4
$\begingroup$

Let $p$ be a prime and let $F$ be a free group of rank $d\geq 1$. Kostrikin [1] proved that the $d$-generated Burnside group $B=B(d,p)=F/F^p$ of exponent $p$ has a maximal finite quotient $\overline{B}=\overline{B(d,p)}=B/N$ where $N$ is the intersection of all the subgroups of $B$ of finite index. I understand that Kostrikin proved that the nilpotency class $c(d,p)$ of $\overline{B(d,p)}$ has an upper bound.

Question 1: Are lower bounds known for $c(d,p)$?

Question 2: Is $c(d,p)\geq d$ for $d\leq 4$ and $p>d$?

Question 3: For which $n$ is the $n$th section of the lower exponent-$p$ central series for $\overline{B(d,p)}$ isomorphic to the $n$th Lie power $L^n(V)$ where $V=({\mathbb F}_p)^d$?

[1] Kostrikin, A. L: On Lie rings with Engel's condition. Dokl. Akad. Nauk SSSR 108, no. 4 (1956) 580-582.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

For Questions 1, 2. We know $c(d,2)=1$ for all $d$ and $c(1,3)=1$, $c(2,3)=2$ and $c(d,3)=3$ for all $d>2$. So suppose if necessary $p\geq 5$.

Take $G(d,p)$ to be the unitiriangular matrices of size $(d+1)\times (d+1)$ over the field of size $p$, where $p>d$. Then $G(d,p)$ is nilpotent of class $d$ and the exponent is $p$. The group can be generated by $d$ elements.

$\endgroup$
2
  • $\begingroup$ Thanks Alireza. A lower bound can be established with a `witness': if $G$ is $d$-generated, has exponent $p$, and class $c$, then $c(d,p)\geq c$. For example $c(2,p)\geq 2$ if $p>2$; take $G$ to be extraspecial of exponent $p$. Your group $G(d,p)$ does not have exponent $p$ if $d>3$, and its class is $d-1$ not $d$. For Question 2 it suffices to find witnesses $G_{d,p}$ for $d=3,4$ and all $p\geq5$. $\endgroup$
    – Glasby
    Commented Jul 1, 2015 at 14:36
  • 1
    $\begingroup$ Sorry Alireza. I think you wanted $G(d,p)$ to be $(d+1)\times(d+1)$ upper triangular matrices over ${\mathbb F}_p$ (not $d\times d$). Then $G(d,p)$ is a witness: $d$-generated, has exponent $p$ if $p> d$, and class $d$. Thus $c(d,p)\geq d$ for $p>d$. This answers Questions 1 and 2. $\endgroup$
    – Glasby
    Commented Jul 1, 2015 at 15:35

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.