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Say I have a free group on the generators $X = \{ x_1, x_2, ... , x_k \}$ with $k \geq 2$. I read (in an article) that if I have a word $w$ written in the generators and their inverses, and the exponents of all $x_i$ for $i \in \{ 1, ..., k \}$ in this word sum up to zero, then the word is an element of the commutator subgroup of $F(X)$, usually referred to as $F'$ or $[F,F]$.

I presume it is a rather trivial task to show that commutators and their products have this property, since it applies to commutators by definition and cancellation of generators and their inverses will also uphold this property for any cancellations in products of commutators.

However, I struggle to find an elegant way to prove the other direction, i. e.: All words $w$ in $F(X)$ with the property as mentioned above are themselves elements of the commutator subgroup $F'$.

I assume I can show this by looking at a word with the above mentioned property, say $w=x u x^{-1} v$, where $x$ is a generator of the free group in question (or its inverse respectively) and $u, v$ are words written in its generators or their inverses. I get

$$ x u x^{-1} v = xux^{-1}u^{-1}uv = [x, u] uv. $$

Apparently $uv$ again has the above mentioned property of exponent sum zero with regards to the used generators, and the word for which we have to show to be an element of the commutator subgroup $F'$ has been shortened by $2$ compared to our original word $w$. So my assumption is that repetition of this process should result in a product of commutators.

My question being:

  • Is there a more elegant proof, which is not algorithmic?
  • Furthermore: Any hint where to find more information on such a proof and this property of elements of the commutator subgroup would be appreciated.
Andreas Faltin
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1 Answers1

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Your proof is completely correct. There is, however, also a more conceptual (to my eye) way to view this characterization of the commutator subgroup:

Let $F_n$ be the free group on $n$ generators $x_1,...,x_n$. Then the quotient $F_n/F_n'$ is the free abelian group $\mathbb{Z}^n$ on $n$ generators with the quotient map $\pi: F_n\to \mathbb{Z}^n$ sending each $x_i$ to $e_i$ (the standard basis vector). It is then clear that for any word $w$, the $i$-th entry of the vector $\pi(w)$ is just the sum of the exponents of $x_i$ in $w$. Hence, a word $w$ has all exponent-sums equal to zero if and only if it lies in the kernel of $\pi$, but that kernel is just the commutator subgroup by definition.

Tim Seifert
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