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14.103 (1999)

Solved

Let $H$ be a proper subgroup of a group $G$ and let elements $a, b \in H$ have distinct prime orders $p, q$. Suppose that, for every $g \in G \setminus H$, the subgroup $\langle a, b^g \rangle$ is a finite Frobenius group with complement of order $pq$. Does the subgroup generated by the union of the kernels of all Frobenius subgroups of $G$ with complement $\langle a \rangle$ intersect $\langle a \rangle$ trivially? The case where all groups $\langle a, b^g \rangle$, $g \in G$, are finite is of special interest.

Contributor: V. P. Shunkov

Progress

No, in general case not necessarily. As a counter-example one can take $G = \langle x, y, z, a, b \mid x^7 = y^7 = [x, y, y] = [x, y, x] = a^3 = b^2 = [a, b] = 1, x^a = x^2, y^a = y^2, x^b = x^{-1}, y^b = y^{-1} \rangle$ with $H = \langle a, b \rangle$. Analogous examples exist also for $p = 2$ and every odd prime $q$. (A. I. Sozutov, Letter of 2002.)

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