# If a universal sentence is satisfied by all groups $(\mathbb{Z}_n,+_n,0)$, how to prove that $(\mathbb{Z},+,0)$ satisfies this sentence?

Let $$\forall x_1\forall x_2\ldots\forall x_n\, \varphi(x_1,x_2,\ldots,x_n)$$ be a universal sentence that is satisfied by all groups $$(\mathbb{Z}_n,+_n,0).$$ Prove that the group $$(\mathbb{Z},+,0)$$ satisfies this sentence.

I would prove: If $$\varphi$$ is a universal sentence and $$\bf A$$ is a structure, then $$\bf A$$ satisfies $$\varphi$$ if and only if for all finitely generated substructures $$\bf B \subseteq A$$, $$\bf B$$ satisfies $$\varphi$$.

$$(\mathbb{Z}_n,+_n,0)$$ is a finitely generated substructure of $$(\mathbb{Z},+,0)$$, and that would be the end of the proof. Is this ok?

• You're using the word "satisfies" backwards: a structure satisfies a sentence, not the other way around. Jan 13, 2020 at 23:42
• More importantly, $\mathbb{Z}_n$ is not a submodel of $\mathbb{Z}$ - think about how addition "loops around" in the former but not the latter. Jan 13, 2020 at 23:44
• Oh, yes. If it were written $(Z,+_n,0),$ it would be true. Can this be proven with $(Z,+,0)?$ Jan 14, 2020 at 9:55
• @ljubinaa What is the structure $(Z,+_n,0)$? i.e. how do you define $+_n$ on all pairs of integers? Jan 15, 2020 at 14:31

Your proposed one-sentence proof is wrong for two reasons. First, as noted by Noah in the comments, $$\mathbb{Z}_n$$ is not a substructure of $$\mathbb{Z}$$. Second, it seems you're appealing to the quoted theorem with $$\mathbf{A} = \mathbb{Z}$$ and $$\mathbf{B} = \mathbb{Z}_n$$. Even if $$\mathbf{B}$$ were a substructure of $$\mathbf{A}$$, you would only be able to conclude that if $$\mathbf{A}$$ satisfies $$\varphi$$, then $$\mathbf{B}$$ satisfies $$\varphi$$, which is the converse of what you want to prove. To use the theorem to show that $$\mathbf{A}$$ satisfies $$\varphi$$, you need to check that every finitely generated substructure of $$\mathbf{A}$$ satisfies $$\varphi$$. And since $$\mathbb{Z}$$ is itself finitely generated, this approach isn't going to work.
Here's an alternative approach: Try to find a structure $$\mathbf{M}$$ satisfying $$\varphi$$ such that $$\mathbb{Z}$$ is isomorphic to a substructure of $$\mathbf{M}$$. Then, since $$\varphi$$ is universal, you can conclude that $$\mathbb{Z}$$ satisfies $$\varphi$$.
How to find $$\mathbf{M}$$? You could use an ultraproduct or compactness. Let $$\theta_k(x)$$ be the formula $$\underbrace{x+x+\dots + x}_{k \text{ times}} \neq 0.$$
Ultraproduct proof: Let $$\mathcal{U}$$ be a non-principal ultrafilter on the natural numbers, and let $$\mathbf{M} = \prod_{n\in \mathbb{N}} \mathbb{Z}_n / \mathcal{U}$$. Since $$\mathbb{Z}_n$$ satisfies $$\varphi$$ for all $$n$$, also $$\mathbf{M}$$ satisfies $$\varphi$$, by Łoś's theorem. Similarly, $$\mathbf{M}$$ is a group. And the element $$[(1,1,1,\dots)]\in \mathbf{M}$$ has infinite order (by Łoś's theorem, since for each $$k$$, $$\theta_k(1)$$ is true in all but finitely many of the $$\mathbb{Z}_n$$). So it generates an infinite cyclic subgroup, which is isomorphic to $$\mathbb{Z}$$.
Compactness proof: Add a new constant symbol $$c$$ to the language of groups, and consider the theory $$T = T_\text{grp}\cup \{\varphi\}\cup \{\theta_k(c)\mid k\in \mathbb{N}\}$$, where $$T_\text{grp}$$ is the theory of groups. Any finite subset of $$T$$ is satisfied by $$\mathbb{Z}_n$$, interpreting $$c$$ as $$1$$, for $$n$$ larger than the largest $$k$$ such that $$\theta_k(c)$$ appears in the finite subset. So $$T$$ has a model, $$\mathbf{M}$$, which is a group satisfying $$\varphi$$. The interpretation of $$c$$ in $$\mathbf{M}$$ has infinite order, so it generates an infinite cyclic subgroup, which is isomorphic to $$\mathbb{Z}$$.