6.19 (1978)

Solved

Let $R$ be a nilpotent associative ring. Are the following two statements for a subgroup $H$ of the adjoint group of $R$ always equivalent: 1) $H$ is a normal subgroup; 2) $H$ is an ideal of the groupoid $R$ with respect to Lie multiplication?

Progress

Not always. Let $R$ be the free nilpotent of index 3 associative algebra over $\mathbb{F}_2$ on the free generators $x, y$. Let $S$ be the subalgebra generated by the elements $[x, y^2] = x \ast y^2$, $[x, xy] = x \ast (xy) = x(x \ast y)$, $[x, yx] = x \ast (yx) = (x \ast y)x$, where $[a, b]$ denotes the commutator in the adjoint group with multiplication $a \circ b = a + b + ab$, and $a \ast b = ab - ba$ is Lie multiplication. Let $M, N$ be the subalgebras generated by $S$ and the elements $x \ast y$ and $[x, y]$, respectively. The minimal subgroup of $(R, \circ)$ that contains $x$ and is an ideal of the groupoid $(R, \ast)$ equals $\langle x \rangle \circ M = \langle x \rangle + M$, where $\langle x \rangle = \{0, x, x^2, x+x^2+x^3\}$, while the minimal normal subgroup containing $x$ equals $\langle x \rangle \circ N = \langle x \rangle + N$. Neither is contained in the other. (E. I. Khukhro, Letter of July, 23, 1979.)

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