Groups of exponent 60 with prescribed orders of elements (Q2746915)

Recall that if \(G\) is a group then \(\omega(G)\) denotes the set of orders of elements of \(G\). If \(\omega(G)=\{1,2\}\) then \(G\) is known to be locally finite. It is also known that if \(\omega(G)=\{1,2,3,4\}\), \(\{1,2,3,6\}\), \(\{1,2,5\}\), or \(\{1,2,3,5\}\), then \(G\) is locally finite. In [Bull. Aust. Math. Soc. 60, No. 2, 197-205 (1999; Zbl 0939.20043)], \textit{N.~D.~Gupta} and \textit{V.~D.~Mazurov} proved that if \(\omega(G)\) is a proper subset of \(\{1,2,3,4,5\}\), then \(G\) either is locally finite, or contains a nilpotent normal Sylow subgroup \(S\) such that \(G/S\) is a 5-group.NEWLINENEWLINENEWLINEIn the article under review, the author proves that if \(\omega(G)=\{1,2,3,4,5\}\) then \(G\) is locally finite.