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I have a finite system of multiplicative equations with a finite number of variables, over a noncommutative group (can assume it is finitely generated if need to). for example, find elements $x,y,z \in G$ for a given group $G$ such that:

$$xyxz=1$$ $$xzxy=1$$ $$yz^3x=1$$

The thing is, my system has a lot of equations and a lot of variables, that also have a lot of dependencies. Threfore, I am looking for an automatic way for reducing the dependencies as much as possible, with the use of a mathematical programming language.

I know how to do this if the group is commutative: then any equation can be translated to the form $\prod{x_{i}^{\alpha_{ij}}}=1$, and it is enough to solve the homogenous linear system with the matrix $(\alpha_{ij})$, over the integers $\mathbb{Z}$.

Can something like this be done for a noncommutative group? I remind that the numbers of equations and variables are finite, and the group is finitely generated.

Thanks

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    No, it cannot be done automatically. Unless you know more about the class of groups/equations you are dealing with. For example for free groups (of at least 2 generators) it is impossible to algorithmically determine whether two given representations represent the same element (which is equivalent to what you are asking for). This is also know as [the word problem](https://en.wikipedia.org/wiki/Word_problem_(mathematics)) which is undecidable. – 2017-02-18
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    @freakish why is the world problem equivalent to solving general equations as above? – 2017-02-19
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    Not solving. You are trying to reduce a system of equations to a simplier version. If that simplier version is unique (or at least easy to compare) then this would give you a way to compare any two words: you would just reduce them to this simplier unique version. So by the word problem such reduction cannot exist (as an algorithm). – 2017-02-19
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    Actually it is related to solving as well. You can always have an equation $wu^{-1}=1$. If you can solve that, then you found an algorithm that determines whether $w==u$. So no, that's not going to work unfortunately. – 2017-02-19

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