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

Question

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?

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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 $(z+1)y = z$.

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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.

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Thanks

Answer 1

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.

Answer 2

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.

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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 $\left\langle \dfrac ab \right\rangle = \mathbb Q$. We want to show that it is not the case that every rational number is an integral multiple of $\dfrac ab$.

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

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.

Answer 3

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

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

Answer 4

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

Edit: This proof exploits the fact that $\mathbb Q$ 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 $M\not\cong V$ 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).