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5.10 (1976)

Solved

Let $E = H \ast K$ and denote the augmentation ideals of the groups $E, H, K$ by $\mathfrak{e}, \mathfrak{h}, \mathfrak{k}$, respectively. If $I$ is a right ideal in $\mathbb{Z}E$, let $d_E(I)$ denote the minimum number of generators of $I$ as a right ideal. Assuming $H$ and $K$ finitely generated, is it true that $d_E(\mathfrak{e}) = d_E(\mathfrak{h}E) + d_E(\mathfrak{k}E)$? Here $\mathfrak{h}E$ is, of course, the right ideal generated by $\mathfrak{h}$; similarly for $\mathfrak{k}E$.

Contributor: K. W. Gruenberg

Progress

Not always (P. A. Linnell, unpublished). A simple example: $H$ is elementary abelian group of order 4, $K$ is elementary abelian of order 9. Here $d_E(\mathfrak{e}) = 3$, although $d_E(\mathfrak{h}E) = d_E(\mathfrak{k}E) = 2$. (K. W. Gruenberg, Letter of July, 27, 1995.)

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