# Solvability of group presentations with 2 “almost disjoint” relations

I am interested in a certain type of $$2$$-relator group presentations arising in algebraic topology which have two relators that only contain a single generator in common. Specifically, suppose I have group presentation of the form

$$G \cong \langle\{S_1\} \cup \{S_2\} \cup \{b\} \mid P_1 = b^{n_1}, P_2 = b^{n_2}\rangle$$

Where $$S_1 = \{x_1, \dots , x_n\}$$, $$S_2 = \{y_1, \dots, y_m\}$$ are sets of generators (one or both possibly empty). In the first relation, $$P_1$$ is a word which is either a product of commutators of the form $$P_1 = [x_1,x_2][x_3,x_4]\dots [x_{n-1},x_n]$$, OR is of the form $$P_1 = x_1^2 x_2^2 \dots x_n^2$$. Similarly, $$P_2$$ takes on one of those two forms but in the $$y_i$$ generators.

Finally, the $$n_{1,2}$$ may be any non-zero integers (in particular, they may be positive or negative).

We therefore have that $$P_1$$ is a word only in the generators of $$S_1$$ and $$P_2$$ is a word only in the generators of $$S_2$$.

My Question:

I am aware that in general, two relator group presentation word/conjugacy/isomorphism problems are not tractable. However, given the restricted form that the group relations have (they are formed from disjoint sets of generators, except for $$b$$), are there any theorems or methods which provide additional power to work with these groups?

Observations:

If $$d = gcd(n_1, n_2) = 1$$, then the $$b$$ generator can be eliminated.

If both of $$S_{1,2}$$ are empty, then the group is cyclic (and trivial if $$d=1$$).

If one of $$S_{1,2}$$ is empty, there are certain simplifications that can be made, but everything seems to be case by case depending on d and the form of $$P_1$$, $$P_2$$.

Are there any results which would allow me to exactly classify when this group presentation can be reduced to a one relatiion group and when two relations is the minimum number possible?

Playing around with different special cases, it intuitively seems obvious that this should be the case (but intuition may be incorrect), but I am constantly getting stuck trying to prove that there could not possibly exist any other isomorphic presentations which have a single relation.

Any strategies, tactics, useful lemmas/theorems would be greatly appreciated!

• It looks like a free product of two groups with the subgroup $\langle b \rangle$ amalgamated. I don't know if that helps! – Derek Holt Nov 9 '16 at 16:48
• In fact, that is exactly what it is! This is the fundamental group of a class of spaces. However, I am new to the field. Are there any standard tools/results from the theory of amalgamations which would address the types of problems I mentioned? – wanderingmathematician Nov 9 '16 at 19:57
• Small cancellation theory might help. – Shaun Dec 3 '18 at 0:33
• I'm not sure, though. I wonder what @DerekHolt thinks :) – Shaun Dec 3 '18 at 0:37
• @Shaun It's been a couple years since I posted this--I forgot it had never been answered. This question was a peripheral one to the primary question I was interested in at the time, and I used small cancellation theory to address my primary question in the end, bypassing this one. However, I would still be interested to know the result here out of curiosity. Small cancellation theory together with Sela's solution to the isomorphism problem for hyperbolic groups would definitely answer this question in some instances, but wouldn't say anything if there were too many $b$'s in the relations. – wanderingmathematician Dec 3 '18 at 1:42