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I read a book about free groups, it says a word is zero exponent sum, but it wasn't defined before. So what is a word which is zero exponent sum?

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It means that if you write a word in the free group $F(X)$ over $X$, say, $w(x)=x_1^{e_1}\cdots x_r^{e_r}$ that $e_1+\cdots +e_r=0$. The point is, that a word in $F(X)$ belongs to the commutator group $F(X)'$ if and only if the exponent sum of each element $x_i$ in $X$ is zero. For example, the word $x^2yx^{−1}zx^{−1}y^2z^{−1}y^{−3}$ is in $F(x,y,z)'$ (e.g., the exponents of $x$ are $2,-1,-1$), but $xy^2z^{-1}x^{-1}y^{-2}$ is not.

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    $\begingroup$ This is one of the most delicious, overwhelming facts in free groups: this characterization of their commutator group. +1 $\endgroup$
    – DonAntonio
    May 2, 2014 at 14:55
  • $\begingroup$ Hello. Where can I find a proof of this theorem you mentioned. That is, a word in $F(X)$ belongs to the commutator group $F(X)'$ if and only if the exponent sum of each element $x_i$ in $X$ is zero. Thanks. $\endgroup$
    – bfhaha
    Oct 14, 2018 at 4:48
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    $\begingroup$ @bfhaha See question 9 here, page $9$, with solution in the references, Lyndon, Schupp. $\endgroup$ Oct 14, 2018 at 8:15
  • $\begingroup$ @Dietrich Burde Could you provide some reference for "a word in $F(X)$ belongs to the commutator group if and only if the exponent sum of each element is zero"? Thanks in advance! $\endgroup$
    – 1123581321
    Feb 28, 2023 at 9:37
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    $\begingroup$ @bfhaha I had a similar question regarding the proof of this characterization of the commutator group. For an answer you can also refer to this post. $\endgroup$ Feb 1 at 19:23

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