Let $G = H_1 * H_2 * \dots * H_n$ such that $H_i$ is cyclic for $1 \leq i \leq n$. So far, I proved that every abelian subgroup of $G$ is either cyclic or conjugate to a $H_i$. Next, I want to classify the solvable subgroups of $G$. The hint I received is to look at the normaliser of abelian subgroups.
However, I am totally stuck at this point. The abelian subgroups themselves are solvable, but how is their normaliser related to the solvable subgroups of $G$?
In a free product, any subgroup that is freely indecomposable and not infinitely cyclic, is conjugate in one factor.
Hence any solvable subgroup is either conjugate into one factor, with the possible exception of an infinite dihedral group when two of the free factors have 2-torsion, and of infinite cyclic subgroups.