by f_1 (x ) = px . Its kernel is ker f_1 = (\mathbb{Z}_p ) ^{b_1} \bigoplus (\mathbb{Z}_p) ^ {b_2} ....
But how does it help me? Am I right in my calculation of the kernel? Hope someone will be able to help me

Thanks in advance !

 

[morphism]

The hint is essentially leading you to consider a proof by induction on k. Notice that $$ \ker f_1 = (\mathbb Z/p)^{b_1} \oplus (\mathbb Z/p)^{b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots = (\mathbb Z/p)^{b_1+b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots. $$

 

[Combinatorics]

The hint is essentially leading you to consider a proof by induction on k. Notice that $$ \ker f_1 = (\mathbb Z/p)^{b_1} \oplus (\mathbb Z/p)^{b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots = (\mathbb Z/p)^{b_1+b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots. $$

Can you please explain how did you get that this is the kernel?

Thanks a lot !

 

[morphism]

Well, if px=0 mod p^i, then p^i divides px, hence p^{i-1} divides x, i.e. x=0 mod p^i.