I want to prove the following result:
Let $p$ be a prime number and $A$ an abelian $p-$group. for $k\ge 1$ define
$$\phi_k(A)=\bigcap_{n\ge k}p^{n-k}A[p^n]$$
where $A[m]=\{a\in A: am=0\}$ If $A$ is artinian we have the quivalence:
$\phi_1(A)=\left\{0\right\}$ if and only if $A$ is finite
This is what i think to use $$A[p]⊃pA[p^2]⊃p^2A[p^3]⊃…$$ but i don't know how to answer using the fact that A is artinian, would you please help me to do so?
Thanks for your help.
For the first implication (only if), I guess you could use the fact that $A$ is a $p$-group. Let $k$ be the rank at which the sequence you mentionned is staionnary ($A$ is artinian). Then $p^{k-1}A[p^k]=\{0\}$. Suppose now $A$ has a element of order $p^n$ with $n > k$ ($p^n a = 0$ and $ma \neq 0$ for $m < p^n$), we have of course $a\in A[p^n]$. Since $p^{n-1}A[p^n] \subset p^{k-1}A[p^k]$, then $p^{n-1}a \in p^{k-1}A[p^k] = \{0\}$, which means $p^{n-1}a = 0$ which contradicts the fact that the order of $a$ is $p^n$.
For the second implication, we have that $p^{n}A[p^{n+1}]=\{0\}$ where $p^n$ if the order of $A$, which proves that $\phi_1(A)=\{0\}$.
[EDIT] My "proof" is incomplete, I've only shown that $A$ has a finite exponent, which allows to conclude it is a direct product of cyclic groups. But it is not sufficient to conclude it's finite.