Prove that the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic.

Is my proof below correct? What specific property of rationals did I exploit in my proof? It looks like the property I exploited is the following: Given any positive rational, I can always write it as sum of arbitrary number of positive rationals, whereas given any positive integer I cannot write it as a sum of arbitrary number of positive integers. Has it got to do with the fact that $\mathbb{Q}$ is a field?

Problem Prove that the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic.

Solution Let there exist an isomorphism between $\mathbb{Z}$ and $\mathbb{Q}$. Now consider the element $1_{\mathbb{Q}}$. We then have $\phi(1_{\mathbb{Q}}) = z \in \mathbb{Z}$. Since $\phi$ has to be a bijection, $z$ cannot be zero, since $\phi(0) = 0$.

Now consider the element $\left(\dfrac1{z+1}\right)_{\mathbb{Q}}$. We now have $$z = \phi(1_{\mathbb{Q}}) = \phi\left(\underbrace{\left(\dfrac1{z+1}\right)_{\mathbb{Q}} + \left(\dfrac1{z+1}\right)_{\mathbb{Q}} + \cdots + \left(\dfrac1{z+1}\right)_{\mathbb{Q}}}_{z+1 \text{ times }} \right) = (z+1) \phi\left(\left(\dfrac1{z+1}\right)_{\mathbb{Q}}\right)$$ However, there is no element in $y \in \mathbb{Z}$ such that $(z+1)y = z$.

First update

Actually I realize that I complicated it unnecessarily. Instead, we can do like this. Since $\phi$ is an isomorphism, we have $\phi(q_{\mathbb{Q}}) = 1_{\mathbb{Z}}$ for some $q \in Q$. However, $$\phi(q) = \phi(q/2+q/2) = 2\phi(q/2)$$ And there is no $y \in Z$, such that $2y=1$. Hence, $\phi(q/2)$ remains unmapped.

Thanks

• If $z=-1$? You have to exclude this case. – Test123 Dec 28 '13 at 16:23
• But for the case $\;z=-1\;$ it looks fine. Deal with the particular case separatedly. – DonAntonio Dec 28 '13 at 16:24
• Have you thought that additive group of $\mathbb Q$ is not cyclic – Babak Miraftab Dec 28 '13 at 16:25
• You may unconsciously assume that $z>0$. You get problems not only with $z=-1$, but also with $z=-2$. - I rather suggest you have a look at $m:=\phi(\frac12\phi^{-1}(1))\in\mathbb Z$, which should have the property $m+m=1$. – Hagen von Eitzen Dec 28 '13 at 16:28
• @HagenvonEitzen Actually, I realized my mistake. I have now added an update. Can you tell me if this works? – John Smith Dec 28 '13 at 16:30

Another proof is as follows:

Suppose that $\phi : \mathbb{Q} \to \mathbb{Z}$ is an isomorphism. Then there is some $r \in \mathbb{Q}$ such that $\phi(r) = 1_{\mathbb{Z}}$.

So what is $\phi(r/2)$? We would have to have

$$1_{\mathbb{Z}} = \phi(r) = \phi\big(2(r/2)\big) = 2\phi(r/2)$$

or equivalently that $\phi(r/2) = \frac{1}{2}$. But this is not in $\mathbb{Z}$, so there can be no such morphism.

• ... Which I see you just added to the question. – Simon Rose Dec 28 '13 at 16:32
• Thanks. I added this to the proof about a minute back. So is my proof a proof by contradiction? – John Smith Dec 28 '13 at 16:33
• Yes. You are assuming the existence of an object, and then deriving a conclusion that is at odds with that very object's existence. – Simon Rose Dec 28 '13 at 16:35
• Ok. Thanks. This is helpful. – John Smith Dec 28 '13 at 16:35
• @Leslie: You can also do a direct proof as follows. Let $\phi:\Bbb Q\to\Bbb Z$ be any homomorphism. Since there is no $z\in\Bbb Z$ such that $2z=1_{\Bbb Z},$ then for all $q\in\Bbb Q,$ we have $1_{\Bbb Z}\ne 2\phi(q)=\phi(2q).$ In particular, then, since $\frac r2\in\Bbb Q$ for all $r\in\Bbb Q$ and $r=2\left(\frac r2\right),$ then for all $r\in\Bbb Q$ we have $\phi(r)\ne1_{\Bbb Z}.$ Hence, $\phi$ is not surjective. Since $\phi:\Bbb Q\to\Bbb Z$ was an arbitrary homomorphism, then no homomorphism $\phi:\Bbb Q\to\Bbb Z$ is surjective, and so $\Bbb Q$ and $\Bbb Z$ are not isomorphic. – Cameron Buie Dec 28 '13 at 16:51

Note that the additive group $\mathbb Z = \langle 1\rangle$ is generated by one element (and hence is cyclic), whereas $\mathbb Q$ is not cyclic, nor can it be finitely generated. In any case, being cyclic is a structural property of groups that is preserved by any isomorphism.

Just to make sure you understand that the additive group $\mathbb Q$ is not cyclic, we want to show $\mathbb Q$ is't generated by some element $\dfrac ab$, where $a, b \in \mathbb Z$ i.e., that there is no element $\dfrac ab\in \mathbb Q$ such that $\langle \dfrac ab \rangle = \mathbb Q$. We want to show that it is not the case that every rational number is an integral multiple of $\dfrac ab$.

Suppose $\langle\dfrac ab \rangle = \mathbb Q$.

Observe that, under this assumption $\dfrac a{2b} \in \mathbb Q$, being a rational number, should then be an integral multiple of $\dfrac ab$, which it clearly isn't; it is $\dfrac 12 \dfrac ab.$

Hence the assumption that $\mathbb Q$ is generated by $\dfrac ab$ cannot be true. Since $\dfrac ab$ is arbitrary, this shows $\mathbb Q$ is not generated by any single element in $\mathbb Q,$ i.e., $\mathbb Q$ is not cyclic.

• The OP asked what property is being exploited. – Namaste Dec 28 '13 at 16:32
• @DeitrichBurde This has been dealt with. – Namaste Jun 11 '18 at 16:37
• All right! It took me a while to remember what we did 5 years ago:) – Dietrich Burde Jun 11 '18 at 18:05
• I hadn't seen your comment, until the answer was recently upvoted, and I was like, Dah! why did I not add that?; it would certainly make a better answer! So, @DeitrichBurde, it was more of a "closure" notification, though I missed expressing my "thanks" for pointing that out! – Namaste Jun 11 '18 at 19:27

Suppose there is an isomorphism $f:\mathbb Z\to \mathbb Q$.

Let $a=f(1)$; then $f(\mathbb Z)=\{na,n \in\mathbb Z\} \ne \mathbb Q$.

The group $\mathbf Q$ has the property that for any $x\in \mathbf Q$ and any integer $n\geqslant 1$, there exists $y\in \mathbf Q$ such that $n\cdot y=x$. In other words, $\mathbf Q$ is divisible. The group $\mathbf Z$ is not divisible, so since "being divisible" is invariant under isomorphism, $\mathbf Z\not\simeq\mathbf Q$.

Edit: This proof exploits the fact that $\mathbf Q$ is a field in the following way (best seen after generalization). Let $A$ be an integral domain(not a field) with field of fractions $K$. A similar proof shows that if $M$ is any finitely generated $A$-module and $V$ is any finite-dimensional $K$-vector space, then $M\not\simeq V$ as $A$-modules. The reason is that because $K$ is a field, $V$ is a divisible $A$-module, and $M$ is not. (The fact that finitely generated $A$-modules are not divisible is probably most easily seen by localizing at a maximal ideal and using Nakayama's lemma).

• I didn't say it is easier - it is just different approach. That said, the fact that divisibility is invariant under isomorphism requires no thought to check - just do the obvious thing. – Daniel Miller Dec 28 '13 at 16:34
• So essentially I exploited the fact that I can divide in rationals. So in essence I exploited the fact that $\mathbb{Q}$ is a field. Is my understanding right? – John Smith Dec 28 '13 at 16:37
• Is it more or is it less? – John Smith Dec 28 '13 at 16:39
• Ok. I do not know about integral domains, modules, Nakayama lemma etc. Hopefully, when I read those I will be able to appreciate your answer. I am still in the first chapter of Dummit and Foote. Thanks for the answer though. Good to know that the key property being exploited is $\mathbb{Q}$ is a field. – John Smith Dec 28 '13 at 16:51