Can we write down a sentence $\sigma$ which will characterize all groups in which any two non-identity element have the same order?

Let $\mathcal{L} = \{e, \cdot\}$ be the language of group theory. Can we write down a sentence $\sigma$ in this language such that $\mathcal{G} \vDash \sigma$ if and only if $\mathcal{G}$ is a group in which any two non-identity element have the same order?

Note that if we were to find a theory $\Sigma$ instead of a sentence, we can do the following: Let $$\Sigma = \text{the group axioms} \cup \{\forall x \forall y \ (x \neq e \land y \neq e) \rightarrow (x \cdot x = e \leftrightarrow y \cdot y = e), \\ \forall x \forall y \ (x \neq e \land y \neq e) \rightarrow (x \cdot x \cdot x = e \leftrightarrow y \cdot y \cdot y = e), \ldots \}.$$ Now $\mathcal{G} \vDash \Sigma$ if and only if $\mathcal{G}$ is a group with the required property.

However, since $\Sigma$ is infinite, we cannot turn it into a single sentence $\sigma$, so I am asking whether we can find a finite $\Sigma$ that will characterize all such groups, thank you.

Solution 1:

This is not a full answer, just a long comment.

Let $\Sigma_n$ be the axiom scheme $\Sigma$ truncated to powers of order $<n$.

To prove that there is no consistent sentence $\sigma$ such that $\sigma\leftrightarrow\Sigma$, it suffices to show that $\Sigma_n$ does not imply $\Sigma$. That is,

● for every $n$ there is a group $\mathcal G$ such that $\mathcal G\models\Sigma_n\wedge\neg\Sigma_m$ for some $m$.

In fact, if $\sigma\leftrightarrow\Sigma$ then, by compactness, $\Sigma_n\rightarrow\sigma$ for some $n$ and therefore $\Sigma_n\rightarrow\Sigma$.

To prove ● let $\mathcal G$ be as suggested by Tan in the comments below. That is, $\mathcal G$ is a group of order $p\cdot q$ where $p$ and $q$ are primes greater than $n$. Then ● follows from the theorems of Cauchy and Lagrange.