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I am not sure exactly how to phrase this problem so I appologise if it is not clear, also this is somewhat long but I wanted to explain exactly where I was with the problem. If you have any questions feel free to ask.


Description of Problem

Given a set of variables $\{x,y,z,...\}$ and a variable $o$ is it possible to define a finite product of these variables and their inverses $\sigma(x,y,z,...,x^{-1},y^{-1},z^{-1},...,o,o^{-1})$ (i.e. a finite sequence made up of these variables and their inverses) such that;

1) For any group $G$ and any assignment of values from $G$ to $\{x,y,z,...\}$ there exists a unique element $g$ of $G$ such that if $o$ is set to $g$;

$\sigma(x,y,z,...,x^{-1},y^{-1},z^{-1},...,o,o^{-1})=1$

and

2) There does not exist a finite product $\gamma(x,y,z,...,x^{-1},y^{-1},z^{-1},...)$ such that;

$o=\gamma(x,y,z,...,x^{-1},y^{-1},z^{-1},...)$ for all groups $G$


Rough explaination as to why I am asking here

My intuition is no but I am unsure how to prove this. There are clearly examples of equations like these solvable in all groups (i.e. $xo=1$) but these have algebraic solutions (in that example $o=x^{-1}$) and there are examples of these equations which are solvable in wide classes of groups (i.e. $o^{n!+1}x=1$ is solvable in any group of order less than $n$ with $o=x^{-1}$) but these are not solvable in all groups. In addition some equations are solvable in all groups but not uniquely (i.e. $o^2=1$ has many solutions in groups with elements of order 2 but can always be solved with $o=1$)


Progress on proof (or proof of falsehood)

It can be shown that $\sigma$ must contain exactly $\pm1$ total occurences of $o$ (where $o^{-1}$ counts as $-1$ occurence of o) using the following argument.

If $G$ is abelian then $\sigma$ can be written as $Ao^n$ for some $A$ which is a product of the other variables. For this to be solvable $o^n=A^{-1}$ must be solvable in every abelian group. If $A=1$ $o$ is not uniquely defined for groups of order $|n|$. If $A\neq1$ and $|n|\neq1$ then $o$ is not defined for groups of order $|n|$ or $n=0$ and so $o$ is not unique. Therefore $|n|=1$ and so the total number of occurences of $o$ in $\sigma$ must be $\pm1$.

In addition it is clear that there must be an odd number of occurences of $o$ greater than $1$ (this time counting $o^{-1}$ as $1$ occurence). This follows as otherwise there is a clear definition of $\gamma$ (if there is $1$ occurence) or the observation above is violated (if there are an even number of occurences).

This is where I am and I am not sure how to proceed. Appologies again for this being overly long. Any information or advice would be appreciated.

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  • $\begingroup$ Your 2nd question is unclear. For the first question, consider the equation $\sigma(x,x^{-1},o,o^{-1})= xx^{-1} ooo^{-1}=1$. It has unique solution $o$, namely $o=1$. $\endgroup$ – Moishe Kohan May 27 at 3:15
  • $\begingroup$ It perhaps wasn't clear that the two statements must both be simultaneously true. I am not sure how to clarify the 2nd statement $\endgroup$ – Fishbane May 27 at 3:30
  • $\begingroup$ I see. The second part is unclear since you did not quantify $x, y$, etc.: Do you mean to say that there is no word $\gamma$ in formal variables ..., such that for each group $G$ the equation $o=\gamma(...)$ has a solution in $G$? $\endgroup$ – Moishe Kohan May 27 at 3:39
  • $\begingroup$ I suppose it would best be described as given $\sigma(x,...,o,o^{-1})$ there is no function given as a word $\gamma(x,...)$ s.t $\sigma(x,...\gamma(x,...),(\gamma(x,...))^{-1})=1$ in any group under any assignment of values of x,y,... $\endgroup$ – Fishbane May 27 at 3:42
  • $\begingroup$ Maybe another way to explain it is to give the intuition behind the idea. If when we define a word we don't allow inverses then the word$\sigma(x,o)=xo$ would be a valid solution as there is always a unique solution by definition of a group but there is no way of writing this solution without using $x^{-1}$ and so in terms of words only using $x$ we cannot write $\gamma(x)$ s.t $x\gamma(x)=1$ in all groups. Maybe this has made things more clear, maybe not. To clarify this is not a solution of the problem as in the problem we are allowed $x^{-1}$ $\endgroup$ – Fishbane May 27 at 3:48
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If you require uniqueness of your solution, then I don’t believe this is possible.

To shorten notation, for a set $X$ I’ll write $\sigma(X)$ to denote a word in elements of $X$. Let $X$ be a set, let $o$ be a variable, and let $\sigma(X, o)$ be any word. Then the group

$$G = \langle X, s, t~|~\sigma(X, s) = \sigma(X, t) = 1\rangle$$

fails the uniqueness condition for solutions to $\sigma(X, o)$.


So, let’s consider the case where we don’t assume uniqueness. In this case, it is (uninterestingly) possible.

The requirement that $\sigma(X, o) = 1$ for all groups means that, in particular, this must be true for the free group $F(X\cup\{o\})$. This means that $\sigma(X,o)$ must reduce to a trivial word by the definition of free groups.

The only next requirement is that $o\notin \langle X\rangle$. Thus, we can produce a situation you want in the following way: let $G$ be any group with identity element $e$, and $\sigma(X,o)$ be any word that reduces to the trivial word.

Then, the assignment $x\mapsto e$ for all $x\in X$ and $o\mapsto g$ for any $g\ne e$ in $G$ satisfies your requirements.

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  • $\begingroup$ Firstly thank you for the answer. I will first tackle the second part of your answer, I did not make it clear that the word should be solvable for all assignments of variables to X. i.e. we cannot simply say that all of them are e unless the group is only e. The first part of your answer is more of interest to me as I just want to prove it impossible. I'm not sure how one would explicitly construct the group you describe. As in for some particular $\sigma$ (for example $xox$) how would we construct the desired group. It would be nice if you could respond, so9rry again for how unclear I was. $\endgroup$ – Fishbane May 27 at 3:35
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Okay so I have no answered my question. If this doesn't seem like an answer it is probably due to me being unable to clearly describe the question I wanted to ask so sorry. Also I'm not sure how best to explain the answer.


It can be shown that no such equation exists by contradiction. First assume there exists $\sigma$ as is required by the problem. Let $n$ be the number of variables it takes (excluding $o$ and only counting $x$ not $x^{-1}$ for example). Take the group $F_n$ (the free group with $n$ generators) and assign to each of the variables that $\sigma$ takes $x_1,x_2,...$ a different one of the generators $g_1,g_2,...$. Find the value of $o$ corresponding to this situation. By definition of $F_n$, $o$ can be written as a finite product of $g_1,g_2,...$. Write $\gamma$ as this representaion of $o$ but where the generators are replaced by the corresponding variables. Therefore by definition of the free group in all groups $o=\gamma$ is a solution.


Sorry if this wasn't clear. Also thanks to Santana Afton for mentioning using the free group in there answer.

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