# Show if an abelian group $G$ has a $\mathbb Q$-vector space structure, then it is unique.

Show if an abelian group $G$ has a $\mathbb Q$-vector space structure, then it is unique.

Hint: show every element must have infinite order. Also, show the unique ring homomorphism $\mathbb Z\to \mathbb Q$ is an epimorphism.

I have been able to prove both hints, but I can't put two and two together? Maybe use the fact that every abelian group is a $\mathbb Z$ module in exactly one way, and since the elements must have infinite order, the module structures of $\mathbb Z$ and $\mathbb Q$ must coincide?

In that case, it would amount to show that the diagram created by the homomorphisms $\phi:\mathbb Z\to\mathbb Q$, $\sigma_1:\mathbb Z\to \mathrm{End}(G)$, and $\sigma_2:\mathbb Q\to \mathrm{End}(G)$ commutes, given that $\phi$ and $\sigma_1$ are unique.

• How can the unique ring homomorphism $\mathbb Z\to \mathbb Q$ be surjective?
– lhf
Sep 20, 2017 at 1:34
• @lhf Epimorphisms need not be surjective in $\mathsf{Ring}$, if you use the categorical definition of epimorphism (as my book does) Sep 20, 2017 at 1:40

## 2 Answers

We know that the $\mathbb{Z}$-module structure is unique, because we must have $1x=x$, $2x=x+x$, $3x=x+x+x$, etc., and similarly with negatives.

Now we know that $\frac{1}{n}x$ must be an element $w$ satisfying $nw = x$. If there were 2 such elements $w$ and $w'$, then $n(w-w')=0$ and multiplying both sides by $\frac{1}{n}$ gives $w=w'$. Therefore there is only one element that can be $\frac{1}{n}x$. And of course $\frac{m}{n}x$ is $\frac{1}{n}x$ added to itself $m$ times (for $m>0$), and similarly with negative rationals.

So the $\mathbb{Q}$-module structure is unique.

For the other hint, note that two different $$\mathbb{Q}$$-module structures correspond to two different ring homomorphisms $$\alpha : \mathbb{Q} \longrightarrow \text{End}_{\text{Ab}}(G)$$ and $$\beta : \mathbb{Q} \longrightarrow \text{End}_{\text{Ab}}(G)$$. Then, the inclusion $$i : \mathbb{Z} \longrightarrow \mathbb{Q}$$ tells us that $$(\alpha \circ i )(1) = (\beta \circ i)(1)$$ so that $$\alpha = \beta$$.