Equation over finite group with presentation $G=\langle a,b|R\rangle$

Question

This question arised like a curiosity, I've been trying to find out information about the solution (after trying to solve it by myself) with no success. The question is: Given a finite group with presentation $G=\langle a,b|R\rangle$ (Where R are relations on the letters $a$ and $b$), does the equation $a^kb=e$ always has solutions for $k\in\mathbb{Z}$? If no, which conditions must be imposed?

I know the answer is not trivial, so I'm primarly requesting for references: articles, authors, especific theories, etc.

Edition:

It is clear by the comments that the answer is certainly no, and that this leads to a classical word problem. Let me slightly modify the question. For a fixed integer $k$ and a finite group $G=\langle a,b|R\rangle$ (Where R are relations on the letters $a$ and $b$), which conditions must be imposed on k, a, b or $G$ in order to the equation $a^nb^k=e$ to have solutions for $n\in\mathbb{Z}$?

Answer 1

As mentioned in the comments, there is not necessarily a solution to $a^kb=e$. This equation will have a solution if and only if $a^k=b^{-1}$ for some integer $k$.

Supposing that $a^k=b^{-1}$, one can rewrite the relations $R$ so that $b$ does not occur anywhere in $R$. Specifically, one can replace every occurrence of $b$ in $R$ with $a^{-k}$. This is to say that if $G$ is a finite group given by $\langle a,b~|~R\rangle$ where $R$ is some set of relations on $a$ and $b$, and if $a^kb=e$ for some integer $k$, then $G=\langle a~|~R'\rangle$ where $R'$ is the re-written set of relations.

More generally, if $G=\langle x_1,x_2,\ldots,x_m|R\rangle$ and if $x_i^kx_j=e$ for some integer $k$, then there must be some redundancy in the group presentation; one can write $G=\langle x_1,x_2,\ldots,x_{j-i},x_{j+1},\ldots,x_m~|~R'\rangle$ where $R'$ is obtained from $R$ by replacing every occurrence of $x_j$ with $x_i^{-k}$.

Answer 2

The condition you impose is equivalent to the condition that $G$ is a cyclic group with $a$ as a generator: for if $a^kb=e$ for some $k$, then this means that $b=a^{-k}$, so that $G$ is generated by $a$, and conversely.

Nevertheless, given a presentation of $G$, it is not necessarily easy to decide whether this condition is satisfied. I'm not sure about this exact problem, but many similar problems have been proven to be computationally undecidable: for example, even deciding whether a group given by a finite presentation is the trivial group or not is impossible; for references, look up the "word problem for groups".