13.63 (1995)
SolvedLet $\pi_e(G)$ denote the set of orders of elements of a group $G$. For $\Gamma \subseteq \mathbb{N}$ let $h(\Gamma)$ denote the number of non-isomorphic finite groups $G$ with $\pi_e(G) = \Gamma$. Is there a number $k$ such that, for every $\Gamma$, either $h(\Gamma) \leqslant k$, or $h(\Gamma) = \infty$?
Progress
No, there is no such number: $h(\pi_e(L_3(7^{3^r}))) = r + 1$ for any $r \geqslant 0$ (A. V. Zavarnitsine, J. Group Theory, 7, no. 1 (2004), 81–97).
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