Why do we not lose any generality by proving it only for finitely generated groups [closed]

Question

In the proof of following theorem, in a paper by Farkas-

Here $\Delta(G) = \{ g \in G : |G:C_G(g)| < \infty \}$ and $U_1(\mathbb{Z}G) $ is the set of normalized units of the integral group ring $\mathbb{Z}G$ i.e. units with augmentation $1$

In the proof, author first reduces to the case that $G=\Delta(G)$ (which is fine) but after that he says that as we are assuming $G= \Delta {G}$, there is no loss in generality in taking $G$ to be finitely generated.

Now for $G=\Delta(G)$, why we do not lose any generality by proving it only for finitely generated groups.

Is there some result like, there is no infinitely generated group which has all its conjugacy classes of finite cardinality? I do not think I am aware of any such result. Also $\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \dots$ being abelian has only singelton conjugacy classes.

Title of Paper- "Trivial Units in Group Rings"

Authors- Daniel R. Farkas and Peter A. Linnell

Appeared in- Canad. Math. Bull. Vol. 43 (1), 2000 pp. 60–62

Answer 1

The reduction to the finitely generated case does not have anything to do with $\Delta(G)$. This is a completely separate (and very elementary) reduction.

Fix any $u \in U$. Let $H$ be the subgroup generated by the support of $u$, so $u \in \mathfrak{U}_1(\mathbb{Z}H)$. Since the support of $u$ is finite, we know that $H$ is finitely generated. Also, since $G$ has finite index in $U$, we know that $H = G \cap \mathfrak{U}_1(\mathbb{Z}H)$ has finite index in $U \cap \mathfrak{U}_1(\mathbb{Z}H)$. Therefore, if the theorem is known to be true for finitely generated groups, then $u \in H \subseteq G$. Since $u$ is an arbitrary element of $U$, this implies $U \subseteq G$.