How to formulize some group is cyclic in first order logic? How should o formulize a group  is cyclic?
To write this i need to use the set of natural numbers but the universe is just the set of elements of the group.
If the group is (Z,+) i can use the positive part of Z to take needed n but just if < is in language... If not i don't know what to do...
 A: We can show that "being cyclic" is not first-order expressible, that is, 
Proposition: 

Proposition: There is no first-order sentence $\varphi$ in the language of groups $L=\{+\}$ such that $G\models \varphi$ if and only if $G$ is cyclic.

Proof: Suppose $\varphi$ is such sentence. For every $n\geq 1$, let $\phi_n$ be the sentence $$\displaystyle{\phi_n:=\exists x_1,\ldots,x_n\left(\bigwedge_{1\leq i<j\leq n}x_i\neq x_j\right),}$$ expressing the existence of at least $n$ distinct elements. Consider the set of sentences 
$$\Gamma:=\{\sigma_{ab-gr},\varphi, \forall x \exists y (y+y=x)\}\cup\{\phi_n:n\in\mathbb{N}\}$$ where $\sigma_{ab-gr}$ is a sentence expressing that $(G,+)$ is an abelian group. We show now that $\Gamma$ is finitely consistent.
Let $\Gamma_0$ be a finite subset of $\Gamma$. Then, $\Gamma_0$ uses finitely many sentences of the form $\phi_n$, and by taking an odd prime $p>\max\{n:\phi_n\in \Gamma_0\}$ we have $\mathbb{Z}/p\mathbb{Z}\models \Gamma_0$, as it is a cyclic group, every element "can be divided by 2" and has more than $n$ elements for every $n$ with $\phi_n\in \Gamma_0$.
Therefore, by compactness theorem, $\Gamma$ is consistent, and by completeness theorem, it must have a model $(M,+)$. 
However, if $(M,+)\models \Gamma$, then $(M,+)$ is an infinite cyclic group, and since the only infinite cyclic group (up to isomorphism) is $(\mathbb{Z},+)$ we would have $(\mathbb{Z},+)\models \forall x \exists y (y+y=x)$, a contradiction as $x=1$ is not "divisible by $2$".
