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In an answer to a 2014 MSE post here, user DonAntonio states the following "cute" fact, which I've restated here in simpler terms for ease-of-reading:

Lemma: Let $F_n$ be the free group of rank $n$ with basis $\{ w_1 \dots w_n\}$. Then $\prod w_{a_i}^{b_i} \in [F,F] \iff$ for each $w_i$, the sum of the powers to which $w_i$ is raised to in this word equals zero.

So for example, $w_1^{-2}w_2w_3w_1^2w_3^{-1}w_2^{-1} \in [F_3,F_3]$ but $w_1^4w_2^{-3}w_1^3w_2^3 \not \in [F_3,F_3]$.

I'm trying to prove this, but I've only started getting into group theory this semester and am a bit uneasy with all...the words.

(Originally, I had a proof attempt here. It was nowhere near complete nor insightful. I like shorter posts, so I removed it.)

user1729
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Descartes Before the Horse
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    I rolled back your edit as after it the question made no sense (just the statement of a lemma and some background, but no suggestion that you were after a proof). Also, the paragraph you removed was nice. It's okay to be uneasy with proofs, and it is the why we ask questions :-) – user1729 Jul 09 '21 at 20:04

2 Answers2

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Denote the right-hand side property by (P). Then it is easy to see that

(1) if two elements of $F_n$ satisfy (P), then so does their product,
(2) each element of the form $[x,y]$ (i.e. each generator of $[F_n,F_n]$) satisfies (P).

This establishes "$\implies$".

For the other direction, your attempt at induction over the word length is a good idea:
The start of the induction is trivial.
For the induction step, use the (likewise trivial) formula $$w_k^b\,u\,w_k^c\,x=[w_k^{-b},u^{-1}]\,u\,w_k^{b+c}\,x$$ and the fact that if $w_k^b\,u\,w_k^c\,x$ satisfies (P), then so does $u\,w_k^{b+c}\,x$, and the length of the latter is one less than that of the former.

jpvee
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  • Sorry, shouldn't the implication be 'if $uw_k^{b+c}x$ satisfies P, then so does $w_k^buw_k^cx$'? Since the latter is a product with the commutator of the former. – Descartes Before the Horse Mar 11 '21 at 22:36
  • Sorry, with A commutator. Not THE. – Descartes Before the Horse Mar 11 '21 at 22:44
  • @DescartesBeforetheHorse: No, the implication is meant as stated. We are trying to prove that if an element $g\in F_n$ satisfies (P), then it lies in the commutator subgroup of $F_n$. By definition of $F_n$ and (P), every such $g$ can be written in the form $w_k^b,u,w_k^c,x$. Now apply the above transformation and write $g$ as a product of a commutator with $u,w_k^{b+c},x$. And then, since the latter factor also satisfies (P), it must be an element of the commutator subgroup by induction hypothesis. – jpvee Mar 12 '21 at 05:20
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For an arbitrary group $G$, we have $g \in [G, G]$ if and only if $\bar{g} = 1$ in the abelianization $G/[G, G]$. For the free group, $F_n/[F_n, F_n]$ is just $\mathbb{Z}^n$ (take your pick of arguments). The stated property follows immediately.

(Edit: Hah! I just noticed my link equals yours. Well, it's another approach, anyway.)

Joshua P. Swanson
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