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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
Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, many find the use of imperative ("Prove") to be rude when asking for help; please consider rewriting your post. – Zev Chonoles May 17 '12 at 0:42
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
up vote 10 down vote accepted

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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