For an abelian group how can i write a first order sentence to show an element has finite order? For an Abelian group $\langle G,+,0\rangle$ how can I write a first order sentence to say that there exists an element whose order is finite?
Since the natural numbers are not in this universe?
For example, if I write "there exists $x \in  G$ and a natural number $n$  s.t. $nx=0$,since $n$ doesn't belong to $G$...
 A: Compactness of first-order logic implies that no such sentence can exist. Suppose there is such a sentence $\psi$. Then $\lnot\psi$ says there is no element of finite order. (A so-called torsion-free Abelian group.)
Let $\phi_n$ be the sentence that there is no element of order $n$. For every $n$ there is such a sentence within the language of groups (show this!). $\lnot \psi$ would show that the Abelian groups axioms + $\{\phi_n: n \in \mathbb{N}\}$ would be finitely axiomatisable, and then compactness implies there is a finite subset $\{\phi_{n_1}, \ldots, \phi_{n_k}\}$ of the $\phi_n$ that (with the Abelian group axioms) also would axiomatise the theory of groups without an element of finite order. But you can define $\mathbb{Z}_m$ where $m = 1+\prod_{i=1}^k {n_i}$ which is a group that has no elements of order $n_1,\ldots,n_k$ (as element orders divide $m$), but does have $1$ as an element of finite order. Contradiction, and so such torsion-free groups (or their complement) aren't finitely axiomatisable. 
