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Suppose $M$ is a $\mathbb{Q}$-module. Why is the given action of $\mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $\mathbb{Q}$-module?

I know that an abelian group can only be made into a $\mathbb{Z}$-module in a unique way, since $1\cdot m=m$ for any $m\in M$. So for $p/q\in\mathbb{Q}$, $(p/q)\cdot m=\frac{1}{q}(pm)$. But $p\cdot m$ is necessarily the sum of $m$ a total of $p$ times. Also, $\frac{1}{q}(qm)=(\frac{1}{q}q)\cdot m=1\cdot m=m$. With this, does it somehow follow that there is only a unique way to make an abelian group $M$ a $\mathbb{Q}$-module when it is possible?

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The short answer is "yes." – Rasmus Oct 24 '11 at 7:28

Let $M$ be a $\mathbb{Q}$-module. Then the additive group $(M,+)$ is torsion-free, because $nm=0$ for some $n\in\mathbb{N}$ implies $m=\frac{1}{n}nm=0$.

In a torsion-free abelian group $G$ an equation of the form $nx=g$, $n\in\mathbb{N}$, $g\in G$ has at most one solution.

In the $\mathbb{Q}$-module $M$ the element $\frac{1}{n}m$ thus is the unique solution of the equation $nx=m$, $m\in M$. Note that this holds whatever the operation of $\mathbb{Q}$ on $M$ is like, while the equation $nx=m$ does only depend on the operation of $\mathbb{Z}$ on $M$, which is unique as we already know.

Hence in your case the equation $qm=pm$ has at most one solution $m$ -- and it has one.

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Let $M$ be an abelian group, then

$M$ is a $\mathbb{Q}-$module iff $M$ is torsion free and divisible.

the $\mathbb{Q}-$module structure is $(p/q).m=y$ where $p.m = q.(y)$ such a $y$ exists by divisbility requirement and can be checked to be well defined by torsion freeness.

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