FC groups with infinite derived subgroup which are not constructed by direct product of finite groups. A group $G$ is called FC (finite conjugacny) if every conjugacy class $C$ of $G$ has a finite order. It is called FD if the derived subgroup (constructed by commutators) is finite. It is clear that every FD group is also FC, but not vice versa. For example, the restricted (external) direct product of a non-abelian finite group with itself infinitely many times is an FC group but not FD.
I am looking for another example of FC groups which is not FD and is not in the form of restricted direct product of finite group. Can you please help me with that?  
 A: One way to get such an example is to look at a subgroup of a direct product of finite groups.
Let $P = \prod_{n = -\infty}^{\infty}G_{n}$ where, for each integer $n$,
we let $G_{n}$ be the dihedral group
$$G_{n} = \langle a_{n}, b_{n}, c_{n}\mid
a_{n}^2, b_{n}^2, c_{n}^2, [a_{n},c_{n}], [b_{n},c_{n}], c_{n} = [a_{n},b_{n}]\rangle ,$$
of order $8$.
For an integer $n$, let $g_{2n-1} = a_{2n-1}a_{2n}$ and $g_{2n} = b_{2n}b_{2n+1}$.
Now let $G = \langle g_{k}\mid k\in\mathbb{Z}\rangle$ be the subgroup of $P$
generated by all the $g_{k}$.
Since $P$ is a direct product of finite groups, it is an FC-group, and hence,
so is its subgroup $G$.
But $G$ is not an FD-group, because its derived subgroup is the infinite group generated by the $c_{n}$
(in fact, their direct product).
To see this, note that
\begin{align}
[g_{2n}, g_{2n+1}]
& = (a_{2n-1}a_{2n})^{-1}(b_{2n}b_{2n+1})^{-1}a_{2n-1}a_{2n}b_{2n}b_{2n+1} \\
& = a_{2n}a_{2n-1}b_{2n+1}b_{2n}a_{2n-1}a_{2n}b_{2n}b_{2n+1} \\
& = a_{2n}b_{2n}a_{2n}b_{2n} \\
& = c_{2n}.
\end{align}
Similarly, the commutator $[g_{2n},g_{2n+1}] = c_{2n+1}$.
The other pairs of $g_{k}$ commute, so in fact,
$[G,G] = Z(G) = \langle c_{n}\mid n\in\mathbb{Z}\rangle$.
Next, use the commutation relations to note that any non-central element $g$ of $G$
can be written in the "normal form"
$$g = g_{i_{1}}g_{i_{2}}\cdots g_{i_{k}}z ,$$
where $z\in [G,G] = Z(G)$ and the indices are sorted:
$i_{1} < i_{2} < \cdots < i_{k}$.
We'll use this to show that $G$ cannot be a direct product of finite groups;
in fact, $G$ is directly indecomposable.
To this end suppose, for an eventual contradiction, that $G = A\times $B,
with both $A$ and $B$ non-trivial.
Let $a$ and $b$ be non-central elements of $A$ and $B$, respectively, and write
$$a = g_{i_{1}}g_{i_{2}}\cdots g_{i_{k}}z_{a}
\;\;\text{and}\;\;
  b = g_{j_{1}}g_{j_{2}}\cdots g_{j_{m}}z_{b},$$
in normal form.
If $g_{i_{1}} = g_{j_{1}} = g_{s}$, say, then $[g_{s-1},a] = c_{s} = [g_{s-1},b]$.
But $[g_{s-1},a]\in A$ and $[g_{s-1},b]\in B$ and $A\cap B = 1$, which contradicts $c_{s}\neq 1$.
Therefore, $g_{i_{1}} \neq g_{j_{1}}$, and a similar argument shows that $g_{i_{k}} \neq g_{j_{m}}$.
Consequently, when $ab$ is written in normal form
$$ab = g_{k_{1}}g_{k_{2}}\cdots g_{k_{r}}z,$$
as above, we must have $r > 1$.
Perforce, none of our original generators $g_{n}$ can be written as $ab$ for
non-central elements $a\in A$ and $b\in B$.
It follows that, for each integer $n$, the generator $g_{n}$ belongs
either to $A[G,G]$ or to $B[G,G]$.
Therefore there exists, for each integer $n$, an element $z_{n}\in[G,G] = Z(G)$
such that $g_{n}z_{n}\in A\cup B$.
Now we find that the commutator
$$[g_{n}z_{n}, g_{n+1}z_{n+1}] = [g_{n}, g_{n+1}] = c_{n+1}\neq 1,$$
which implies that $g_{n}z_{n}$ and $g_{n+1}z_{n+1}$ belong to the same factor $A$ or $B$,
(say $A$) which implies that $G = A[G,G]$.
But $c_{n} = [g_{n-1},g_{n}]$ also belongs to $A$, so $[G,G]\leq A$ and thus,
$G = A$, a contradiction.
(According to my notes, this example is due to P. Hall.)
