Let $G = \langle S \mid R \rangle$ be a finite presentation, $G$ is $\delta$-hyperbolic.

Prove that if $H,K \leq G$ where $H,K \cong C_2 \times C_2$, we can decide if $H$ and $K$ are conjugate.

I am having trouble seeing how to do this. We are trying to decide if there is a $g \in G$ such that $gHg^{-1} = K$.

Obviously, due to the finite presentation, we can list possible $g$ and check them and we will eventually find the conjugating element if it exists. But when can we stop?

Since $G$ is hyperbolic, it has solvable conjugacy problem. If $H$, $K$ were cyclic, we could just use this to check if the generators are conjugate. The problem is that we have 2 generators. We can check conjugacy individually, but not all at the same time I don't think. The proof I know for the solvability of the conjugacy problem puts a bound on the length of the words we need to check, so we can stop after a finite time. Is there such a bound for checking conjugacy of these subgroups?

Or maybe there is an alternate approach.


By the main result of this paper, the conjugacy problem for finite lists of elements is solvable in hyperbolic groups. This answers your question, but it might be overkill!

$\mathit{Added\ later}$: An earlier reference for the conjugacy of finite lists of elements in hyperbolic groups is Theorem A of this paper by Bridson and Howie. The main result of their paper is that the problem is solvable in quadratic time provided that some of the elements in the lists are non-torsion, and in the paper by Buckley and myself, we extended that result to torsion elements, which you need in your situation. But the solvability of the problem (in exponential time) was proved by Bridson and Howie, by bounding the length of the elements that you need to consider as possible conjugating elements.

  • $\begingroup$ There's no "might" about it, definitely overkill haha. But surely there must be an easier way, $C_2 \times C_2$ is the smallest non cyclic group, there must be something there we can use to make this easy to show... $\endgroup$ – pizzaroll May 26 at 20:26
  • $\begingroup$ @pizzaroll But you can't say that it's definitely overkill without suggesting an alternative. I think conjugacy problems involving torsion elements of hyperbolic groups are slightly more complicated than for non torsion elements. $\endgroup$ – Derek Holt May 26 at 20:48
  • $\begingroup$ I guess this is true. Hopefully you will forgive me for saying I don't like your answer very much, but I will accept it after a while if I (and others) can't come up with anything better, it certainly is a correct answer to the question. $\endgroup$ – pizzaroll May 26 at 20:52

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