$A\cong \Bbb Z_{(p)}\otimes A$ if multiplication by $n$ is iso 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.
 A: 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 $A$.
A: Instead of trying to prove that $A$ has the universal property of the tensor product, why not use the universal property to produce a map $\mathbb{Z}_{(p)} \otimes_\mathbb{Z} A \to A$ that is inverse to $a\mapsto 1\otimes a$?
Since $x \mapsto nx$ is bijective for each $n$ with $p \nmid n$, then there is some $m \in A$ such that $nm = 1$.  We may as well call this element $1/n$.  Now there are a few steps to complete.
1) Show that $x \mapsto nx$ is bijective for all $n \in \mathbb{Z}$ with $p \nmid n$ (not just $n \in \mathbb{N}$).
2) Define a $\mathbb{Z}$-bilinear map $\mathbb{Z}_{(p)} \times A \to A$.
3) Use the universal property of the tensor product to produce a map $\mathbb{Z}_{(p)} \otimes_\mathbb{Z} A \to A$ and show that it is inverse to $a\mapsto 1\otimes a$
