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?

$\begingroup$ Drike, can you give an example of at least one infinite group like this? $\endgroup$– Ashot MinasyanOct 18 '11 at 10:17

$\begingroup$ @Ashot: Take the direct sum $G=\oplus_{i \in I} G_i$, where $I$ is infinite and each $G_i$ is a finite centerless group. $\endgroup$– GuntramOct 18 '11 at 15:35

$\begingroup$ Any infinite FC group with trivial centre would do. Would you like a concrete example ? (I just know they exist) $\endgroup$– DrikeOct 18 '11 at 16:03

1$\begingroup$ I was rather thinking of a finitely generated infinite example. Such an example would not be an FCgroup. $\endgroup$– Ashot MinasyanOct 19 '11 at 11:11
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 FCgroup, 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.

$\begingroup$ @Gumtram: you probably mean that G is the cartesian product of $G_i$, not direct product? In this case G is uncountable; it would be interesting to find a countable example. $\endgroup$ Oct 19 '11 at 9:46

$\begingroup$ @Ashot: The cartesian product is a FCgroup, so I have to take the direct product. $\endgroup$– GuntramOct 19 '11 at 9:58

$\begingroup$ @Gumtram: I guess we do not agree on the definition of direct and cartesian products. For me the former only involves elements with finite support and the latter can have elements of arbitrary support. $\endgroup$ Oct 19 '11 at 10:12
A modification of Guntram's example could produce a countable group with the required property, which is not an FCgroup. Let $G$ be the direct product of nonabelian 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 FCgroup. 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.

$\begingroup$ @Ashot : thanks for spotting the link with residually finite groups. But there is something I do not understand. "My" property is preserved by inverse image of embeddings, so you seem to be claiming that it is equivalent to being residually finite. $\endgroup$– DrikeOct 21 '11 at 6:35

$\begingroup$ @Drike  no it's not. For instance, $\mathbb{Z}$ doesn't have your property. $\endgroup$– HJRWOct 21 '11 at 7:33

$\begingroup$ @Drike, I am not sure what you mean by saying that the "property is preserved by inverse image of embeddings". The construction above shows that a group with your property has a normal subgroup N (consisting of elements with finite conj. classes), but you have almost no control of what happens outside of N. $\endgroup$ Oct 21 '11 at 9:43

$\begingroup$ @Ashot  my mistake, I thought "my" property was preserved when going to a subgroup, but it is not. $\endgroup$– DrikeOct 21 '11 at 10:25