Ulm and Frattini Subgroups Let $A$ be an abelian group. We define $U(A)=\cap (nA), n\in \mathbb N$ be the Ulm subgroup of $A$. The Frattini subgroup of $A$ is $\Phi(A)=\cap(pA)$ ($p\in \mathbb P$). I was trying to show that taking repeatedly (a countable times at most) the Frattini subgroup of $A$, let's call this $\Phi^\omega(A)$, lead us to the Ulm subgroup of $A$.
Clearly, $\Phi^\omega(A)\leq U(A)$, since the $n$th application of $\Phi$ is contained in $\cap (pq...tA)$, $p,q,...,t\in \mathbb P$.
I have trouble showing the other inclusion. If, for instance, $\cap (pqA)$ is contained in $\cap (q(\cap(pA)))$, then all my trouble would be solved.
Any hints or ideas?
Edit: First page,  Second page
 A: Proposition
$\cap\{(pqA)\,|\,p, q\in\mathbb{P}\}=\cap\{q(\cap\{pA\, |\,p\in \mathbb{P}\})\, |\, q\in \mathbb{P}\}$
Proof
Assume for a contradiction that
$\cap\{q(\cap\{pA\, |\,p\in \mathbb{P}\})\, |\, q\in \mathbb{P}\}<\cap\{(pqA)\,|\,p, q\in\mathbb{P}\}$
and let $b\in\cap\{(pqA)\,|\,p, q\in\mathbb{P}\}$ such that $b\notin \cap\{q(\cap\{pA\, |\,p\in \mathbb{P}\})\, |\, q\in \mathbb{P}\}$.
There are $r,q\in \mathbb{P}$ such that $(b=rqc\, \, (c\in A) \implies qc\notin\Phi(G))$ 
or in other words $(b=rqc \implies \exists s \in \mathbb{P} : qc\notin sA)$. 
Obviously $q\neq s$ and by hypothesis we can write $b=rsx=rqc\,\, (x,y\in A)$.
Then $r(qc-sx)=0$, therefore $(qc\notin sA)$ $(qc-sx)\in A[r]-\{ 0\}$.
If $r\neq s$, $A[r]=sA[r]$ and there is a $z\in A[r]$ such that $qc-sx=sz$. So $qc\in sA$, a contradiction.
Now, $r=s$ and the following implication works
$(*)$ $b=qrc\,\, (c\in A) \implies qc\notin rA$
Now, since $r\neq q$, $A[r]=qA[r]$ and it exists an $a\in A[r]$ such that $qc-rx=qa$ (take in mind that $(qc-sx)\in A[r]-\{0\}$).
So we know that $q(c-a)=rx\in rA$ and $b=r^2 x=rq(c-a)$ and this contradicts $(*)$.
$\square$
Theorem
For all $n\in \mathbb{N}$ we have $\Phi^n(G)=U^n(G)$, where
$U^n:= \cap\{(p_1p_2...p_n)A\, |\, (p_1,...,p_n)\in \mathbb{P}^n\}$ $(n\geq1)$
Proof
We argue by induction on $n$. The above proposition proves the case when $n=2$ (it works fine also if we assume $n=1$ as starting point).
Let $n\geq 3$
So we have $\Phi^{n-1}(G)=U^{n-1}(G)$. Then $\Phi^n(G)=\Phi(U^{n-1}(G))$.
Assume now that $\Phi^n(G)$ is strictly contained in $U^n(G)$. Hence it exists an $b\in U^n(G)$ and $b\notin \Phi^n(G)$.
In other words, we have that exists a prime $q$ such that the following implication hold:
$b=qc\, \, (c\in G) \implies c\notin U^{n-1}(G)$ ($\exists q_1, q_2, ...,q_{n-1}: c\notin (q_1...q_{n-1})G$).
However $b$ is in $U^n(G)$ and so we can write $b=q(q_1...q_{n-1}x)$, a contradiction. $\square$
Corollary
$\Phi^\omega(G) = U(G)$
