Abelian groups and $\mathbb{Z}$-modules We know that any Abelian group is a $\mathbb{Z}$-module. Is the converse true?
 A: Yes, there is a bijective correspondence between $\mathbb Z$-modules and abelian groups.
From module to group, just forget the scalar multiplication; the module laws directly require that the module's addition constitutes an abelian group.
From group to $\mathbb Z$-module, declare $n\cdot a$ to be $\underbrace{a+a+\cdots+a}_n$ and so forth.
A: Let $A$ be a commutative ring with unit. To give yourself an $A$-module $M$ is equivalent to giving yourself an abelian group $M$ plus an action of $A$ on $M$, that is, a ring morphism $\varphi : A \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$ where $\textrm{Hom}_{\textrm{Ab.}}(N',N)$ is the set of morphism of abelian groups from $N'$ to $N$ if $N,N'$ are abelian groups.
Note that $\textrm{Hom}_{\textrm{Ab.}}(M,M)$, also noted $\textrm{End}_{\textrm{Ab.}}(M)$, is a ring with unit, but is not necessarily commutative. Now as you know, for any ring $R$ with unit, you have a unique ring morphism $\mathbf{Z}\rightarrow R$. So that in particular if $M$ is an abelian group, you have a unique ring morphism $\varphi : \mathbf{Z} \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$, that is, a unique structure of $\mathbf{Z}$-module on $M$. Inversely, having one structure of $\mathbf{Z}$-module on $M$ does not give very much, that is, not more that the already existing unique ring morphism $\varphi : \mathbf{Z} \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$.
In a pedantic way : the forgetful functor from the category of $\mathbf{Z}$-modules to the category of abelian groups is an equivalence of category (and indeed worth mentionning, even an isomorphism of categories).
