Given some group $G$ that is generated by $a$ and $b$, each of which has infinite order, and some free subgroup $N$ generated by $a^k$ and $b^k$, is there any algorithm that tells me if some $x \in G$ is also in $N$? If not generally, what about for $k=2,3,4$?



The generalized word problem on groups
The answer is "no". Take the free group of rank 2 $F=<a,b>$. Take an infinite recursively enumerable non-recursive set $X=\{u\_1,u\_2,...\}$, a subset of a recursive set $X'$ of words in $a,b$ satisfying the small cancelation condition $C'(1/12)$ and containing neither $a^2$ nor $b^2$. Take another infinite set of words $Y$ in $a^2,b^2$ satisfying $C'(1/12)$, $Y=\{v\_1,v\_2,...\}$. Consider the factor group $G$ of $F$ by relations $u\_1=v\_1,u\_2=v\_2,...$. Then the subgroup $H$ of $G$ generated by the images of squares $a^2,b^2$ is free and the membership problem for $H$ is undecidable.


The group $G$ is infinitely presented, the question does not ask for a finitely presented example, which is more difficult and I do not know the answer in that case. If rank can be assumed bigger than 2, the question might be manageable.


In the above construction one can replace $a^2, b^2$ by $a^m,b^n$ for any $m,n>1$.


