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2.43 (1966)

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A group $G$ is called an FN-group if the groups $\gamma_i G / \gamma_{i+1} G$ are free abelian and $\bigcap_{i=1}^\infty \gamma_i G = 1$, where $\gamma_{i+1} G = [\gamma_i G, G]$. A variety of groups $\mathfrak{M}$ is called a $\Sigma$-variety (where $\Sigma$ is an abstract property) if the $\mathfrak{M}$-free groups have the property $\Sigma$.
$\qquad$ a) Which properties $\Sigma$ are preserved under multiplication and intersection of varieties? Is the $FN$ property preserved under these operations?
$\qquad$ b) Are all varieties obtained by multiplication and intersection from the nilpotent varieties $\mathfrak{N}_1, \mathfrak{N}_2, \dots$ (where $\mathfrak{N}_1$ is the variety of abelian groups) $FN$-varieties?

Contributor: A. I. Mal’cev

Progress

a) The property $FN$ is preserved by multiplication of varieties (A. L. Shmel'kin, Trans. Moscow Math. Soc., 29 (1973), 239–252). The property $FN$ is not preserved by the intersection of varieties. The following example is due to L. G. Kovács. Let $\mathfrak{U}$ and $\mathfrak{V}$ be the varieties of all nilpotent groups of class at most 4 satisfying the identities $[x, y, y, x] \equiv 1$ and $[[x, y], [z, t]] \equiv 1$, respectively. Then $\mathfrak{U}$ and $\mathfrak{V}$ are $FN$-varieties because 1) both of them are nilpotent of class at most 4 and contain all nilpotent groups of class $\leqslant 3$, and 2) relatively free groups in $\mathfrak{U}$ and $\mathfrak{V}$ are torsion-free (well-known for $\mathfrak{V}$ and follows for $\mathfrak{U}$ from (P. Fitzpatrick, L. G. Kovács, J. Austral. Math. Soc. Ser. A, 35, no. 1 (1983), 59–73)). On the other hand, $\mathfrak{U} \cap \mathfrak{V}$ is not an $FN$-variety because the relatively free group in $\mathfrak{U} \cap \mathfrak{V}$ of rank 3 is not torsion-free. Indeed, every torsion-free group in $\mathfrak{U} \cap \mathfrak{V}$ is of class $\leqslant 3$ (follows from the paper by Fitzpatrick and Kovács cited above), and there is a 3-generated (exponent 2)-by-(exponent 2) group of class precisely 4 in $\mathfrak{U} \cap \mathfrak{V}$. (A. N. Krasil'nikov, Letter of July, 17, 1998.)

b) Yes (Yu. M. Gorchakov, Algebra i Logika, 6, no. 3 (1967), 25–30 (Russian)).

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