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11.21 (1990)

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Let $\mathfrak{N}_p$ denote the formation of all finite $p$-groups, for a given prime number $p$. Is it true that, for every subformation $\mathfrak{F}$ of $\mathfrak{N}_p$, there exists a variety $\mathfrak{M}$ such that $\mathfrak{F} = \mathfrak{N}_p \cap \mathfrak{M}$?

Contributor: A. F. Vasil’yev

Progress

No, it is not true. There is a natural one-to-one correspondence between formations of finite $p$-groups and varieties of pro-$p$-groups: for every variety $\mathcal{V}$ of pro-$p$-groups the class of all finite groups in $\mathcal{V}$ is a formation of $p$-groups and every formation of $p$-groups arises in this way. There are continuously many varieties of nilpotent pro-$p$-groups of class at most 6 (A. N. Zubkov, Siberian Math. J., 29, no. 3 (1988), 491–494) and only countably many varieties of nilpotent groups of class at most 6. (A. N. Krasil’nikov, Letter of July, 17th, 1998.)

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