Kourovka Notebook Problem 16.91

Definitions and mathematical statement

Let $F$ be a free group with an infinite basis, with no restriction to countable rank. Write $[F,F]$ for the subgroup generated by all commutators $x^{-1}y^{-1}xy$, where $x,y\in F$, and let $F_{\mathrm{ab}}=F/[F,F]$ be its abelianization. Define the group of $IA$-automorphisms by \[ \operatorname{IA}(F)=\ker\!\left(\operatorname{Aut}(F)\longrightarrow\operatorname{Aut}(F_{\mathrm{ab}})\right). \] Thus an automorphism belongs to $\operatorname{IA}(F)$ precisely when it induces the identity on the abelianization. For $\alpha\in\operatorname{Aut}(F)$, its normal closure in $\operatorname{Aut}(F)$ is \[ \langle\!\langle\alpha\rangle\!\rangle_{\operatorname{Aut}(F)} =\left\langle\beta\alpha\beta^{-1}:\beta\in\operatorname{Aut}(F)\right\rangle. \] Is it true that for every infinitely generated free group $F$ there exists $\alpha\in\operatorname{IA}(F)$ such that \[ \langle\!\langle\alpha\rangle\!\rangle_{\operatorname{Aut}(F)}=\operatorname{IA}(F)? \]

Short English statement

For every free group of infinite rank, does there exist a single automorphism acting trivially on its abelianization whose conjugates by all automorphisms generate the entire group of automorphisms acting trivially on the abelianization?

Sources

• Supplied local UnsolvedMath registry record, permanent ID $701282$, exact source code catalogue entry, title Kourovka Notebook Problem 16.91. The supplied registry labels the problem open; this status has not been independently verified.
• V. Tolstykh, proposer, 2006. The Kourovka Notebook (Unsolved Problems in Group Theory), original problem 16.91, 16th issue (2006). The supplied background also identifies the 21st issue edition (2026), arXiv:1401.0300v48.
• The supplied collection description attributes editing of The Kourovka Notebook, Issues 1--21 (1965--2026), to E. I. Khukhro and V. D. Mazurov.