Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.