Is $(\mathbb{Q},+)$ isomorphic to $(\mathbb{Q}^n,+)$? Is easy to show that $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$ as $\mathbb{Q}$-vector spaces and then $(\mathbb{R},+)$ is isomorphic to $(\mathbb{R}^n,+)$ as abelian groups. This result might suggest that we also have $(\mathbb{Q},+) \cong (\mathbb{Q}^n,+)$ as abelian groups, but a similar approach to this dosen't work.
How can I prove or disprove this?
 A: For every nonzero $x, y\in\mathbb{Q}$, there are nonzero integers $m, n$ such that $$mx=ny$$ (where $mx, ny$ are interpreted in the obvious way). Now, any homomorphism $f$ between $(\mathbb{Q}, +)$ and another structure $(G, *)$ must preserve multiplication by integers: $$f(mx)=mf(x).$$ So if $(\mathbb{Q}, +)\cong(\mathbb{Q}^n, +)$, then $(\mathbb{Q}^n, +)$ would also have to have the property that for every nonzero $x, y$ there are nonzero integers $m, n$ such that $mx=ny$. But this is clearly false: set $x=(1, 0), y=(0, 1)$.
As a fun exercise, see if you can generalize this to show that $(\mathbb{Q}^m, +)\cong (\mathbb{Q}^n, +)\iff m=n$.

It may appear that something similar holds for $\mathbb{R}$: for every nonzero $x, y\in\mathbb{R}$, there are nonzero reals $a, b$ such that $ax=by$. But the corresponding statement is false in $\mathbb{R}^n$, for obvious reasons. So, what gives?
The answer is that homomorphisms need not preserve multiplication by reals. In fact, multiplication of a group element by a real is in general not something that makes sense! For example, what is $\pi$ times the permutation $(1, 3, 6)\in S_{17}$? By contrast, three times $(1, 3, 6)$ is naturally interpreted as $$(1, 3, 6)\circ(1, 3, 6)\circ(1, 3, 6)$$ which is just the identity permutation.
One rather abstract way to say this is that every group is naturally a $\mathbb{Z}$-module, but not naturally an $\mathbb{R}$ module; see https://en.wikipedia.org/wiki/Module_(mathematics). (In fact, some groups are not even the additive group of any $\mathbb{R}$-module - for example, any non-divisible group has this property.)
A: They aren't isomorphic.
If $n>1$ then $\mathbb{Q}^n$ has a free finitely generated subgroup of rank at least $2$ (namely $\mathbb{Z}^n$). Now, every finitely generated subgroup of $\mathbb{Q}$ is cyclic, so they can't be isomorphic.
