On regular automorphisms of order \(3\) and Frobenius pairs (Q1587001)

An automorphism of a group \(G\) is regular if the neutral element of \(G\) is the only fixed point of this automorphism. A group \(A\) of automorphisms of a group \(G\) is regular if every nonidentity automorphism in \(A\) is regular. The author gives an answer to question 14.57(b) of the Kourovka notebook posed by V.~D.~Mazurov [\textit{V.~D.~Mazurov} and \textit{E.~I.~Khukhro} (ed.), The Kourovka notebook. Unsolved problems in group theory, Institute of Mathematics, Novosibirsk (1999; Zbl 0943.20003)]. More precisely, he proves the following theorem: Let \(A\) be a nontrivial regular group of automorphisms of an Abelian group \(G\) which is generated by elements of order 3. Suppose any of the following two conditions is satisfied: (i) \(A\) is a periodic group; (ii) \(G\) has a nontrivial element of finite order and the product of any two elements of order 3 in \(A\) has finite order. Then the group \(A\) is finite and isomorphic to a group of order 3 or to the group \(\text{SL}_2(3)\) of order 24 or to the group \(\text{SL}_2(5)\) of order 120. The author also gives an affirmative answer to V.~P.~Shunkov's conjecture 6.56 on nilpotency of the kernel of a Frobenius group whose complement contains an element of order 3 stated in the Kourovka notebook. Recall that a Frobenius group \(G\) with kernel \(F\) and complement \(H\) is a semidirect product \(FH\) with the following conditions: (i) \(H\) is a proper subgroup of \(G\) and \(H\cap H^f=1\) for every nontrivial \(f\in F\); (ii) \(F\setminus\{1\}=G\setminus\{H^f\mid f\in F\}\).