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11.64 (1990)

Solved

Let $\pi(G)$ denote the set of prime divisors of the order of a finite group $G$. Are there only finitely many finite simple groups $G$, different from alternating groups, which have a proper subgroup $H$ such that $\pi(H) = \pi(G)$?

Contributor: V. S. Monakhov

Progress

No, there are infinitely many such groups. If $G = S_{4k}(2^s)$ and $H \cong \Omega^-_{4k}(2^s)$, then $\pi(G) = \pi(H)$ (V. I. Zenkov, Letter of March, 10, 1994).

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