8.73 (1982)

Solved

We say that a finite group $G$ separates cyclic subgroups if, for any cyclic subgroups $A$ and $B$ of $G$, there is $g \in G$ such that $A \cap B^g = 1$. Is it true that $G$ separates cyclic subgroups if and only if $G$ has no non-trivial cyclic normal subgroups?

Progress

Not always. For $p_1 = 2, p_2 = 5, p_3 = 11, p_4 = 17$ let $R_i$ be an elementary abelian group of order $p_i^2$ and $\varphi_i$ a regular automorphism of order 3 of $R_i$, $i = 1, 2, 3, 4$. In the direct product $R_1\langle \varphi_1 \rangle \times R_2\langle \varphi_2 \rangle \times R_3\langle \varphi_3 \rangle \times R_4\langle \varphi_4 \rangle$ let $G$ be the subgroup generated by all the $R_i$ and the elements $\varphi_2\varphi_3\varphi_4$ and $\varphi_1\varphi_3\varphi_4^{-1}$. Then $G$ has no non-trivial cyclic normal subgroups and every (cyclic) subgroup of order $2 \cdot 5 \cdot 11 \cdot 17$ intersects any of its conjugates non-trivially. (N. D. Podufalov, Abstracts of the 9th All-Union Group Theory Symp., Moscow, 1984, 113–114 (Russian).)

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