TITLE: Groups with prescribed $\text{C}(x)$, $\text{C}(y)$, $\text{C}(xy)$.
QUESTION [5 upvotes]: My question is inspired by this question, but I want to ask something much more specific.
Let $X,Y,Z$ be finite groups with $1\ne x\in \text{Z}(X)$, $1\ne y\in \text{Z}(Y)$, and $1\ne z\in \text{Z}(Z)$.
Under what conditions does there exist a group (not necessarily finite)  $G\geqslant X,Y,Z$ such that $xyz=1$ and $\text{C}_G(x)=X$, $\text{C}_G(y)=Y$, $\text{C}_G(z)=Z$?
It seems to me that there may be tight restrictions if $(o(x),o(y),o(z))$ is one of $(2,2,k)$, $(2,3,3)$, $(2,3,4)$, $(2,3,5)$ but  beyond that I have no intuition.

REPLY [3 votes]: There is lots of research on this kind of question, but unfortunately the best results are reserved for when the identity "$xyz=1$" is instead of the form $R^n(X,Y,Z)=1$ for $n>1$, e.g. $(xyz)^2=1$ or $[x,y,z]^{47}=1$, or when the groups $X,Y,Z$ are "locally indicable", i.e. all finitely generated subgroups surject onto the infinite cyclic group. Neither situation is the case here.
Basically, if such a group $G$ exists then the subgroup $\langle X,Y,Z\rangle \leq G$ is a quotient of the group $X∗Y∗Z/\langle\langle xyz\rangle\rangle$, which is the free product of $X,Y,Z$ quotiented by the relator $xyz=1$. Therefore, just consider this quotient group, and you are asking if $X,Y,Z$ each embed into this group. The literature refers to $X∗Y∗Z/\langle\langle xyz\rangle\rangle$ as a "one-relator product", and embedding results like you are after is a "Freiheitssatz". See, for example, this paper of Fine, Howie & Rosenberger, and a more recent paper is this preprint of Howie and Short.