Kourovka Notebook Problem 20.36

Definitions and mathematical statement

Let $F_n=\langle x_1,\ldots,x_n\rangle$ be the free group of rank $n$, for each positive integer $n$. For a subset $S$ of a group $F$, write $\langle\!\langle S\rangle\!\rangle_F$ for its normal closure in $F$. Define \[ N_n=\langle\!\langle w^4:w\in F_n\rangle\!\rangle_{F_n}, \qquad B(n,4)=F_n/N_n. \] Thus $B(n,4)$ is the free Burnside group on $n$ generators subject to the identity $g^4=1$ for every element $g$.

For $F=F_3$, put $N=N_3$ and define \[ \mu=\min\left\{r\in{\mathbb{Z}}_{\geq0}:\ \exists w_1,\ldots,w_r\in F,\quad N=\langle\!\langle w_1^4,\ldots,w_r^4\rangle\!\rangle_F\right\}, \] with $\mu=\infty$ if this set is empty. The equality of normal subgroups means that \[ \langle x_1,x_2,x_3\mid w_1^4=1,\ldots,w_r^4=1\rangle \longrightarrow B(3,4) \] is an isomorphism under the natural map sending each displayed generator to its canonical image. Each fourth-power relator is counted once in the displayed list, irrespective of its word length.

Questions (catalogue entry).

1. (a) Is $\mu<10^6$?
2. (b) Is $\mu<10^3$?
3. (c) What is the exact value of $\mu$?

The supplied source reports that $|B(3,4)|=2^{69}$, attributing this to Bayes, Kautsky, and Wamsley (1974), whose proof uses Sanov's (1940) finiteness theorem for $B(n,4)$. It further reports that Sanov's proof for $B(3,4)$ uses more than $2^{32}$ fourth powers, explaining this by the fact that the subgroup generated by two canonical generators and the square of the third has order $2^{32}$. Finally, it attributes to Havas and Newman (1980) the lower bound of $105$ defining relations for $B(3,4)$; in particular, the reported bound gives $\mu\geq105$.

The formulation above interprets "defined with fourth powers" as a presentation on the three free generators with relators that are fourth powers of words. The source does not explicitly specify the presentation conventions; this interpretation requires human confirmation. All historical claims and the open status are supplied source assertions, not independently verified conclusions.

Short English statement

For the free Burnside group $B(3,4)$, determine the smallest number of fourth-power relations needed on its three free generators. In particular, can fewer than $1{,}000{,}000$, or fewer than $1{,}000$, such relations suffice? The supplied record labels these questions open and reports a lower bound of $105$ defining relations.

Sources

• Local UnsolvedMath registry, permanent ID $701694$, exact source code catalogue entry; supplied complete source record, upstream revision d371ee917ac9647fbb06cd6fc529deec6f3dda82.
• The Kourovka Notebook (Unsolved Problems in Group Theory), Problem 20.36; proposed by G. Havas and M. F. Newman (2022). The supplied background identifies the 20th issue (2022) and the 21st issue edition (2026), with identifier arXiv:1401.0300v48. The supplied collection description names E. I. Khukhro and V. D. Mazurov as editors.
• Bayes, Kautsky, and Wamsley (1974): cited in the supplied statement for the order $2^{69}$ of $B(3,4)$. No title, publication details, or link is supplied.
• Sanov (1940): cited in the supplied statement for finiteness of $B(n,4)$, with the additional reported qualification concerning more than $2^{32}$ fourth powers in the proof for $B(3,4)$. No title, publication details, or link is supplied.
• Havas and Newman (1980): cited in the supplied statement for the lower bound of $105$ defining relations. No title, publication details, or link is supplied.