# Power of commutator formula

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 $n \geq 3$.

• I believe the formula people were thinking of is: $(xy)^m=x^my^m[y, x]^{m/2}$. This holds in a $2$-step nilpotent group, so is in your "uninteresting" category. See p141 of Robinson, A Course in the Theory of Groups. However, I am sure that there is a similar formula used in Leedham-Green and MacKay's book on $p$-groups which they use in the chapter on $p$-groups of maximal class. Thus, nilpotency class more than $2$... – user1729 Jun 14 '13 at 19:18
• $m/2$ is actually the binomial coefficient $m$ choose 2, though – Jack Schmidt Jun 14 '13 at 19:21
• @MathsLover: sounds good, put it the proof in the answer box. :-) – Jack Schmidt Jun 14 '13 at 19:21
• Or check Gorenstien "Finite Groups" Harper an Row, chapter 5 – Geoff Robinson Jun 14 '13 at 19:24
• I did finding “Cullen's” formula for $n=3$ new and deeply disturbing. Thanks @exitingcorpse. :-) – Jack Schmidt Jun 14 '13 at 19:29

## 1 Answer

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.

## 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 $|G'| \leq 3[G:Z(G)]^2$.

• +10. Very nice Jack. I have searched for you to find the result I mentioned yesterday. I hope you are to upset cause of me but I want to be sure of it, not just posting an answer. See please Rotman's book. Page 114, Theorem 5.32( Schur). As I thought last night, he assumed there $G/Z$ is finite. Sorry again. I am one one of your fan. :-) – mrs Jun 15 '13 at 5:01
• @BabakS. thanks, this has been very educational. I haven't had a chance to read about strengthening of Schur's inequality, but I suspect I'll now understand them! – Jack Schmidt Jun 15 '13 at 14:26
• I am not sure how related this is to your actual aims, but I miss the paper in your long list: Bavard improved on Culler's results in this paper. He shows that $[a,b]^n$ can be written as a product of $\lfloor \frac{n}{2}\rfloor + 1$ commutators and shows that this is actually best possible. He gives an explicit formula in section 1.4. – Martin Jun 15 '13 at 14:51
• @Martin thanks! That is very similar to the original question, which asks about $(ab)^{-n} (ba)^n [a,b]^n$. I think $[a,b]^n$ by itself is probably a more natural thing to want to know about. :-) Culler, Bravard, and Calegari all appear to be using the same basic technique that apparently I ought to learn. – Jack Schmidt Jun 15 '13 at 15:11
• I should have mentioned this in my previous comment: much of Bavard's paper (only one r, by the way :-)) can be found in chapter 2 of Calegari's recent scl monograph. – Martin Jun 15 '13 at 15:21