please help me to solve the following problem.

Let $F$ be a non-abelian free group and $w(x)=1$ be an equation in one variable $x$ ($w(x)$ may contain elements of $F$ as constants). Clearly, one can consider $w(x)$ as an element of free product $F\ast \langle x\rangle$.

Suppose $w(a)=1$ for all $a\in F$.

Is it true that $w(x)$ equals $1$ in $F\ast \langle x\rangle$?

Do you know a simple proof? Probably, you can remember the papers which can be useful for this propblem?

  • $\begingroup$ These are called "mixed identities". See Anašin, V. S. Mixed identities in groups. Mat. Zametki 24 (1978), no. 1, 19–30, 141, references therein, and papers referring to it. The fact that the answer is "yes" should be in the literature. $\endgroup$ – user6976 Apr 29 '18 at 18:07

Yes. Denote by $a,b$ the first two free generators of $F$.

The case $w\in F\cup\langle x\rangle$ is clear. So we can suppose, after conjugation, that $w=u_1x^{n_1}\dots u_kx^{n_k}$ with $k\ge 1$, $u_i\in F\smallsetminus\{1\}$, and $n_i\in\mathbf{Z}\smallsetminus\{0\}$.

Choose $n$ large enough such that the reduced form of $a^nu_ia^{-n}$ starts with $a$ and finishes with $a^{-1}$ for all $u_i$ that is not a power of $a$. In all cases, the reduced form $v_i$ of $a^nu_ia^{-n}$ starts and finishes with $a^{\pm 1}$. Then $q=\prod v_ib^{n_i}$ is a reduced form and hence $\neq 1$. We have $$a^{-n}qa^n=a^{-n}\left(\prod a^nu_ia^{-n}b^{n_i}\right)a^n=\prod u_ia^{-n}b^{n_i}a^n=w(a^{-n}ba^n)\neq 1.$$

Added: This actually shows that the canonical map $F\ast\langle x\rangle\to F^F$, $w\mapsto (t\mapsto w(t))$ is injective. Indeed this is a group homomorphism ($F^F$ being a group for the target operation: infinite power of the group $F$ by the index set $F$). Of course this fails when $F$ has rank $r\le 1$: for $r=0$, $x$ is the kernel, while for $r=1$ $axa^{-1}x^{-1}$ is in the kernel. The above shows injectivity of the projection to $F^X$, where $X=\{a^nba^{-n}:n\in\mathbf{Z}\}.$

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  • $\begingroup$ Dear @YCor, thank you for the clear answer. The proof is really easy! $\endgroup$ – Artem May 2 '18 at 5:14

If $F=\langle a_1,\ldots,a_n\rangle=\langle\underline{a}\rangle$ is finitely generated, we can replace $w$ with a coefficient-free word $w'(x,y_1,\ldots,y_n)=w'(x,\underline y)$ such that $w(x)=w(x,\underline{a})$. Now consider the first-order sentence $\phi$: $$\exists y_1,\ldots,y_n\forall x\ (w'(x,\underline{y})=1)$$ Then the result follows from Sela's proof of Tarski's problem:

Theorem [Sela, Theorem 3]: All finitely generated non-abelian free groups have equivalent first-order theories.

In particular, $F$ satisfies $\phi$ (where $\underline{y}=\underline{a}$ works to satisfiy the existential part of $\phi$) if and only if $F*\langle a_{n+1}\rangle$ satisfies $\phi$.

I'm not sure about the case where $F$ is not finitely generated, but because at most finitely many of the generators of $F$ could appear as coefficients in $w(x)$, you might be able to reduce to this case, depending on what you need this result for.

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  • 2
    $\begingroup$ What's the point in using a very hard and deep theorem to prove an elementary and easy fact? $\endgroup$ – YCor May 1 '18 at 14:09
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    $\begingroup$ That's a good point, but I suppose that Sela's papers and those which he references could possibly provide some useful results and techniques for whatever question @Artem had in mind. I work on things related to this, so I mostly wanted to point towards some possible references. $\endgroup$ – Christopher Perez May 1 '18 at 16:54

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