# Are countable FC-groups maximally almost periodic?

An FC-group is a group in which every element has a finite conjugacy class. A group G is said to be maximally almost periodic if there is an injective homomorphism from G into a compact Hausdorff group.

My question is whether every discrete countable FC-group G is maximally almost periodic.

In general an FC-group G is isomorphic to a subgroup of the direct product of a locally finite FC-group and a torsionfree abelian group. Since discrete abelian groups are maximally almost periodic, this reduces the general question to the case where G is additionally assumed to be locally finite.

• Obviously $G/Z(G)$ is residually finite, hence MAP, so it works when $Z(G)$ is trivial. In general it sounds tricky.
– YCor
Aug 7, 2014 at 9:52
• Why is $G/Z(G)$ residually finite when $G$ is FC?
– Andy
Oct 8, 2018 at 9:42
• Ah ok, just figured this out. If $g\in G$ is not in the center, there is some $h\in G$ with which it does not commute. Then $C_h$ is a finite index normal subgroup avoiding $g$.
– Andy
Oct 8, 2018 at 9:48

Here's a counterexample. Consider a non-abelian finite, 2-step nilpotent group $F$. Let $Z$ be its derived subgroup. Define $G$ as the quotient of the direct sum $F^{(\mathbf{N})}=\bigoplus F_n$ of copies of $F$ indexed by $\mathbf{N}$ by identification of all copies of $Z$. Thus, writing $H=F/Z$, the group $G$ lies in a central extension, with central kernel $Z$ and quotient the infinite direct sum $H^{(\mathbf{N})}$. I claim that $G$ is not maximally almost periodic, and more precisely that any homomorphism of $G$ into a compact Hausdorff group kills $Z$.
0) $G$ is an FC-group: this is because it's quotient of a direct sum of finite groups.
1) I claim that for every finite index subgroup $M$ of $G$ we have $Z\subset [M,M]$. Indeed, let $M_n$ be the image of $M$ in the $n$-th copy $H_n$ of $H$ in $H^{(\mathbf{N})}$. Then the image of $M$ in $H$ is contained in $\bigoplus M_n$. Since it has finite index, this implies that for all but finitely many $n$, we have $M_n=H_n$. Let $M'$ be the inverse image of $M$ in $F^{(\mathbf{N})}$ and $M'_n$ its projection on $F_n$. Then for all but finitely many $n$, we have $M'_nZ_n=F_n$. Since $F_n$ is nilpotent and $Z_n=[F_n,F_n]$, this implies that for those $n$, we have $M'_n=F_n$, and hence for those $n$, we have $Z_n\subset [M'_n,M'_n]$. By fixing such an $n$ and projecting into $G$, we deduce that $[M,M]$ contains $Z$.
2) To conclude, it's enough to show that every homomorphism $f$ from $G$ to a compact Lie group $L$, with dense image, kills $Z$. Fix $f$. Since $G$ is nilpotent, $L^0$ being a compact nilpotent connected Lie group, it is abelian. Define $M=f^{-1}(L^0)$. Then $M$ has finite index, hence $Z\subset [M,M]$. Since $f$ maps $M$ into the abelian group $L^0$, we deduce that $[M,M]\subset ker(f)$, hence $Z\subset ker(f)$.
Besides, as I mentioned, the result is true for FC-groups $G$ with trivial center: indeed these groups act faithfully on themselves by conjugation, and since the orbits are finite, the closure of the image of $G\to Sym(G)$ is compact. This is more generally true for residually finite FC-groups. However since as you mention, it is also true for abelian groups (which may fail to be residually finite), it's unclear what it the greatest generality.