Let, for $p$ prime integer, $\Bbb{Q},\Bbb{Z}_{(p)},\Bbb{Z}_p, \Bbb{Q}_p$ be the field of rationals, the ring of rationals whose denominator is prime to $p$, the ring of $p$-adic integers and the field of $p$-adic numbers, respectively.
Recall that an abelian group is a $\Bbb{Z}$-module and that such a group is torsion-free iff every nonzero element has infinite order.
As far as I understand, $\Bbb{Z}$-linear independence and $\Bbb{Q}$-linear independence are the same, for torsion-free $A$.
The following fact holds for every prime integer $p$ and every torsion-free abelian group $A$: $$(\Bbb{Q}\otimes A)\cap (\Bbb{Z}_p\otimes A)=\Bbb{Z}_{(p)}\otimes A$$
My questions are about the proof of this fact.
Proof. Let $x \in \Bbb{Z}_p\otimes A$, so that we can write down $x=\sum_{i=1}^n\pi_ia_i$ where $\pi_i\in\Bbb{Z}_p$ and $a_i\in A$. Wlog we may assume $a_1,\dots,a_k$ $\Bbb{Z}$-linearly independent and $\pi_{k+1},\dots,\pi_n$ rational integers (I got this).
If $x\in \Bbb{Q}\otimes A$, then $x$ depends on a maximal $\Bbb{Z}$-linearly independent system $\{a_1,\dots,a_k,a'_{k+1},\dots\}$ of $A$, say $$x=\sum_{i=1}^km_ia_i+\sum_{i=k+1}^rm_ia'_i$$ for rational numbers $m_i$.
(1) A maximal independent system of $A$ becomes, after identifying $1\otimes a$ with $a$, a maximal independent system in the $\Bbb{Q}_p$-module $\Bbb{Q}_p\otimes A$,
then
(2) there is essentially only one dependence relation for $x$.
Thus
(3) $\pi_1,\dots,\pi_n$ must be rationals
and
(4) $x\in \Bbb{Q}_p\otimes A$.
This proves $(\Bbb{Q}\otimes A)\cap (\Bbb{Z}_p\otimes A)\le \Bbb{Q}_p\otimes A$. QED
I was not able to prove statements (1),(2),(3),(4). May you help me, please?