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10.75 (1986)

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Suppose that a group $G$ contains an element $a$ of prime order $p$ such that the normalizer of every finite subgroup containing $a$ has finite periodic part and all subgroups $\langle a, a^g \rangle$, $g \in G$, are finite and almost all of them are soluble. Does $G$ possess a periodic part if $p > 2$?

Contributor: V. P. Shunkov

Progress

It was proved in (V. P. Shunkov, Groups with involutions, Preprints no. 4, 5, 12 of the Comput. centre of SO AN SSSR, Krasnoyarsk, 1986 (Russian)) that if $a$ is a point, then the answer is affirmative; on the other hand, a group with a point $a$ of order 2 satisfying the given hypothesis, which has no periodic part, was exhibited in the same works. For the definition of a point see (V. I. Senashov, V. P. Shunkov, Algebra and Logic, 22, no. 1 (1983), 66–81).

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