# Showing that a Particular Word Problem is Decidable

I need to give an algorithm to show that the word problem in the group $\langle x,y \mid \mid x^{1984} = y^{2014} = 1 \rangle$ is decidable. How do I show this? I'm not too sure where to start.

• What sort of course is this question from? How much group theory do you know? Can you see how to show this group is virtually-free? May 20, 2014 at 12:12
• @SamJones I know a fair amount of group theory, but I don't see how to show that this group is virtually-free. This is from a course in computability theory, specifically it is a question from when we were focusing on Kolmogorov complexity.
– Newb
May 20, 2014 at 13:59
• This group is a free product of two finite groups which makes it virtually-free (see here math.stackexchange.com/questions/95695/…). You can then invoke the Muller-Schupp theorem to show that it has context-free word problem. However, this might not be the intended approach for that course. Yuval Filmus' hint below is probably more like what was intended. May 20, 2014 at 14:16
• Indeed --- I don't think I'm meant to take the group-theoretic approach. I will ponder Yuval's hint some more.
– Newb
May 20, 2014 at 14:42

• Imagine the presentation was $<x, y \; || >$ i.e. the free group of rank 2. How do you know when two words over the alphabet $\Sigma = \{x, y, x^{-1}, y^{-1} \}$ represent the same element of the free group? Then can you see how to apply this to the presentation that you have? May 21, 2014 at 12:28