TITLE: Power of commutator formula
QUESTION [12 upvotes]: A few people remember a commutator formula of the form $[a,b]^n = (a^{-1} b^{-1})^n (ab)^n c$ where $c$ is a product of only a few commutators (say $n-1$) of them.  Here $a,b$ are in a (free) group and $[a,b] := a^{-1} b^{-1} a b$.
Does anyone remember such a formula with proof?
Some such formula must exist where $c$ is in the commutator subgroup of $\langle a,b\rangle$, but my recollection is that $c$ is a product of something more like $n^2$ commutators.
Answers that only work for $n=2$ are less interesting to me. There should be a radical difference for $n \geq 3$.

REPLY [18 votes]: Positive results are fairly nice. I explain them in this note for commutators of powers (see the last pages for pretty pictures and crazy formulas). I didn't get around to the special case of powers of commutators, but Culler and Bavard give definitive results.
Negative results: These are just well known true formulas that have “$c$” being way too long, even allowing ourselves to omit longer commutators. If you've never done it, try to write $(ab)^3 = a^3 b^3 c$ and actually get a formula for $c$ only involving commutators (of commutators) of $a$ and $b$.

Robinson, page 137, 5.3.5, $(xy)^m = x^m y^m [y,x]^{\binom{m}{2}} \mod \gamma_3$ has $c$ of quadratic length
Hall, Chapters 11 and 12. I don't actually see the formula (!) but the algorithms used to derive the formula and their $\mod \gamma_4$ brethren are there, as well as the application to regular $p$-groups, which says that you can at least choose $c$ to be a product of $n$th powers, but no bound on how many. This is mod a more difficult to describe subgroup, as well.
Gorenstein, Chapter 5.6 has the technique (again modulo a hard to describe subgroup, due to the $p$ versus $c$ of $\gamma_c$ relation). It also has applications of the formula in 5.3.9, but not the formula proper.
Leedham-Green–McKay, Corollary 1.1.7
$$[x,y]^n = (x^{-1} y^{-1})^n (xy)^n [y,x]^{\binom{n}{2}} [[y,x],x]^{\binom{n}{3}} \cdots$$
$$[y,x]^n = [y^n,x] [[x,y],y]^{\binom{n}{2}} [[[x,y],y],y]^{\binom{n}{3}} \cdots$$
modulo the subgroup generated by commutators containing at least 2 $x$s. Again quadratic (cubic, quartic, etc. if don't mod out by $\gamma_3$) not linear.
Culler's formula is $$[a,b]^3 = [b^a,a^{-1} b^a b^{-1}][bab^{-1},b^2]$$ express a product of three commutators as a product of two commutators, which is disturbing. In fact Culler showed that the commutator length of $[a,b]^n$ is less than or equal to $\tfrac{n}{2} + 1$, and Bavard showed one has equality (with the greatest integer less than or equal to $\tfrac{n}{2}+1$). Danny Calegari and Alden Walker have improved the algorithms used in these papers while using the same basic topological idea. I would also mention the diagrams used in these works are on the inside covers of Rotman's group theory textbook.

Culler, Marc.
“Using surfaces to solve equations in free groups.”
Topology 20 (1981), no. 2, 133–145.
MR605653
DOI:10.1016/0040-9383(81)90033-1
Bavard, Christophe.
“Longueur stable des commutateurs.”
Enseign. Math. (2) 37 (1991), no. 1-2, 109–150.
MR1115747
DOI:10.5169/seals-58734


Two versus three
$n=2$ is special and has a finite formula with everything sorted:
$$(ab)^2 = abab = aab[b,a]b = a^2 b^2 [b,a] [[b,a],b]$$
$n=3$ is more like the rest, and has no finite formula if you try sort the commutators by weight. Even the unsorted formula is pretty long:
$$\begin{array}{ll}
(ab)^3 &= ababab \\
&= aab[b,a]bab \\
&= aab[b,a]ab[b,a]b \\
&= aaba[b,a][[b,a],a]b[b,a]b \\
&= aaab[b,a]^2[[b,a],a]b[b,a]b \\
&= a^3b[b,a]^2 b [[b,a],a] [[[b,a],a],b] [b,a] b \\
&= a^3b^2 [b,a][b,a,b][b,a][b,a,b][[b,a],a] [[[b,a],a],b] [b,a] b \\
&= a^3 b^3 [b,a][b,a,b][b,a,b][b,a,b,b][b,a][b,a,b][b,a,b] \\
&\quad [b,a,b,b][b,a,a][b,a,a,b][b,a,a,b][b,a,a,b,b][b,a][b,a,b]
\end{array}$$
Of course if we go mod $\gamma_4$ then we lose all that $[b,a,b,b]$ nonsense and are left with:
$$(ab)^3 = a^3 b^3 [b,a]^3 [b,a,b]^5 [b,a,a]^1 \mod \gamma_4$$
The powers on those commutators are called Hall polynomials if you let $n$ vary.
$$(ab)^n = a^n b^n [b,a]^{\binom{n}{2}} [b,a,a]^{\binom{n}{3}} [b,a,b]^{2\binom{n}{3}+\binom{n}{2}} \mod \gamma_4$$
has cubic growth. The Hall polynomial for $[b,a,a,\ldots,a]$ is always $\binom{n}{k}$. There are also three variable and higher versions.
Commutator length
Actually, the commutator length may be much shorter than the formulas indicate.
$$(ab)^3 = a^3 b^3 [ {(ab)}^{-1} b^2, b^{-1} (ab)^2 ]$$
$$(ab)^3 = (ba)^3 [aba,bab]$$
The first can be applied with $a=x^{-1} y^{-1}$ and $b=xy$ to answer the main question.
Guided by scallop by Alden Walker and Danny Calegari, I found $$(ab)^4 = a^4 b^4 
 [a^{-1}b^2, a^{-2}ba][ ba^{-2}ba,(ba)^2b ]$$
I think these shorter formulas lose some of the theoretical importance that the “nicer” formulas had, but I worry this sort thing will give a positive answer to the question. I continue to be interested in a positive answer.
Positive result
To prove Schur's theorem that if $[G:Z(G)]$ is finite then so is $G'$, Ornstein showed this in a way very similar to Cullen's formula and the commutator length ideas. The first step was the remembered claim $[a,b]^n = (ba)^{-n} (ab)^n u$ where $u$ is a product of $n-1$ commutators. This follows from induction on $n$ with $n=1$ being clear. $$\begin{array}{ll}
[a,b]^n
&= [a,b] [a,b]^{n-1} \\
&= [a,b] (ba)^{1-n} (ab)^{n-1} u_{n-2} \\
&= (ba)^{-1}(ab) (ba)^{1-n} (ab)^{n-1} u_{n-2} \\
&= (ba)^{-1}(ba)^{1-n} (ab)^{n-1} (ab) [ (ab), (ba)^{1-n} (ab)^{n-1} ] u_{n-2} \\
&= (ba)^{-n} (ab)^n u_{n-1}
\end{array}$$
Thanks to Babak for finding this simple proof. This is used with $n=[G:Z(G)]$ since then $(ba)^{-n} (ab)^n = 1$ because $(ab)^n \in Z(G)$ and $(ab)^n = ((ba)^b)^n = ((ba)^n)^b = (ba)^n$. This gives that $[a,b]^n$ is a product of $n-1$ commutators, rather than $n$. In any product of commutators of minimal length, no commutator appears to a power higher than $n$. Since $xyx = x^2 y^x$ and the commutator length of a conjugate is the same as the original, we can sort any such expression to bring all copies of a commutator into a power. Hence no commutator appears anywhere in a minimal expression $n$ or more times. Since there are only at most $n^2$ commutators, that is a total of less than $n^3$ expressions, so $|G'| \leq [G:Z(G)]^3$. I suspect the other formulas we have give that $|G'| \leq 3[G:Z(G)]^2$.