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3.8 (1969)

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Let $G$ be the free product of free groups $A$ and $B$, and $V$ the verbal subgroup of $G$ corresponding to the equation $x^4 = 1$. Is it true that, if $a \in A \setminus V$ and $b \in B \setminus V$, then $(ab)^2 \notin V$?

Contributor: V. G. Vilyatser

Progress

No, it is not. If $A$ and $B$ are free groups with bases $a_1, a_2$ and $b_1, b_2$, respectively, then, for example, the commutators $a = [[a_1, a_2, a_2], [a_1, a_2]]$ and $b = [[b_1, b_2, b_2], [b_1, b_2]]$ do not belong to $V$, but $(ab)^2 \in V$. Indeed, it follows from (C. R. B. Wright, Pacif. J. Math., 11, no. 1 (1961), 387–394) that $a^2 \in V$, $b^2 \in V$, and $[a, b] \in V$. (V. A. Roman’kov, Talk at the seminar Algebra and Logic, March, 17, 1970.)

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