1
$\begingroup$

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

enter image description here

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

$\endgroup$
4
  • $\begingroup$ Instead of giving a dropbox link, please provide proper bibliographic details of the paper (journal, year, title, etc) $\endgroup$
    – Yemon Choi
    May 5, 2015 at 19:57
  • $\begingroup$ For other readers of this question: the paper in question is available freely with DOI 10.4153/CMB-2000-008-0 $\endgroup$
    – Yemon Choi
    May 5, 2015 at 20:19
  • 2
    $\begingroup$ I gave a brief but correct answer to this question when you asked it on math.stackexchange, but you deleted the question! $\endgroup$
    – Derek Holt
    May 6, 2015 at 8:31
  • $\begingroup$ Yes Derek. Thanks for your answer. I was not fully convinced then and was wondering the role of FC group in it. Now it makes it clearer except one doubt I still have but I ll think on it for a while before asking Dave. $\endgroup$ May 6, 2015 at 12:58

1 Answer 1

4
$\begingroup$

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 $U \subseteq G$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.