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3.43 (1969)

Open

Let $\mu$ be an infinite cardinal number. A group $G$ is said to be $\mu$-overnilpotent if every cyclic subgroup of $G$ is a member of some ascending normal series of length less than $\mu$ reaching $G$. It is not difficult to show that the class of $\mu$-overnilpotent groups is a radical class. Is it true that if $\mu_1 < \mu_2$ for two infinite cardinal numbers $\mu_1$ and $\mu_2$, then there exists a group $G$ which is $\mu_2$-overnilpotent and $\mu_1$-semisimple?

Contributor: B. I. Plotkin

Progress

Remark of 2001: In (S. Vovsi, Sov. Math. Dokl., 13 (1972), 408–410) it was proved that for any two infinite cardinals $\mu_1 < \mu_2$ there exists a group that is $\mu_2$-overnilpotent but not $\mu_1$-overnilpotent.

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