# Proving a commutator identity for groups

From Wikipedia:

If the derived subgroup is central, then $(xy)^n = x^ny^n{\left[y,x\right]}^{n \choose 2}$.

I was able to prove this for the case $n = 2$: \begin{align*} (ab)^{-2}a^2b^2[b,a]&= b^{-1}a^{-1}b^{-1}a^{-1}a^2b^2b^{-1}a^{-1}ba\\ &= b^{-1}a^{-1}b^{-1}aba^{-1}ba\\ &= b^{-1}[a,b]a^{-1}ba\\ &= [a,b]b^{-1}a^{-1}ba\\ &= [a,b][b,a] = 1. \end{align*} For $n = 3$ and further my naive approach isn't working, though. I have no idea where the "$n$ choose $2$" comes from, it seems weird. How to prove this for the general case?

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Hint: first prove $y^nx = xy^n[y,x]^n$ by induction on $n$ and then prove the main result by induction on $n$. – Derek Holt Oct 30 '11 at 11:32
When $n<m$ one generally makes the convention that $\binom{n}{m}=0$ (indeed, you can choose $4$ elements out of a $2$ element set in exatcly zero ways!). So the statement is also true in the cases $n\leq2$. – Mariano Suárez-Alvarez Oct 30 '11 at 12:51
@Mariano: Thanks, I didn't think about that. – Mikko Korhonen Oct 30 '11 at 17:15
@Mariano: Actually, if we interpret $\binom{n}{2}$ as meaning $\frac{n(n-1)}{2}$ for all integers $n$, then the formula is true for any integer $n$, positive or negative. – Arturo Magidin Oct 30 '11 at 19:06
@Arturo: indeed! I should have been more precise – Mariano Suárez-Alvarez Oct 30 '11 at 19:36

Your proof is fine. I'll give a version of the proof that is a bit briefer and a conceptual explanation for the result that indicate how the hypothesis is used.

To me this is just a simple induction. $$\begin{array}{rl} (ab)^{n+1} &= ab(ab)^{n} \\ \\ &= ab \left( a^n b^n [b,a]^{\binom{n}{2}} \right) \\ \\ &= a a^n b [b,a]^n b^n [b,a]^{\binom{n}{2}} \\ \\ &= a^{n+1} b^{n+1} [b,a]^{\binom{n+1}{2}} \end{array}$$ This first line just separates the power. The second line is the induction hypothesis. The third line commutes n copies of a past a single copy of b, resulting in n copies of $[b,a]$. The fourth line groups the $a$s together, the $b$s together, and the commutators together, using that $$\binom{n}{2} + n = \binom{n}{2} + \binom{n}{1} = \binom{n+1}{2}.$$

A conceptual way to see this is that $(ab)^n$ requires moving the first $a$ past 0 $b$s, the next $a$ past 1 $b$, the next $a$ past 2 $b$s, etc. until the last $a$ needs to move past $n-1$ of the $b$s, so that we get a total of $$0+1+2+\dots+(n-1) = \binom{n}{2}$$ commutators. The hypothesis that $[b,a]$ commutes with $a$ and $b$ is only used to avoid higher order commutators. No matter what, we get those $\binom{n}{2}$ copies of $[b,a]$.

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Thanks! Now I can see why it must be true, this is much better than my own proof. So it all just boils down to $ba = ab[b,a]$, and then moving the $a$'s past $b$ this way. Which was exactly the hint Derek commented.. – Mikko Korhonen Oct 30 '11 at 19:17
@m.k.: Yes; this is the first (and simplest) case of what is called the "collection process". You "collect" all the $a$s on the left; then all the $b$s; then all the $[b,a]$s, and so on. – Arturo Magidin Oct 30 '11 at 21:56

By the way, the formula holds in more generality. Using the commutator identities: $$[x,yz] = [x,z]z^{-1}[x,y]z,\qquad [xy,z] = y^{-1}[x,z]y[y,z],$$ which can be verified by direct computation.

Theorem. If $a$ and $b$ commute with $[b,a]$, then:

1. For all integers $n$, $[a^n,b] = [a,b]^n = [a,b^n]$.
2. For all integers $n$, $\displaystyle (ab)^n = a^nb^n[b,a]^{n(n-1)/2}$.

Proof.

1. If $n=0$ or $n=1$, the identity is trivial. Assuming it holds for $k$, we have \begin{align*} [a^{k+1},b] &= [aa^k,b]\\ &=a^{-k}[a,b]a^k[a^k,b]\\ &= [a,b][a^k,b]\\ &= [a,b][a,b]^k = [a,b]^{k+1}. \end{align*} So the identity $[a^n,b]=[a,b]^n$ holds for all nonnegative integers. If $k\gt 0$, then \begin{align*} 1 &= [a^{k}a^{-k},b] = a^{k}[a^{k},b]a^{-k}[a^{-k},b]\\ &= a^k [a,b]^k a^{-k}[a^{-k},b]\\ &= a^ka^{-k}[a,b]^k[a^{-k},b]\\ &= [a,b]^k[a^{-k},b]. \end{align*} So $[a^{-k},b] = ([a,b]^k)^{-1} = [a,b]^{-k}$, hence the identity holds for all integers.

Finally, $$[a,b^n] = [b^n,a]^{-1} = ([b,a]^n)^{-1} = ([b,a]^{-1})^{n} = [a,b]^n.$$ This proves 1.

To prove 2, the result holds for $n=0$ and $n=1$. If the result holds for $k$, then \begin{align*} (ab)^{k+1} &= (ab)(ab)^k\\ &= aba^kb^k[b,a]^{k(k-1)/2}\\ &=aa^kb[b,a^k]b^k[b,a]^{k(k-1)/2}\\ &=a^{k+1}bb^k[b,a^k][b,a]^{k(k-1)/2}\\ &= a^{k+1}b^{k+1}[b,a]^k[b,a]^{1+2+\cdots+(k-1)}\\ &= a^{k+1}b^{k+1}[b,a]^{1+2+\cdots+k}\\ &= a^{k+1}b^{k+1}[b,a]^{(k+1)k/2}. \end{align*} And if $k\gt 0$, then \begin{align*} (ab)^{-k} &= ((ab)^k)^{-1}\\ &= \left(a^kb^k[b,a]^{k(k-1)/2}\right)^{-1}\\ &= [b,a]^{-k(k-1)/2}b^{-k}a^{-k}\\ &= b^{-k}a^{-k}[b,a]^{-k(k-1)/2}\\ &= a^{-k}b^{-k}[b^{-k},a^{-k}][b,a]^{-k(k-1)/2}\\ &= a^{-k}b^{-k}[b,a]^{(-k)(-k)}[b,a]^{-k(k-1)/2}\\ &= a^{-k}b^{-k}[b,a]^{-k(-k + (k/2)- (1/2)}\\ &= a^{-k}b^{-k}[b,a]^{-k(-k-1)/2}, \end{align*} proving the formula for all integers. $\Box$

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Nice. And I guess if you want to be really general, I noticed when $G'$ is Abelian (or just when a, b, c and the commutators each commute with each other): $(abc)^n = a^nb^nc^n[b,a]^{n \choose 2}[c,a]^{n \choose 2}[c,b]^{n \choose 2}$ and so on for products of 4, 5, ..., k elements. – Mikko Korhonen Oct 31 '11 at 14:39
@m.k.: Careful! It's not enough for $G'$ to be abelian: it really has to be central for that identity to hold. If $G'$ is abelian but not central, you need other terms. For example, $$(ab)^2 = abab = aab[b,a]b = a^2bb[b,a][[b,a],b] = a^2b^2[b,a][b,a,b].$$ – Arturo Magidin Oct 31 '11 at 15:57
Ah, of course.. – Mikko Korhonen Oct 31 '11 at 17:12

I finally found a proof so I'm answering my own question, but I didn't use Derek's hint. It might be around in disguise, though. And the identity is still a bit mysterious to me. My solution might be a bit silly way to do it, but it seems to work.

Assume $a$ and $b$ are elements of some group such that they both commute with $[a,b]$ (and thus they commute with $[b,a]$ too).

First I prove the following identity by induction: $aba^{n-1}b^{-1}a^{-n} = [b,a]^{n-1}\$ for all $n \geq 1$.

It's pretty clear it holds for $n = 1$. Suppose the statement holds for $n = k$. Then

\begin{align*} aba^{(k+1)-1}b^{-1}a^{-(k+1)}\ &= aba^kb^{-1}a^{-1}a^{-k} \\ &= aba^k[b,a]a^{-1}b^{-1}a^{-k} \\ &= [b,a]aba^{k-1}b^{-1}a^{-k} \\ &= [b,a][b,a]^{k-1} \\ &= [b,a]^{(k+1)-1}\\ \end{align*}

Using this you can prove $(ab)^n = a^nb^n[b,a]^{n \choose 2}$ by induction. Case $n=2$ was proved in my question. Suppose the identity holds for $n = k-1$. Then

\begin{align*} (ab)^{k}\ &= ab(ab)^{k-1} \\ &= aba^{k-1}b^{k-1}[b,a]^{k-1 \choose 2} \\ &= (aba^{k-1}b^{-1}a^{-k})a^{k}b^{k}[b,a]^{k-1 \choose 2} \\ &= [b,a]^{k-1}a^{k}b^{k}[b,a]^{k-1 \choose 2} \\ &= a^{k}b^{k}[b,a]^{{k-1 \choose 2}+(k-1)} \\ &= a^{k}b^{k}[b,a]^{k \choose 2} \end{align*}

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By the way: the formula holds in more generality. If $[b,a]$ commutes with $a$ and $b$, then for all integers $n$ (positive or negative) we have $$(ab)^n = a^nb^n[b,a]^{n(n-1)/2}.$$ – Arturo Magidin Oct 30 '11 at 19:09