I have a question about some quotient groups of the lower central series of a free group.


When there's a free group $F = \langle x\_1,\cdots, x\_n, y\_1, \cdots, y\_m\rangle $,


let $A$ be the subgroup generated by elements of the forms $[x\_i, a]$ in $F\_{k+1}/F\_{k+2}$, with $1\leq i\leq n$ and $a\in F\_{k}/F\_{k+1}$; 
 and $B$ be the subgroup generated by those $[y\_j, b]$ of $F\_{k+1}/F\_{k+2}$, with $1\leq j\leq m$ and $b\in F\_k/F\_{k+1}$. 


(May consider when $k\geq 2$.)


In this case, $A \cap B = 1$?


add.
What is the rank of $A$ (and $B$) in the free abelian group $F\_{k+1}/F\_{k+2}$?


Thank you for your reading, considering and answering.



quotient groups of the lower central series of a free group
I don't believe so. 


I will associate my commutators on the left, so that $[a,b,c]=[[a,b],c]$; and use $[x,y]=x^{-1}y^{-1}xy$; I'm not sure it matters, but that's what I'm used to.


Say $k=2$; in $F/F\_{4}$, the Hall-Witt identity becomes
$$ [r,s,t][s,t,r][t,r,s]\equiv 1\pmod{F\_{4}}.$$


Now consider the case where $a\_1=[y\_2,y\_1]$, $b\_1=[y\_2,x\_1]$, $b\_2=[y\_1,x\_1]$. Then
$$ [a\_1,x\_1] \equiv [y\_2,y\_1,x\_1] \equiv [y\_2,x\_1,y\_1][y\_1,x\_1,y\_2]^{-1} \equiv [b\_1,y\_1][b\_2,y\_2]^{-1}\pmod{F\_4}.$$
Now note that $[a\_1,x\_1]\in A$, but $[b\_1,y\_1][b\_2,y\_2]\in B$. 


Similar problems are likely to arise with other values of $k$.


**Edited:** It seems your question was different from what I understood; rather, your $A$ is generated by all commutators of the form $[x\_i,a]$ with $a\in F\_k$ (modulo $F\_{k+2}$), and your $B$ by all commutators of the form $[y\_j,b]$ with $b\in F\_k$ (modulo $F\_{k+2}$).


$F\_{k}/F\_{k+1}$ is free abelian, with basis given by the basic commutators of weight $k$ on $x\_1,\ldots,x\_n,y\_1,\ldots,y\_m$; by the Witt Formula, the rank is
$$M\_{n+m}(k)=\frac{1}{k}\sum\_{d|k}\mu(d)(n+m)^{k/d}$$
where $\mu$ is the Möbius function. 
The rank of $A$ is at most $nM\_{n+m}k$, and the rank of $B$ is at most $mM\_{n+m}(k)$. But in general they will be much smaller, since commutators of the form $[c,x\_j]$ and $[c,y\_j]$, where $c$ is a basic commutator of weight $k$, are seldom basic themselves; so you will have a fair amount of "collision". It would take some rather careful analysis with basic commutator collection to figure it out exactly. I would suggest looking at


* Ward, James. *Basic commutators.* Philosoph. Trans. Roy. Soc. London Ser. A **264** (1969), 343-412, MR 0251148


to see if by any chance Ward has already computed these (or for information on techniques for computing it). 




---


**Added.**


Let's look at the case of $k=2$. $F\_2/F\_3$ is freely generated by commutators of the forms $[x\_j,x\_i]$, $1\leq i\lt j\leq n$, $[y\_s,y\_r]$, $1\leq r\lt s\leq m$, and $[y\_t,x\_k]$, $1\leq k\leq n$ and $1\leq t\leq m$.


On the other hand, $F\_3/F\_4$ is freely generated by commutators of the forms 


1. $[x\_j,x\_i,x\_k]$ with $1\leq i\lt j\leq n$, $i\leq k\leq n$;
2. $[x\_j,x\_i,y\_t]$ with $1\leq i\lt j \leq n$, $1\leq t\leq m$;
3. $[y\_t,x\_k,x\_i]$ with $1\leq k\leq i\leq n$, $1\leq t\leq m$;
4. $[y\_t,x\_k,y\_v]$ with $1\leq k\leq n$, $1\leq t\leq m$, $1\leq v\leq m$; and
5. $[y\_s,y\_r,y\_t]$ with $1\leq r\lt s\leq m$, $r\leq t\leq m$.


For $A$, we will obtain all commutators of type 1 and all commutators of type 3; no commutators of type 5; when we consider a commutator of the form $[y\_s,y\_r]$ and take the commutator with $x\_i$, we obtain $[y\_s,y\_r,x\_i] \equiv [y\_s,x\_i,y\_r][y\_r,x\_i,y\_s]^{-1}$. So commutators of type 4 are paired off, except we do not get the ones in which $t=v$. Finally, we obtain all commutators of type 2 because when we take $[y\_t,x\_j]$ for $a$, and take the commutator with $x\_i$, we will obtain $[y\_t,x\_j,x\_i]\equiv [y\_t,x\_i,x\_j][x\_j,x\_i,y\_t]^{-1}$; and since $[y\_t,x\_i,x\_j]\in A$, we obtain the commutator $[x\_j,x\_i,y\_t]$. 


So, what is the rank of $A$? We have $\binom{n}{3}+2\binom{n}{2}$ commutators of type 1; we have $m\binom{n}{2}$ commutators of type 2 and $m\binom{n}{2}+mn$ of type 3; and we have 
$n\binom{m}{2}$ generators corresponding to commutators of type 4. This gives
$$\mathrm{rank}(A) = \binom{n+1}{3} + (2m+1)\binom{n}{2}+n\binom{m+1}{2}.$$


As for $B$, the argument is similar: we get no commutators of type 1, all commutators of type 5 (of which there are $\binom{m}{3}+2\binom{m}{2}$); all commutators of type 2 (of which there are $m\binom{n}{2}$); we also get all of type 4, which gives $nm^2$ commutators; as for type 3, these cannot occur from expressions of the form $[b,y\_j]$ with $b\in F\_2/F\_3$. So the rank of $B$ will be
$$\mathrm{rank}(B) = \binom{m+1}{3}+\binom{m}{2} + m\binom{n}{2} + nm^2.$$


Finally, for the overlap: according to the Witt formula, the rank of $F\_3/F\_4$ is
$$\frac{1}{3}(n+m)\left( (n+m)^2 - 1\right).$$
So this allows you to compute the rank of the overlap.


For larger values of $k$ the computations become more difficult, because not all basic commutators are of the form $[c,x\_i]$ or $[c,y\_j]$. But this should give you an idea of the kind of thing that is going on (and why I don't want to try doing this for $k=3$).


