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I want use semi-formal language to describe the following four points.

(1) The group axioms with signature $\{*\}$

(2) The property "linear order" with signature $\{<\}$

The following properties are not able to formulate:

(3) Torsion group : $\forall x\in G \exists n\in \mathbb N: g^n=e$

(4) $\not\exists$ a normal subgroup

My approach:

(1) Associativity: $(\forall x,y,z) [(x*y)*z=x*(y*z)]$

Neutral element: $\forall x\exists e [x*e=e*x=x]$

Inverse element: $\forall a\exists b[a*b=b*a=e]$

Is this correct?

(2) Antisymmetry: $(\forall x,y,z) [x<=y \vee y<=x\rightarrow x=y]$

Transitivity: $(\forall x,y,z) [x<=y \vee y<=z\rightarrow x<=z]$

Totality: $(\forall x,y) [x<=y \vee y<=x]$

(3) I think it is not possible to formulate it because we have no information about the $n$. The teacher told me it is difficult formal proof and we wont discuss it, but I would be interested in a formal proof for this.

(4) Same here, normal subgroups are subsets, no elements, therefore we would need second-order-logic.

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Neutral element is not stated correctly. For torsion, would use Compactness Theorem. If you have not yet met that, there is a lot of ground to cover. –  André Nicolas Apr 17 '13 at 18:54
    
Ok I edited the neutral element. Well I have of the compactness theorem as a corollary of Gödel's completeness theorem. Still I would be very thankful if somebody could show me a proof using compactness theorem. –  Voyage Apr 17 '13 at 19:03
    
Often we have a constant symbol for the neutral element. If we don't, write $\exists e\forall x (e\ast x=x\ast e=x)$. –  André Nicolas Apr 17 '13 at 19:29

2 Answers 2

up vote 2 down vote accepted

We solve the torsion group part of the question. Suppose that there is a set $S$ of sentences that can be added to the other axioms of group theory such that the models of the resulting theory $T$ are precisely the torsion groups.

Add a constant symbol $c$ to the language. Add to $T$ the special axioms $\phi_2,\phi_3,\phi_4,\dots$, where $\phi_k$ says that $c^k$ is not equal to the identity. It is not difficult to write down the $\phi_k$. Let the resulting theory be $T'$.

We claim that the theory $T'$ is consistent. If it is not, some finite subset $T_0$ of $T'$ is inconsistent. Such a finite subset can include only finitely many of the $\phi_k$. Suppose all $k$ such that $\phi_k$ is in $T_0$ are $\lt N$. It is easy to produce a model of $T_0$: a cyclic group of order $N$ will do the job.

We conclude that $T'$ has a model $G$. If $g$ is the interpretation of the constant symbol $c$ in $G$, then $g$ satisfies all the special axioms, so $g$ has infinite order. This contradicts the assumption that the only models of $T$ are torsion groups.

Remark: The question essentially asked whether there is a single sentence that "says" we have a torsion group. The solution shows that in fact we cannot even produce a set (possibly infinite) of sentences that will do the job.

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Thank you, two questions. First, am I right you used the compactness theorem to state there has to be a finite subset $T_0$ of $T'$... ? Secondly, $T_0$ only consists of $\phi_k$ where $k<N$ ? –  Voyage Apr 17 '13 at 19:58
    
Compactness was used to prove consistency of $T'$. We showed that every finite subset of $T'$ has a model. Compactness then shows $T'$ has a model. (But later this leads to a contradiction, so we conclude there cannot be a theory $T$ with the desired properties.) And $T_0$ is any finite subset of $T'$. The theory $T_0$ could have many sentences in it, most of them unconnected to the $\phi_k$. We concentrated on the $\phi_k$ in it because that's what leads to the result. –  André Nicolas Apr 17 '13 at 20:04
    
I should add that the argument in the post is of a standard nature, there was no creativity on my part. –  André Nicolas Apr 17 '13 at 20:06

If you know of ultraproducts, you can solve both (3) and (4) at the same time.

Each ${\bf Z}_p$ is a simple torsion group, so if you take a group $G=\prod_p {\bf Z}_p/\mathcal U$ (where $\mathcal U$ is a non-principal ultrafilter), you will get a group which is torsion-free (because for almost all $p$ all elements of ${\bf Z}_p$ have order greater than a given $n$), and is also an infinite abelian group, so it can't be simple.

Since each ${\bf Z}_p$ is a simple torsion group, and their ultraproduct is neither simple nor a torsion group, these are not first-order properties.

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