TITLE: Cyclic amalgamation of solvable groups
QUESTION [6 upvotes]: Given a nonempty set of groups $\{G_\lambda|\lambda \in \Lambda\}$ together with a group $H$ which
is isomorphic with a subgroup $H_\lambda$ of $G_\lambda$ by means of monomorphism $\phi_\lambda: H \to G_\lambda$. There is a frequently used term known as the free product of the $G_\lambda$'s with the amalgamated subgroup $H$. Roughly speaking, this is the largest
group generated by the $G_\lambda$'s in which the subgroups $H_\lambda$ are identified with $H$ by $\phi_\lambda$. Such groups are known as generalized free products.
My question concerns a result proved by Kahorbaei and Majewicz claiming the following.



Question. I'm uncertain about this proof. Why must the cosets $a\delta_{m+1}A$ and $b\delta_{n+1}B$ have the same order in its respective factor group? I'm worried about the infinite cyclic subgroup.

Here is the definition for residually solvable group.

Edit: I suspect there are counterexamples to the statement (or maybe I misunderstood their definition of the generalized free product). It seems that it requires extra hypothesis such as cosets of same order or something called compatible filtration in Proposition 1 of Baumslag (as suggested by Carl-Fredrik Nyberg Brodda).

REPLY [3 votes]: The elements do not have to have the same order it would seem.  According to the paper ON TORSION-FREE METABELIAN GROUPS WITH COMMUTATOR QUOTIENTS OF PRIME EXPONENT  by Narain Gupta and Said Sidki there  are torsion-free metabelian groups whose commutator quotients are nontrivial with prime exponent for any odd prime $p$. So choosing examples $A$ and $B$ with different primes, say $3$ and $5$, gives infinite order elements $a\in A$ and $b\in B$ (not in the commutator subgroup) that have orders $3$ and $5$, respectively, modulo the commutator subgroup.