(a) Prove that every finitely generated subgroup of $(\Q, +)$ is cyclic.

Let $G$ be a finitely generated subgroup of $(\Q, +)$ and let $r_1, \dots, r_n$ be nonzero generators of $G$.
Let us express
\[r_i=\frac{a_i}{b_i},\]
where $a_i, b_i$ are integers.

Let
\[s:=\frac{1}{\prod_{j=1}^n b_j} \in \Q.\]
Then we can write each $r_i$ as
\[r_i=\frac{a_i}{b_i}=\left(\, a_i\prod_{\substack{j=1\\j\neq i}}^n b_i \,\right)\cdot \frac{1}{s}.\]

It follows from the last expressions that the elements $r_i$ is contained in the subgroup $\langle s \rangle$ generated by the element $s$.
Hence $G$ is a subgroup of $\langle s \rangle$.
Since every subgroup of a cyclic group is cyclic, we conclude that $G$ is also cyclic.

(b) Prove that $\Q$ and $\Q \times \Q$ are not isomorphic as groups.

Seeking a contradiction, assume that $\Q$ is isomorphic to the direct product $\Q \times \Q$:
\[\Q\cong \Q\times \Q.\]

Then consider the subgroup $\Z\times \Z$ of $\Q\times \Q$.
We claim that the subgroup $\Z\times \Z$ is not cyclic.
If it were cyclic, then there would be a generator $(a,b)\in \Z\times \Z$.

However, for example, the element $(b, -a)$ cannot be expressed as an integer multiple of $(a, b)$.
To see this, suppose that
\[n(a,b)=(b,-a)\]
for some integer $n$.

Then we have $na=b$ and $nb=-a$. Substituting the first equality into the second one, we obtain
\[n^2a=-a.\]
If $a\neq 0$, then this yields that $n^2=-1$, which is impossible, and hence $a=0$.

Then $na=b$ implies $b=0$ as well.
However, $(a,b)=(0,0)$ is clearly not a generator of $\Z\times \Z$.

Thus we have reached a contradiction and $\Z\times \Z$ is a non-cyclic subgroup of $\Q\times \Q$.
This implies via the isomorphism $\Q\cong \Q \times \Q$ that $\Q$ has a non-cyclic subgroup.
We saw in part (a) that this is impossible.
Therefore, $\Q$ is not isomorphic to $\Q\times \Q$.

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