A question about commutators in free groups

Question

Let $F$ be the free group on $X=\{ x_1,\dots, x_n\}$ for some $n\geq2$. Define the lower central series of $F$ inductively: $\gamma_1(F):= F$, $\gamma_{i+1}(F)=[\gamma_i(F),F]$ for $i\geq1$. Is it true that $$[x_{k_1},\dots,x_{k_n}]\in\gamma_{n+1}(F)$$ whenever exist $i,j\in\{1,\dots,n\}$, $i\neq j$, such that $x_{k_i}=x_{k_j}$? In other words, I wonder if the previous condition is sufficient to have $$[x_{k_1},\dots,x_{k_n}]\equiv1 \mod{\gamma_{n+1}(F)}.$$

Answer 1

No, this is false. I will write commutators as $[x,y]=x^{-1}y^{-1}xy$ and left-normed (so $[x_1,x_2,\ldots,x_n] = [\cdots [x_1,x_2],x_3]\cdots ],x_n]$). Similar arguments hold for right-normed commutators and/or the convention $[x,y]=xyx^{-1}y^{-1}$.

A group that satisfies the identity $[x,y,y]=1$ for all $x$ and $y$ is said to be a $2$-Engel group (equivalently, if $[x,y,x]=1$ for all $x$ and $y$).

(More generally, the $n$-Engel word is defined recursively by $e_0(x,y)=x$ and $e_{n+1}(x,y) = [e_{n}(x,y),y]$. A group $G$ is an Engel group if for every $x,y\in G$ there exists $n$, which may depend on $x$ and $y$, such that $e_{n}(x,y)=1$; a group $G$ is $n$-Engel if $e_n(x,y)=1$ for all $x,y\in G$.)

There is a lot of interest in Engel groups because of a long-standing open problem: is every $n$-Engel group locally nilpotent? This is Problem N7 in World of Groups.

But to answer your specific question: there are nilpotent groups of class $3$ that are not $2$-Engel. That is, groups $G$ with $G_4=\{1\}$, but for which there exist $x,y\in G$ with $[x,y,y]\neq 1$. For instance, for an odd prime $p$ let $G$ be the group of order $p^4$ consisting of elements of the form $$x^{a_1}y^{a_2}[y,x]^{a_3}[y,x,x]^{a_4},\qquad 0\leq a_1,\ldots,a_4\lt p,$$ with multiplication given by $$\bigl(x^{a_1}y^{a_2}[y,x]^{a_3}[y,x,x]^{a_4}\bigr)\bigl( x^{b_1}y^{b_2}[y,x]^{b_3}[y,x,x]^{b_4}\bigr) = x^{c_1}y^{c_2}[y,x]^{c_3}[y,x,x]^{c_4},$$ where $$\begin{align*} c_1 &= a_1+b_1\bmod p\\ c_2 &= a_2+b_2\bmod p\\ c_3 &= a_3+b_3 + b_1a_2\bmod p\\ c_4 &= a_4+b_4 + a_3b_1 + a_2\binom{b_1}{2}\bmod p\\ \end{align*}$$ (these formulas can be obtained through the commutator collection process).

In this group, $[y,x,x]\neq 1$. In particular, in the absolutely free group we cannot have $[x_j,x_i,x_i]\equiv 1\pmod{\gamma_{4}(F)}$, because there is a homomorphism of $F$ onto this group where the image of $\gamma_4(F)$ is trivial, but the image of $[x_j,x_i,x_i]$ is not.

More generally, in the free Metabelian group of rank $2$ (the group $\mathscr{F}=F_2/[\gamma_2(F_2),\gamma_2(F_2)]$), the commutators of the form $[y,x,x,\ldots,x]$ are all nontrivial; in fact, the group is not nilpotent, and $\gamma_k(\mathscr{F})/\gamma_{k+1}(\mathscr{F})$ is free abelian, generated by the commutators of the form $[y,{}_ix,{}_jy]$ where $i\gt 0$, $i+j=k-1$, and $[a,{}_nb]$ is defined by letting $[a,{}_{1}b]= [a,b]$ and $[a,{}_{n+1}b] = [[a,{}_nb],b]$, so you can have arbitrarily many repetitions and still have a nontrivial commutator.