8.30 (1982)

Open

Let $\mathfrak{X}$, $\mathfrak{Y}$ be Fitting classes of soluble groups which satisfy the Lockett condition, i. e. $\mathfrak{X} \cap \mathfrak{S}_* = \mathfrak{X}_*$, $\mathfrak{Y} \cap \mathfrak{S}_* = \mathfrak{Y}_*$ where $\mathfrak{S}$ denotes the Fitting class of all soluble groups and the lower star the bottom group of the Lockett section determined by the given Fitting class. Does $\mathfrak{X} \cap \mathfrak{Y}$ satisfy the Lockett condition?

Progress

Editors’ comment (2001): The answer is affirmative if $\mathfrak{X}$ and $\mathfrak{Y}$ are local (A. Grytczuk, N. T. Vorob’ev, Tsukuba J. Math., 18, no. 1 (1994), 63–67).

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