6
$\begingroup$

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?

$\endgroup$
1
  • 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, 2013 at 19:58

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.