$A$ has $\Bbb Z_m$-module structure 
Hilton & Stammbach - Page 15 - Exercise 1.4:
Let $A$ be an abelian group. I want to show that $A$ admits the structure of a $\Bbb Z_m$-module iff $mA=0$.

So first I assume that $A$ has the structure of a $\Bbb Z_m$-module, and then it feels I should be able to say $0$ and $m\in \Bbb Z$ have to act the same way, in which case $mA=0A=0$. But this doesn't seem rigorous, since $m\not\in \Bbb Z_m$.
Instead, I know that one of the axioms of a $\Bbb Z_m$-module is that for each $a\in A$, $(m-1+1)a=(m-1)a+a=0a=0$ so $mA=0$. Is that rigorous?
Conversely, if $mA=0$, I have no idea what to do.
I feel like I am missing something. The exercises around this are about homological algebra. But I can only obtain quotients of the modules, and not the ring.
 A: Let $\mathrm{End}(A)$ denote the ring of additive group endomorphisms of $A$ and $\mu:\Bbb Z\to\mathrm{End}(A)$ be the only ring homomorphism (which maps $1\in\Bbb Z$ into the identity on $A$).
Then for each $n\in\Bbb Z$ we have $\mu(n):x\in A\mapsto nx$.
A $\Bbb Z/m\Bbb Z$-module structure on $A$ is given by a ring homomorphism $\Bbb Z/m\Bbb Z\to\mathrm{End}(A)$ and it makes the following diagram commutative:
$\hspace 6.5cm$
A such ring homomorphism there exists if and only if $\mathrm{Ker}(\mu)\supseteq m\Bbb Z$ and this is equivalent to $m\in\mathrm{Ker}(\mu)$, that's $mA=0$.
A: Let $[n]$ be the class of $n$ in $\mathbb{Z}_m$. Suppose that $A$ has  a $\mathbb{Z}_m$ structure, for every $a\in A, ma=[m]a=0.a$ since $[m]=0$ in $\mathbb{Z}_m$.
On the other and, suppose that $mA=0$, define $b:\mathbb{Z}_m\times A\rightarrow A$ by $b([n],a)=na$  it is well defined, if $[n]=[n']$ there exists $p$ such that $n'=n+pm, b([n'],a)=n'a=(n+pm)a=na$.
A: If $A$ admits the structure of a $\mathbb Z_m$ module, take $a\in A$ and then $ma=a+a+\ldots+a$ ($m$ times) $=(1\cdot a)+\ldots+(1\cdot a)=(1+\ldots +1)\cdot a = 0\cdot a=0$. Since $a$ is arbitrary, it follows that $mA=0$. (I use $1$ to denote the equivalence class of $1$ in $\mathbb Z_m $.)
Conversely, let $mA=0$ i.e. $ma=0$ for all $a\in A$. $A$ is trivially a $\mathbb Z$-module. For $a\in A$ and $[n]\in \mathbb Z_m$ ($[n]$ is the equivalence class of $n\in \mathbb Z$ under congruence mod $m $), define $[n]\cdot a=na$. Now this is well defined because, if $n_1\equiv n_2 \mod m$ then $n_1=n_2+km$ for some $k\in \mathbb Z $ and so $n_1a=n_2a+kma = n_2a$ given that $ma=0$. Now you would have a lengthy (but mostly trivial) process of proving the module axioms - they mostly follow from the equivalent axioms holding for $A$ as a $\mathbb Z $-module, and I will omit the details here.
