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Let $F_n$ be a free group on $n$ generators. Let $w \in F_n$ be a word such that there does not exist any solution in $F_n$ for the equation $w.w(t_1, \ldots, t_n) = 1$, where $t_1, \ldots, t_n$ are variables and $w(t_1, \ldots, t_n)$ is the formal image of word on variables $t_1, \dots, t_n$.

For any $v_1, \ldots, v_n \in F_n$ with the property that they generate rank $n$ subgroup, would the equation $w(v_1, \ldots, v_n).(w(v_1, \ldots, v_n))(t_1, \ldots, t_n) = 1$ still have no solution in $F_n$?

I have tried many examples for words in $F_2$ and it works fine. For partial answer in $F_2$, one may look this link for similar question.

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  • $\begingroup$ Does "unsolved" mean "has no solution"? $\endgroup$
    – YCor
    Commented Sep 28, 2023 at 10:35
  • $\begingroup$ @YCor Yes and thanks. $\endgroup$
    – Shri
    Commented Sep 28, 2023 at 11:12
  • $\begingroup$ Could you please give an example of a word $w$ as in your first paragraph? $\endgroup$
    – Sam Nead
    Commented Sep 28, 2023 at 17:39
  • $\begingroup$ @SamNead Take $w =a^3[a,b]([a^{-1},b])^2$ in $F_2$ with basis $a, b$. As there is a finite group generated by 2 elements where the image of the word map $w$ is not closed with respect to inverse, hence the equation does not have any solution in free group. See this article's lemma 4.2. $\endgroup$
    – Shri
    Commented Sep 29, 2023 at 5:30
  • $\begingroup$ I don't understand the question. You can study equations over groups with or without coefficients. Since you say that the $t_i$ are all variables, your equation does not seem to have any coefficients. Such an equation always has a solution, namely the trivial one: setting $t_i=1$ for all $i$ gives the solution $w(1,\ldots,1)=1$. Please clarify: does your inclusion also include coefficients, are you hypothesising there are no non-trivial solutions, or something else? $\endgroup$
    – HJRW
    Commented Sep 29, 2023 at 15:32

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