# Varieties of groups with certain properties

Is there an example of a periodic variety $\mathbf{V}$ of groups that satisfies all of the following properties?

1. $\mathbf{V}$ is finitely based
2. $\mathbf{V}$ contains finitely many subvarieties
3. $\mathbf{V}$ is not finitely generated

Recall that a variety of groups is periodic if it satisfies the identity $x^n = 1$ for some $n \geq 1$; finitely based if all its identities are deducible from some finite set of its identities; non-finitely generated if it cannot be generated by a single finite group.

• You might add a few terms for the nonspecialists as well as the specialists with short memories. I am guessing that periodic means there is a positive integer n so that $x^n = x$ is an identity of V, finitely based means there is a finite set of identities that defines the desired variety V, and not finitely generated means no finite subset C of finite groups satisfies V=HSP(C). Every variety (at least of finite type) is generated by an infinitely generated free algebra, so 3 needs some clarification. Gerhard "Minds Some Threes And Twos" Paseman, 2017.06.26. – Gerhard Paseman Jun 26 '17 at 20:26
• I suspect (my interpretation of your) 3 implies there are infinitely many subvarieties, even ignoring 1 and ignoring restrictions on the type. If I can, I will post a proof after I consult references. If you want to try it, check out Chapter 4 of Algebras, Lattices, Varieties by McKenzie, McNulty, Taylor. Gerhard "Memory Not So Long Anymore" Paseman, 2017.06.26. – Gerhard Paseman Jun 26 '17 at 20:40
• There exist non-finitely generated varieties of groups with only 3 subvarieties (Kozhevnikov 2012). Therefore I am curious to know if there is a finitely based example. – E W H Lee Jun 26 '17 at 20:54
• Note that the condition that $\mathbf{V}$ is periodic follows from 1-2-3. Indeed, a non-periodic variety contains, for every $k$, the variety of abelian groups of exponent dividing $k$ as a subvariety. – YCor Jun 27 '17 at 7:48