First Order Logic and groups: Prove that no $T$ exists such that $M \vDash T \cup T_{grp}$ **IFF** I'm learning First Order Logic by myself using a University textbook, and it has the following question in it (as a self exercise):
Let $L = \{\cdot, e \}$ the language of Groups and let $T_{grp}$ be the theory of groups:
A group which has 3 formulas that guarantee


*

*The associative property

*The existence of a neutral multiplication constant

*The existence of an inverse element for every element


Prove that no $T$ exists such that $M \vDash T \cup T_{grp}$ IFF for every $g\in|M|$ there exits an $n\in \mathbb{N}$ such that $g^{n} = e^{M}$ (where $g^n$ is multiplying g by itself $n$ times)
I've been sitting on this question for over an hour and can't seem to figure out what I'm supposed to do.
Can anyone shed some light?
Any help is appreciated, Thank you
 A: I believe that you want to prove that the theory of groups in which every element has finite order is not axiomatizable. 
Assume that such $T$ exists. Consider new language $L'=L\cup\{c\}$, and a theory $T'=T\cup T_{grp}\cup\{c^n\neq e\mid n\geq 1\}$. Note that $T'$ doesn't have model, since its model $M$ would be a model of $T\cup T_{grp}$, hence every element of $M$ is of finite order, and $T'$ says that $c^M$ is of infinite order.
By compactness, some finite subset $T_0'\subseteq T'$ doesn't have model, hence $T_0=T\cup T_{grp}\cup T_0'$ doesn't have model. But $T_0$ contains only finitely many sentences $c^n\neq e$, i.e. there exists $N\geq 1$ such that $c^n\neq 1$ is not in $T_0$, for $n\geq N$. Now, consider group $M=(Z_N,+_N,0,c^M=1)$, where $Z_N=\{0,1,2\ldots,N-1\}$ and $+_N$ is addition modulo $N$. Every element of $M$ is of finite order, hence $M$ is a model for $T\cup T_{grp}$. Also $c^n\neq e$, for $n<N$, are satisfied, since $c^M=1$ is of order $N$. Therefore $M\vDash T_0$, a contradiction.
A: Another argument:
Let $T$ be a theory such that $G \models T$ for any torsion group $G$. Now, let $P$ denote an ultraproduct $\prod\limits_{n \geq 2} \mathbb{Z}/ n \mathbb{Z}$. By assumption, $\mathbb{Z}/ n \mathbb{Z} \models  T$ for any $n \geq 2$ so $P \models T$. But the order of the element in $P$ corresponding to the sequence $(1,1,1, \ldots)$ is clearly infinite.
Therefore, any theory satisfied by all torsion groups is also satisfied by a group with an element of infinite order.
