Are there any verbal subgroups in a rank 2 free group $F(a,b)$ arising as normal closures $\langle\langle r \rangle\rangle$ of a (nontrivial) element $r \in F(a,b)$ other than the commutator subgroup $[F,F]=\langle\langle [a,b] \rangle\rangle$?

More generally, are there any subgroups of the given type $\langle\langle r \rangle\rangle$ containing a verbal subgroup?

  • 1
    $\begingroup$ It it may help: a subgroup of a free group is verbal if and only if it is stable under all group endomorphisms (aka fully characteristic). $\endgroup$ – YCor Nov 21 '13 at 19:58

1) The answer is no: namely, suppose that $n\ge 2$, $r\in F_n$ and $N=\langle\langle r\rangle\rangle$ is verbal, then either $r=1$, or $n=2$ and $r$ is conjugate to $[x_1,x_2]$. Indeed $F_n/N$ is a 1-relator group; it was established; if $r\neq 1$ it satisfies a law, and Magnus [a,b] proved that this only happens when $F_n/N$ is a solvable Baumslag-Solitar group. On the other hand, if $N$ is verbal then so is $N[F_n,F_n]$, and this implies that the abelianization of $F_n/N$ is the square of some finite group. This excludes all solvable Baumslag-Solitar groups except $\mathbf{Z^2}$.

2) The answer is yes, and precisely (by the above argument) the only examples are when $n=2$ and the relator defines a solvable Baumslag-Solitar group (in which case $N$ contains the second derived subgroup of $F_n$, which is a verbal subgroup).

[a] Magnus W. Das Identitätsproblem für Gruppen mit einer definierenden Relation, Math. Ann. (1932) 106. 295–307.

[b] Moldavanskii D. I. On a theorem of Magnus (in Russian) Uch. zap. Ivanovsk. gos. ped. inst. 1969. T.44. S.26–28.


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