FC groups and finite quotients Let $G$ and $H$ be finitely generated FC-groups such that $G\times \mathbb{Z}\cong H\times\mathbb{Z}$. Can we say that $G\cong H$? If $G$ is finitely generated FC-group and $$\mathcal{F}(G)=\mathcal{F}(H)$$ can we say that $H$ is FC-group too?
 A: The answer to your first question is no.  Take the group
$$ \langle a,b,c\ |\ [a,b]=[a,c]=b^{11}=1,b^c=b^4\rangle.$$
Consider the subgroups $G=\langle b,c\rangle$, $H=\langle b,ac^2\rangle$, $K_1=\langle a\rangle$, and $K_2=\langle a^2c^5\rangle$.  One can check that the group is both the direct product $G\times K_1$, and $H\times K_2$.  Both $K_1$ and $K_2$ are infinite cyclic.
Now $G\cong\langle u,v\ |\ u^{11}=1, u^v=u^4\rangle$, and $H\cong\langle x,y\ |\ x^{11}=1, x^y=x^5\rangle$.  These groups are not isomorphic, but both have finite-index centers, and so both are FC groups.
A: About the second part of the question. If $G$ is trivial and $H$ is simple and infinite, they have the same finite quotients, so the answer is "no" in general. If $H$ is residually finite, then the answer should be "yes" because $G$ is a central extension of a finite group, but I do not know the reference now. The answer to the first part should also be "yes". See Some cancellation theorems with applications to nilpotent groups by Hirshon (especially the last section) where a connection between the two parts of the question is established. 
A: Since $\mathbb Z$ is abelian and conjugacy is componentwise in direct products clearly G FC and $G\times\mathbb Z\cong H\times \mathbb Z$ implies H is FC. Since you didn't define $\mathcal F$ I can't answer the rest of the question. 
