Suppose $\mathfrak{U}$ is a group variety. Let’s define $N_{\mathfrak{U}} \subset \mathbb{N}$ as a such set of numbers, that for any finite group $G$, $|G| \in N_{\mathfrak{U}}$ implies $G \in \mathfrak{U}$.


If $\mathfrak{O}$ is the variety of all groups, then $N_{\mathfrak{O}} = \mathbb{N}$.

If $\mathfrak{B}_m$ is the variety of all groups of exponent $m$, then $N_{\mathfrak{B}_m}$ is the set of all divisors of $m$

If $\mathfrak{N}_c$ is the variety of all groups of nilpotency class $c$, then $N_{\mathfrak{N}_c}$ is the set of all numbers $n=p_1^{e_1}\cdots p_m^{e_m}$ with $p_i^k\not\equiv 1(\mod p_j)$ for $i,j\in\{1,\ldots,m\}$ and $1\leqslant k\leqslant e_i$, and $e_i \leq c + 1$ for $i\in\{1,\ldots,m\}$.

If $\mathfrak{U}$ and $\mathfrak{V}$ are two varieties, then $N_{\mathfrak{U}\cap\mathfrak{V}} = N_{\mathfrak{U}} \cap N_{\mathfrak{V}}$

My question is:

Does there exist some number-theoretic characterisation of all such subsets $N \subset \mathbb{N}$, such that $N = N_{\mathfrak{U}}$ for some variety $\mathfrak{U}$?

Any $N_{\mathfrak{U}}$ satisfies the property:

If $a \in N_{\mathfrak{U}}$ and $b | a$, then $b \in N_{\mathfrak{U}}$

Suppose $|G| = b$ and $G \notin \mathfrak{U}$. Then $G \times C_{\frac{a}{b}} \notin \mathfrak{U}$.

If $\exists n \in \mathbb{N}$, such that $\forall k \in \mathbb{N}$ $n^k \in N_{\mathfrak{U}}$, then $\mathfrak{U} = \mathfrak{O}$.

By previous lemma, we can assume without loss of generality, that $n = p$ is prime. The only variety, that contains all $p$-groups for a fixed prime $p$ is $\mathfrak{O}$

However, I am not sure, whether those two conditions are sufficient to characterise all such sets or not.

This question was inspired by this MO question


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