# Finite conjugacy classes

Let $$G$$ be an infinite group. Let $$N_0$$ be the set of all $$x\in G$$ for which the conjugacy class $$\{y^{-1}xy: y\in G\}$$ is a finite set. Clearly $$N_0$$ is a normal subgroup. Iteratively, form an ascending transfinite sequence by

• for $$n$$ a non-limit ordinal, let $$N_n\subseteq G$$ be the set of all $$x\in G$$ for which $$xN_{n-1}\in G/N_{n-1}$$ has finite conjugate class.
• for $$\omega$$ a limit ordinal, $$N_{\omega} = \bigcup_{\alpha<\omega} N_{\alpha}$$.

There are two possibilities: either $$(N_\alpha)$$ stops at $$G$$, or at a proper subgroup $$G^{FC}$$ of $$G$$. For convenience, let's call (only for this post) $$G^{FC}$$ the [FC]-kernel of $$G$$, and $$G$$ a hyper-[FC] group if $$G=G^{FC}$$. A simple observation is that $$G/G^{FC}$$ is an [ICC] group. Let's call $$G^{FC}$$ a minimal [FC]-kernel if whenever $$N_0\subseteq H\subseteq G$$ is another normal subgroup for which $$G/H$$ is [ICC], we have $$H\supseteq G^{FC}$$.

I apologize if the questions below are elementary for MO.

Q0: Could you please point me to a reference where I can find the proper terminology for the made-up names hyper-[FC] group & [FC]-kernel?

Q1: Is $$G^{FC}$$ a hyper-[FC] group in itself?

Q2: Is $$G^{FC}$$ always minimal in the sense above?

edit: Following Sean Eberhard's suggestion, I've just came across the book "Finiteness Conditions and Generalized Soluble Groups" by Robinson. The terminology he used in his book is FC-Hypercenter for $$G^{FC}$$, and FC-Hypercentral if $$G=G^{FC}$$. It is a big deal(!) that the made-up names by yours truly were a close call after all.. The book explains various central series in detail for those who are curious about the similarities in the construction & the properties of X-hypercenters, and more..

• Why is $N_0$ clearly a subgroup? Aug 8, 2023 at 13:54
• @LSpice Let $O(x)=\{gxg^{-1}:g\in G\}$. For $x,y\in N_0$ and all $g\in G$, $gxy^{-1}g^{-1} = gxg^{-1}(gyg^{-1})^{-1}\in O(x)O(y)^{-1}$. So $O(xy^{-1})\subseteq O(x)O(y)^{-1} \Rightarrow |O(xy^{-1})|\leq |O(x)||O(y)|$. Aug 8, 2023 at 14:15
• $N_0$ is called the FC-center of $G$. groupprops.subwiki.org/wiki/FC-center Aug 8, 2023 at 14:32
• @SeanEberhard thanks a bunch. I wonder if $(N_\alpha)$ is called [FC]-central sequence, analogous to central sequences. My made up name hyper-[FC] was inspired by the same anology. I'm asking for the right terminology so that I can use them as keywords for literature search. Aug 8, 2023 at 15:04
• @LSpice the FC-center $\mathrm{FC}(G)$ is a "clearly" subgroup because it consists of elements whose centralizer has finite index.
– YCor
Aug 17, 2023 at 14:46

The answers to Q1 and Q2 are both positive and use a simple observation. Suppose $$N_1 \le N_2$$ are normal subgroups of $$G$$ and consider the natural map $$G/N_1 \to G/N_2$$. Then the FC-center of $$G/N_1$$ maps into the FC-center of $$G/N_2$$. Indeed, if an element has a finite conjugacy class modulo $$N_1$$ then it certainly does so modulo the larger group $$N_2$$.
Q1: Let me write $$F(G)$$ for the FC-center and $$F^\alpha(G)$$ for the iterated variants. Let $$N = \bigcup_\alpha F^\alpha(G)$$ be the limit of the whole process. Then $$N$$ is a characterstic subgroup of $$G$$, as is $$F^\alpha(N)$$ for all $$\alpha$$. I claim inductively that $$F^\alpha(G) \le F^\alpha(N)$$ for all $$\alpha$$. The induction is obvious for zero and limit ordinals, and the case of successor ordinals follows from the observation above applied to $$N/F^\alpha(G) \to N / F^\alpha(N)$$.
Q2: Suppose $$H$$ is a normal subgroup of $$G$$ such that $$G/H$$ is ICC. I claim that $$F^\alpha(G) \le H$$ for all $$\alpha$$. Again it suffices to a consider a successor ordinal $$\alpha+1$$, and now we apply the observation above to $$G / F^\alpha(G) \to G/H$$, noting that $$F(G/H)=1$$.
In light of Q2, an alternative name for the "hyper FC center" might be the "ICC residual". It is the smallest normal subgroup $$H$$ such that $$G/H$$ is ICC.