# Are there equations which have solutions in all groups but which are not algebraicly solvable

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.

• 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 $o=1$. – Moishe Kohan May 27 at 3:15
• It perhaps wasn't clear that the two statements must both be simultaneously true. I am not sure how to clarify the 2nd statement – Fishbane May 27 at 3:30
• 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 $G$? – Moishe Kohan May 27 at 3:39
• 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,... – Fishbane May 27 at 3:42
• 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 $x^{-1}$ – Fishbane May 27 at 3:48

## 2 Answers

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.

• 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. – Fishbane May 27 at 3:35

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.