Prove or disprove: $(\mathbb{Q}, +)$ is isomorphic to $(\mathbb{Z} \times \mathbb{Z}, +)$? Prove or disprove: $\mathbb{Q}$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$. I mean the groups $(\mathbb Q, +)$ and $(\mathbb Z \times \mathbb Z,+).$ Is there an isomorphism?
 A: I assume that you are asking whether we have an isomorphism of additive groups.
In that case, assume that $\phi: \mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ is such an isomorphism. So we have for example that $\phi(0) = (0,0)$. Let $a\in \mathbb{Q}$ be such that $\phi(a) = (1,0)$ and $b$ be such that $\phi(b) = (0,1)$. Then we see that $\mathbb{Q}$ is equal to $\{na + mb \lvert n,m \in \mathbb{Z}\}$. This is a contradiction...
(Hence the argument is that $\mathbb{Q}$ is not finitely generated while $\mathbb{Z}\times \mathbb{Z}$ is.)
(I will leave the details to you.)
A: Another argument that you can construct with the following (be sure you can prove/answer every section):
1) An abelian additive group $\,A\,$ is said to be divisible if $\,\forall a\in A\,\,n\in\mathbb{N}\,\,\exists b\in A\,\,s.t.\,\,nb=a\,$ .
To be sure, $n\neq 0$
2) $\,\mathbb{Q}\,$ is a divisible group
3) Any homomorphic image of a divisible group is a divisible group
4) Is $\,\mathbb{Z}\times\mathbb{Z}\,$ divisible?
A: Let $ \phi: \mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ be a homomorphism. Fix $u/v\in\mathbb{Q}$ and let $(a_n,b_n)=\phi(u/v^n)$. Since $\phi(u/v)=v^{n-1}\phi(u/v^n)$, we get $a_1=v^{n-1}a_n$ and $b_1=v^{n-1}b_n$ for all $n\in\mathbb N$, which is clearly impossible unless $\phi(u/v)=(a_1,b_1)=(0,0)$.
So, the only homomorphism  $\mathbb{Q} \to \mathbb{Z}\times \mathbb{Z}$ is the zero map, and there is no chance of an isomorphism.
A: Another idea: suppose there is an isomorphism $\mathbb Q \to \mathbb Z \oplus \mathbb Z$, then tensor both sides $\otimes_\mathbb{Z} \mathbb Q$, and get a $\mathbb Q$-module isomorphism $\mathbb Q = \mathbb Q \otimes_\mathbb{Z} \mathbb Q \to  (\mathbb Z \oplus \mathbb Z)\otimes_\mathbb{Z} \mathbb Q = \mathbb Q\oplus \mathbb Q$.
A: Yet another way to see the two cannot be isomorphic as additive groups: if $a,b\in\mathbb{Q}$, and neither $a$ nor $b$ are equal to $0$, then $\langle a\rangle\cap\langle b\rangle\neq\{0\}$; that is, any two nontrivial subgroups intersect nontrivially. To see this, write $a=\frac{r}{s}$, $b=\frac{u}{v}$, with $r,s,u,v\in\mathbb{Z}$, $\gcd(r,s)=\gcd(u,v)=1$. Then $(su)a = (rv)b\neq 0$ lies in the intersection, so the intersection is nontrivial.
However, in $\mathbb{Z}\times\mathbb{Z}$, the elements $(1,0)$ and $(0,1)$ are both nontrivial, but $\langle (1,0)\rangle\cap\langle (0,1)\rangle = \{(0,0)\}$.
A: Another way of seeing this (yes, there are many ways!) is to notice that two isomorphic groups have the same quotients. That is, if $G\cong H$ and $G\twoheadrightarrow K$ then $H \twoheadrightarrow K$.
Now, by this question (which was only asked the other day, which is why I am posting this answer!), we know that every proper quotient of $\mathbb{Q}$ is torsion (that is, every element has finite order). On the other hand, $\mathbb{Z}\times\mathbb{Z}$ has a torsion-free proper quotient, $\mathbb{Z}\times\mathbb{Z}\twoheadrightarrow \mathbb{Z}$. Thus, they cannot be isomorphic.
(Indeed, this actually proves that there cannot be a homomorphism from $\mathbb{Q}$ to $\mathbb{Z}\times\mathbb{Z}$, as lhf has already shown - the result we use tells us that the map cannot have non-trivial kernel, while this proves that the kernel cannot be trivial either.)
A: You can show that the group $(\mathbb{Q}, +)$ has no proper subgroup of finite index, but for example $\mathbb{Z} \times 2\mathbb{Z}$ has finite index in ($\mathbb{Z} \times \mathbb{Z}, +)$.
