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17.125 (2010)

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Does every finite group $G$ contain a pair of conjugate elements $a, b$ such that $\pi(G) = \pi(\langle a, b \rangle)$? This is true for soluble groups.

Contributor: P. Shumyatsky

Progress

Comment of 2013: it was proved in (A. Lucchini, M. Morigi, P. Shumyatsky, Forum Math., 24 (2012), 875–887) that every finite group $G$ contains a 2-generator subgroup $H$ such that $\pi(G) = \pi(H)$.

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