Groups with trivial centralizer-connected component

Question

Let $G$ be an infinite group such that for every $g\neq 1$ in $G$, there is some $b$ such that $C(b)$ has finite index in $G$ and $g$ is not in $C(b)$. Is something known about those groups? Are they FC groups, or related to FC groups?

Answer 1

Let $I$ be an infinite index set and for $i \in I$ let $G_i$ be a finite group with trivial centre. Let $G:=\prod_{i \in I} G_i$ be the direct product of these groups. Then $G$ satisfies the property you mentioned $$ \forall g \in G\backslash \lbrace 1 \rbrace \exists b \in G: [G:C(b)]<\infty, g \not\in C(b) $$ but is not an FC-group, i.e., not every conjugacy class is finite.

Indeed, let $g=(g_i)_{i \in I}\neq 1$ and $i_0 \in I$ such that $g_{i_0}\neq 1$. Let $b=(b_i)_{i \in I}$ be such that $b_i=1$ for $i \neq i_0$ and $b_{i_0}$ be such that it does not commute with $g_{i_0}$ (which exists since $G_{i_0}$ is assumed to have trvial centre). Then $g\not\in C(b)$, but $C(b)$ contains $\prod_{i \neq i_0} G_i \times \lbrace 1 \rbrace$ which is of index $|G_{i_0}|$.

On the other hand, every element $g=(g_i)_{i \in I} \in G$ with infinite support, i.e. $g_i\neq 1$ for infinitely many $i$, clearly has infinitely many conjugates.

Answer 2

A modification of Guntram's example could produce a countable group with the required property, which is not an FC-group. Let $G$ be the direct product of non-abelian symmetric groups $G=\times_{n\ge 3} S_n$, where $S_n$ is the symmetric group on $n$ elements. For each $n$, $S_n$ has an inner automorphism $\alpha_n$, of degree $n$ (conjugation by an $n$-cycle). Now consider the semidirect product $H=G \rtimes \langle a \rangle$, of $G$ with an infinite cyclic group $\langle a \rangle$, such that $a$ preserves every direct factor $S_n$ of $G$, acting on it by $\alpha_n$. I think that the group $H$ satisfies Drike's property, but is not an FC-group. The centralizer of $a$ will have infinite index and the conjugacy class of $a$ will be infinite.

Update: in fact, in the above example the cyclic group $\langle a \rangle$ can be replaced with an arbitrary residually finite group $F$. Any residually finite group $F$ is "approximated" by homomorphisms into finite symmetric groups (e.g.,via the natural actions on cosets of finite index subgroups), giving rise to a sequence approximating homomorphisms of $F$ into $Inn(S_n)\cong S_n$ (where $n$ varies). One can define the action of $F$ on the direct product of $\times S_n$ as before and check that the corresponding semidirect product $H=(\times S_n)\rtimes F$ satisfies Drike's property. The base group will consist exactly of the elements with finite conjugacy classes.

Thus any residually finite group can be embedded into a group with the described property. It is easy to see that the property implies that the group is residually finite, so there is no room for improvement.