Necessary and sufficient conditions for an abelian group to be a vector space over $\mathbb{Q}$ I know if $A$ is an abelian group under addition, then the properties of addition of vector spaces are satisfied. But how do I determine necessary and sufficient conditions for $A$ to be an abelian group? In particular, I would appreciate an explanation of how to check a condition is necessary and then sufficient. 
 A: An abelian group $A$ is a vector space over $\mathbb{Q}$ if and only if it torsion free and is divisible. That is, for each $n$ in $\mathbb{N}$ and each $b$ in $A$ (i) $nb= 0 \Rightarrow b=0$ (ii) there exists $a$ in $A$ such that $b=na$ (see below for the definition of $nx$)
Proof: Suppose that $A$ is a vector space over $\mathbb{Q}$. Then trivially for each $n$ in $\mathbb{N}$ and for each $b$ in $A$ we have (i) $nb=0 \Rightarrow 0=\frac{1}{n}(nb)= 1b=b$ and (ii) $n (\frac{1}{n} b) = 1b= b$. 
Conversely suppose that $A$ is torsion free and divisible. For each $\frac{p}{q}\neq 0$ in $\mathbb{Q}$ and $b$ in $A$ we would like to define $\frac{p}{q}b$. It is natural to want to define it (using divisibility) as $a$ such that $qa=pb$, however this would only make sense if there is a unique such $a$. Suppose that $a_1$ and $a_2$ are both such that $qa_1=pb$ and $qa_2=pb$. It follows that $q(a_1-a_2)=0$ which implies that $a_1=a_2$ (since $A$ is torsion free) and so we can define $\frac{p}{q}b$ as the unique $a$ such that $qa=pb$. One easily checks that this makes $A$ into a $\mathbb{Q}$ vector space. For example to prove $$\left(\frac{p}{q} + \frac{s}{t}\right)a=\frac{p}{q}a + \frac{s}{t}a$$ you show that both sides are solutions $x$ to $(qt)x=(tp+qs)a$.
Note 1  Above we use two seemingly different definitions of $nx$. However they are actually the same.
Recall that every abelian group $A$ can be made into a $\mathbb{Z}$-module by definining:
$$na=\begin{cases} 0 & \text{if $n=0$} \\(n-1)a +a& \text{if $n\geq 1$}\\(n+1)a -a & \text{if $n\leq -1$.}\end{cases}$$
Since $\mathbb{Z}$ is the initial object in the category of rings, for each field $\mathbb{F}$ there is a unique ring homomorphism $i:\mathbb{Z} \to \mathbb{F}$. Explicitly $i$ is defined by
$$i(n)=\begin{cases} 0 & \text{if $n=0$}\\ i(n-1) +1& \text{if $n\geq 1$}\\i(n+1) -1 & \text{if $n\leq -1$.}\end{cases}$$
Using this homorphism every vector space $V$ over $\mathbb{F}$ becomes a $\mathbb{Z}$-module by defining $n\cdot x = i(n)x$. One easily shows that for such a vector space the two ways of turning it into a $\mathbb{Z}$-module are actually the same. According to the second definition $0\cdot x=i(0)x=0x=0$ and for $n\geq 1$ in $\mathbb{N}$ we have $n\cdot x =i(n)x=(i(n-1)+1)x=(n-1)\cdot x +x$. Similarly for $n\leq -1$ in $\mathbb{Z}$ we have $n\cdot x =i(n)x=(i(n+1)-1)x=(n+1)\cdot x -x$.
