Finite abelian group as Z-module If $M$ is a finite abelian group then $M$ is naturally a $Z$-module. Can this action be extended to make $M$ into a $Q$-module ?
 A: From basic linear algebra, we know that a module over a field $k$ (i.e. a $k$-vector space) is isomorphic to $\bigoplus_I k$ for some index set $I$.
In the case where $k=\Bbb{Q}$, this implies that every non trivial $\Bbb{Q}$-module is infinite (since $\Bbb{Q}$ is infinite). Therefore, there are no finite $\Bbb{Q}$-module. 
If you don't want to use linear algebra: If $M$ is of order $n$, then for every $a\in M$, $n\cdot a=0$. Now suppose $M$ has a structure of $\Bbb{Q}$-module that extends the $\Bbb{Z}$ action. Then for every $0\ne a\in M$, $$0=n\cdot (\frac 1 n \cdot a)=a$$ which is a contradiction since we assumed $a\ne 0$. Therefore, every such module $M$ must be trivial.
A: We can prove that it is not possible even $M$ is a cyclic group. Assume $M=(a)$ a cyclic group of order of $pq$(obviously abelian group). We know that $M$ is a $\mathbb{Z}$-module, and we define:
$$n.a = a \dots a = a^n$$
If we assume that we can extend the above definition, then we should have for every $m \in \mathbb{Z}$:
$$ \frac{1}{m}.a = a^{t_m} \Rightarrow m.(\frac{1}{m}.a) = (m.\frac{1}{m}).a = m.a^{t_m}$$
$$ a = a^{m.t_m} \Rightarrow m.t_m \equiv 1\pmod {pq}$$
As a result, if we assume $m=p$, then the above equation will not be correct.
