Are $\mathbb Q\times\mathbb Q$ and $\mathbb Q\times\mathbb Q\times\mathbb Q$ isomorphic? I know $\mathbb Q$ is not isomorphic to $\mathbb Q\times \mathbb Q$ as groups. But is it true for $\mathbb Q\times\mathbb Q$ and $\mathbb Q\times\mathbb Q\times\mathbb Q$?
That is, are $\mathbb Q\times\mathbb Q$ and $\mathbb Q\times\mathbb Q\times\mathbb Q$ isomorphic as groups?  Where $\mathbb Q$ is set of rational numbers.  
 A: I would like to flesh out Qiaochu Yuan's comment. It relies on a result that, while not difficult to prove, may not be obvious to beginning students.
Theorem: A map of abelian groups $\mathbb{Q}^m \to \mathbb{Q}^n$ is also a map of $\mathbb Q$-vector spaces.
Proof. Let $T$ be our map. From the definition of a map between abelian groups, we have $T(v+w)=T(v)+T(w)$.  It suffices to show that $T(rv)=rT(v)$ for every $r\in \mathbb Q$ and $v\in \mathbb Q^m$.
By induction, for every $n\in \mathbb N$, we have $T(nv)=nT(v)$.  Combining this with $T(-v)=-T(v)$, we get $T(nv)=nT(v)$ for every $n\in \mathbb Z$.(N.B. This doesn't require more that the definition of group, and for a non-abelian group, using multiplicative notation, $\phi(a^n)=\phi(a)^n$ holds for arbitrary group homomorphisms).
We now have to deal with division.  First, we need to define it, see that $v/k$ is well defined, and observe that it agrees with our usual notion of division.  We say that $v/k=w$ if $kw=v$.  If $ku=v$, then $k(w-u)=0$.  But because there is no torsion in $\mathbb Q$, there won't be in $\mathbb Q^m$, and so division yields a unique answer.  It is straight forward that the usual notion of division by an integer satisfies this property, and so this does exactly what one would expect.
Finally, we show that $T(v/k)=T(v)/k$.  This follows from $kT(v/k)=T(k(v/k))=T(v)$.  We can now write $\frac{p}{q}T(v)=T(\frac{p}{q}(v))$.
