# 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?

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Isomorphic as what? Fields? Rings? $\mathbb{Z}$-modules? –  Neal May 17 '12 at 0:37
Let $\phi: \mathbb{Q}\to \mathbb{Z}\times \mathbb{Z}$ be an isomorphism, and suppose that $\phi(a)=(1,0)$. Then $\phi(\frac{a}{2}+\frac{a}{2})=\phi(\frac{a}{2})+\phi(\frac{a}{2})=(1,0)$. However, there is no element $u$ of $\mathbb{Z}\times \mathbb{Z}$ such that $u+u=(1,0)$. –  André Nicolas May 17 '12 at 1:23
AHH My answer and @André's is basically saying that there isn't an isomorphism. If such an isomorphism were to exist, then that leads to a contradiction. You don't need to know exactly what $\phi$ is for every element in $\mathbb{Q}$, you just need one thing that doesn't make sense. –  Thomas May 17 '12 at 2:00
@DougSpoonwood: In general, if $X$ is an infinite set and $n$ is some finite number then $|X^n|=|X|$ (it is sufficient to prove $|X^2|=X$). It is only when you try to do things like $X^{\aleph_0}$ that you go "up" a level or too. However, the fact that $\mathbb{R}^2=\{(a, b): a, b\in\mathbb{R}\}$ and $\mathbb{C}=\{a+ib: a, b\in\mathbb{R}\}$ have the same cardinality is surely easy, as the map $(a, b)\mapsto a+ib$ is a bijection... –  user1729 May 17 '12 at 9:14

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 $\langle (1,0)\rangle\cap\langle (0,1)\rangle = \{(0,0)\}$.

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$\langle a\rangle\cap\langle b\rangle\neq\{0\}$ $\implies \{0\} \langle a\rangle$ and $\{0\} \langle b\rangle$ $\implies \langle a\rangle$ and $\langle b\rangle$ are not subgroups of $\mathbb{Q}$. –  AHH May 17 '12 at 3:04
@AHH: No; $\langle a\rangle\cap\langle b\rangle\neq\{0\}$ means that the intersection is not just $0$. That could be either because $0$ is not in the intersection, or, in this case, because therre are things other than $0$ that are *also* in the intersection. It does not mean that $0$ is not in the intersection. By definition, $\langle a\rangle$ means "the smallest subgroup that contains $a$", so it must be a subgroup. –  Arturo Magidin May 17 '12 at 3:06

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.)

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Note that: $\forall a \in \mathbf{Q} \implies a = m/n : m,n \in \mathbf{Z} , n \neq zero.$ –  AHH May 17 '12 at 1:43

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?

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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.

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Can we find $\phi: \mathbb{Q} \to \mathbb{Z}\times {0} \varsubsetneqq \mathbb{Z}\times \mathbb{Z}$ "homomrphism"?! –  AHH May 17 '12 at 3:24
@AHH, no, by the same reason. –  lhf May 17 '12 at 3:25
if $\phi(a)= \lfloor a \rfloor ,\forall a \in \mathbb{Q}$, then $\phi$ a homomrphism. –  AHH May 17 '12 at 3:33
@AHH, no, it's not: $\phi(1/2)+\phi(1/2)=0$ but $\phi(1)=1$. –  lhf May 17 '12 at 3:34
good counterexample :) –  AHH May 17 '12 at 3:36

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 $\mathbb Q = \mathbb Q \otimes_\mathbb{Z} \mathbb Q \to (\mathbb Z \oplus \mathbb Z)\otimes_\mathbb{Z} \mathbb Q = \mathbb Q\oplus \mathbb Q$.

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This also works with $\otimes_\mathbb{Z} \mathbb Z /n$ for any $n\ge 2$, giving an isomorphism $0 \to \mathbb Z/n \oplus \mathbb Z/n$, which of course is just a rephrasing of some of the arguments elsewhere. I like applying functors and seeing things are different that way, it's the algebraic topologist in me. –  Justin Young May 19 '12 at 22:10

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 ($\mathbb{Z} \times \mathbb{Z}, +)$.

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That's not true - every proper, non-trivial subgroup of $(\mathbb{Q}, +)$ has finite index! (Remember, every quotient is finite.) –  user1729 May 17 '12 at 10:13
@user1729: If $M$ is a proper subgroup of finite index $d$, then for any $q \in \mathbb{Q}$ you have $d(q + M) = M$ so $dq \in M$. But take some $r \not\in M$ and you get $d(r/d) \not\in M$, a contradiction. –  Mikko Korhonen May 17 '12 at 10:26
Sorry - my brain is slow this morning... –  user1729 May 17 '12 at 10:34

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.)

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In ($\mathbb{Q}$, +) every element has a "square root": for any $q \in \mathbb{Q}$ there is an $x \in \mathbb{Q}$ such that $x + x = q$. But in $(\mathbb{Z} \times \mathbb{Z}, +)$ for any element $(a,b) \in \mathbb{Z} \times \mathbb{Z}$ there is an $x \in \mathbb{Z} \times \mathbb{Z}$ such that $x + x = (a,b)$ if and only if both $a$ and $b$ are even.

This motivates the following proof. If $\phi: \mathbb{Q} \rightarrow \mathbb{Z} \times \mathbb{Z}$ were an isomorphism, then $\phi(q) = (1,1)$ for some $q \in \mathbb{Q}$. Then $(1,1) = \phi(q/2 + q/2) = \phi(q/2) + \phi(q/2)$, but this is a contradiction since $1$ is not even.

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