Proving Non-Isomorphism between $\mathbb{Q}$ and $\mathbb{Z}$ Using POSET Concepts

Question

Hello fellow mathematicians!

I am seeking assistance in proving that the groups $\mathbb{Q}$ (the group of rational numbers under addition) and $\mathbb{Z}$ (the group of integers under addition) are not isomorphic. However, I would like to approach this problem using the concepts of partially ordered sets (POSET).

In particular, I am interested in utilizing the order structure of the groups to establish their non-isomorphism. My intuition tells me that examining the existence of certain elements and their orders might provide valuable insights.

Here's what I have considered so far:

1. In the group $\mathbb{Q}$, we know that the order of an element $\frac{a}{b}$, where $a$ and $b$ are coprime integers, is equal to the absolute value of $b$. How does this relate to the order structure of $\mathbb{Z}$?
2. The group $\mathbb{Z}$ is known to contain elements of arbitrary large order. Can we observe a similar property in $\mathbb{Q}$?
3. Are there any inherent properties of POSETs that can help us establish a rigorous argument for non-isomorphism?

To effectively demonstrate the non-isomorphism between $\mathbb{Q}$ and $\mathbb{Z}$ using POSET concepts, I kindly request your expertise in formulating a well-structured proof. I would greatly appreciate a detailed explanation that highlights the key ideas and logical steps involved.

Thank you in advance for your valuable insights and contributions. Let's unravel the fascinating world of group theory together!

Answer 1

A group does not generally come with a (compatible) poset structure. Even if it did, there is no reason to expect it to tell you about two groups not being isomorphic.

However, you can consider the poset of subgroups $\text{Sub}(G)$ of a group $G$, ordered by inclusion.

If two groups $G_1$ and $G_2$ are isomorphic and $f:G_1\to G_2$ is an isomorphism, then it induces an increasing bijection $\bar f:\text{Sub}(G_1)\to \text{Sub}(G_2)$ where $\bar f(H) := \{f(h) : h\in H\}$ for $H\in \text{Sub}(G_1)$. In other words, $\text{Sub}(G_1)$ and $\text{Sub}(G_2)$ are isomorphic as posets, isomorphism in this case being an increasing bijection.

Now lets look at the posets $\text{Sub}(\mathbb{Z})$ and $\text{Sub}(\mathbb{Q})$. Descriptions for subgroups of both of those groups exist... but the caveat is that $\text{Sub}(\mathbb{Q})$ is way harder to describe.

If $H\in \text{Sub}(\mathbb{Z})$, then there exists unique $n\in \mathbb{N}_0$ such that $H = n\mathbb{Z}$, and so $\text{Sub}(\mathbb{Z})$ can be identified with natural numbers ordered by the relation $n\preceq m$ iff $m|n$. This ordering comes with a property that if $p$ is prime, then $p \prec 1$ and $p$ is a direct predecessor of the maximum $1$. For example $2\prec 1$.

Now suppose we are given $H\in\text{Sub}(\mathbb{Q})$ and $H\neq \mathbb{Q}$. If $x\in \mathbb{Q}\setminus H$ then consider the subgroup $H_0$ generated by $H\cup \{x\}$. Write $x = \frac{p}{q}$ where $p$ and $q$ are co-prime. If it were $\frac{1}{q^2}\in H_0$, then there'd exist $a\in \mathbb{Z}$ and $y\in H$ with $\frac{1}{q^2} = ax+y$ so $\frac{1-apq}{q^2}\in H$. Since $1-apq$ and $q^2$ are co-prime, there exist $c, d\in \mathbb{Z}$ with $c(1-apq)+dq^2 = 1$. Then $\frac{1}{q} = dq+cq\cdot \frac{1-apq}{q^2}\in H$, but that's impossible. So we have to have $\frac{1}{q^2}\notin H_0$.

So we proved that for any $H\in\text{Sub}(\mathbb{Q})$ that isn't a maximum, there exists $H_0$ such that $H\subset H_0\subset \mathbb{Q}$, where inclusions are proper inclusions. Consequently $\text{Sub}(\mathbb{Q})$ doesn't contain elements which are direct predecessors to the maximum of $\text{Sub}(\mathbb{Q})$, so $\text{Sub}(\mathbb{Z})$ and $\text{Sub}(\mathbb{Q})$ are not isomorphic.

There is another way to see that $\text{Sub}(\mathbb{Z})$ and $\text{Sub}(\mathbb{Q})$ are not isomorphic as posets, and that to see they're not even of the same size. Indeed, $|\text{Sub}(\mathbb{Z})| = \aleph_0$ while $|\text{Sub}(\mathbb{Q})| = \mathfrak{c}$ since given any two subsets $\mathcal{P}$ and $\mathcal{Q}$ of prime ideals, the subgroups generated by $\{\frac{1}{p} : p\in \mathcal{P}\}$ and $\{\frac{1}{q} : q\in\mathcal{Q}\}$ are equal iff $\mathcal{P} = \mathcal{Q}$.

However, I think the first method is more illustrative of how could we apply the concept of poset of subgroups to prove that two groups are not isomorphic by carefully analyzing the orders they induce and their properties.

Answer 2

As noted in the comments, there are easy ways of showing that $(\mathbb{Z},+)$ and $(\mathbb{Q},+)$ are not isomorphic. But the question asked for an order-theoretic argument, and the following provides one.

Recall that an ordered abelian group consists of a group $G$ together with a partial order $\leq$ that satisfies the following condition for all $g, h, k \in G$: $$g \leq h \iff g+k \leq h+k$$ Roughly speaking, that means that $\leq$ respects the group-structure of $G$. It is a totally ordered abelian group when, in addition, for all $g, h \in G$, either $g \leq h$ or $h \leq g$.

Note that when $G$ is a totally ordered abelian group, we have the following: $$ 0 < 2g \implies 0 < g <2g$$

Now suppose that $(G,\leq)$ is a totally ordered abelian group, and that $G$ is isomorphic, as a group, to $G'$. Let $\phi \colon G \to G'$ be a group isomorphism (paying no attention to the order structure on $G$), and define an ordering $\leq'$ on $G'$ by $$g' \leq' h' \iff \phi^{-1}(g') \leq \phi^{-1}(h')$$ It is clear that when $G'$ is equipped with $\leq'$ it becomes a totally ordered abelian group.

Now with the standard ordering, $(\mathbb{Z},+,\leq)$ is a totally ordered abelian group. Suppose for a contradiction that $\phi \colon (\mathbb{Z},+) \to (\mathbb{Q},+)$ is a group isomorphism. Equip $(\mathbb{Q},+)$ with the ordering $\leq'$ defined above, so that $(\mathbb{Q},+,\leq')$ becomes a totally ordered abelian group.

Since $0 < 1$, and $\phi(0) = 0$ (as $\phi$ is a group homomorphism), we have $0 <' q$ where $q = \phi(1)$. Since $q \in \mathbb{Q}$, also $q/2 \in \mathbb{Q}$. By the second displayed equation above, $$0<' q/2 <' q$$ That implies $$0 < \phi^{-1}(q/2) < 1$$ But there is no solution in $\mathbb{Z}$ to $0< x < 1$, so we have the needed contradiction, showing that $(\mathbb{Z},+)$ and $(\mathbb{Q},+)$ are not isomorphic as groups.