The expected number of random elements to generate a finite Abelian group (Q5948361)

For a finite group \(G\), let \(r(G)\) denote the minimal number of generators of \(G\). Also let \(E(G)\) denote the expected number of elements from \(G\), independently chosen with the uniform distribution, to generate \(G\). Clearly, \(E(G)\geq r(G)\). Let \(e(G)=E(G)-r(G)\), the excess of \(G\). For the symmetric group \(S_n\) and the alternating group \(A_n\), the results of \textit{J. D. Dixon} [Math. Z. 110, 199-205 (1969; Zbl 0176.29901)] yield that \(E(S_n)=2.5+o(1)\) and \(E(A_n)=2+o(1)\) as \(n\to\infty\). From the results of \textit{W. M. Kantor} and \textit{A. Lubotzky} [Geom. Dedicata 36, No. 1, 67-87 (1990; Zbl 0718.20011)] and \textit{M. W. Liebeck} and \textit{A. Shalev} [Geom. Dedicata 56, No. 1, 103-113 (1995; Zbl 0836.20068)], the numbers \(e(G)\) are unbounded in general but the relation \(E(G)=2+o(1)\) holds for any finite nonabelian simple group \(G\) as \(|G|\to\infty\). In the paper under review the author computes the excess \(e(G)\) for any finite Abelian group \(G\). Corollary 2 states that \(\sup\{e(G):G\text{ a finite Abelian group }\}=2.118456563\dots\). Thus the numbers \(e(G)\) remain uniformly bounded for finite Abelian groups, for finite simple groups, for the symmetric groups, and -- by the remarks at the end of the paper -- for finite nilpotent groups.