Let $A$ be an abelian group and suppose the multiplication by $n$ map $$A\to A, x\mapsto n\cdot x$$
is bijective for all $n\in \Bbb N$ which are not divisible by $p$.
Claim: The map $A\to \Bbb Z_{(p)}\otimes_{\Bbb Z} A$ given by $a\mapsto 1\otimes a$ is an isomorphism.
I tried showing this by first checking the universal property of the tensor product for $A$. That is, given a $\Bbb Z$-bilinear map $f:\Bbb Z_{(p)}\times A\to B$, I want to factor it uniquely as $$\Bbb Z_{(p)}\times A\to A\to B$$ where the first map is $(n,a)\mapsto n\cdot a$. An obvious solution is the map $f(1,-):A\to B$ but then I don't know why it's unique. We need to see that $n\cdot a=m\cdot a'$ implies $f(n,a)=f(m,a')$.
Thank you for your help or alternative approach.
Try showing $A\to\mathbb{Z}_p\otimes_{\mathbb{Z}}A$ is injective and bijective. That is, compute its kernel, and show that any tensor $x/y\otimes a$ may be rewritten as $1\otimes b$ with $b\in A$. You can obviously slide $x$ past the $\otimes$ symbol to get $1/y\otimes xa$, but do you have any idea how to get rid of the $y$? You will have to use the fact that the element $xa\in A$ equals $y$ times some other element of $A$.