Let $M$ be a torsion abelian group. For a prime $p$, let $\Bbb Z_p$ be the $p$-adic integers (seen as an abelian group), and $M[p^{\infty}] := \{m \in M : p^n m= 0 \text{ for some } n \geq 0\}.$
Is it true that $M[p^{\infty}] \cong M \otimes_{\Bbb Z} \Bbb Z_p$ ?
If not, is it still possible to have a natural structure of $\Bbb Z_p$-module on $M[p^{\infty}]$ ?
I know for instance that $S^1[p^{\infty}] \cong \Bbb Q_p / \Bbb Z_p$ which is naturally a $\Bbb Z_p$-module, but I don't know about general $M$.
If $p^n m = 0$ and $a = ([a_k]_{p^k})_{k \geq 0} \in \Bbb Z_p$, we could define $a \cdot m := a_n m \in M$, but I'm not sure whether this is the right thing to do.
Yes, it is true.
For a torsion group $M$ we have $M\cong \bigoplus_q M[q^\infty].$ It is easy to see that $M[q^\infty]\otimes \mathbb Z_p=0$ for $q\ne p.$ Hence, it is enough to prove that $$M[p^\infty]\otimes \mathbb Z_p\cong M[p^\infty].$$
Note that we have a similar isomorphism for the $p^n$-torsion subgroup $M[p^n]$ for a fixed $n$: $$M[p^n]\otimes \mathbb Z_p \cong M[p^n] \otimes \mathbb Z_p/p^n \cong M[p^n] \otimes \mathbb Z/p^n \cong M[p^n] \otimes \mathbb Z \cong M[p^n].$$ Further, we have $M[p^\infty]=\bigcup_n M[p^n].$ Hence $M[p^\infty]$ is the colimit of $M[p^n]$ $$M[p^\infty]={\sf colim} (M[p] \hookrightarrow M[p^2]\hookrightarrow M[p^3]\hookrightarrow \dots).$$ Finally, we combine this with the fact that the functor of tensor product $-\otimes \mathbb Z_p$ comutes with colimits (because it is left adjoint to the functor ${\sf Hom}(\mathbb Z_p,-)$): $$ M[p^\infty]\otimes \mathbb Z_p=({\sf colim}\ M[p^n])\otimes \mathbb Z_p \cong {\sf colim}\ (M[p^n]\otimes \mathbb Z_p ) \cong {\sf colim}\ M[p^n] \cong M[p^\infty].$$