# 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?

• The direct product of $\mathbf{Z}$ by an infinite direct sum of finite groups works... – YCor Jun 19 '14 at 12:28
• Fair enough. I guess I should have asked for an example of FC groups which is not FD and is not decomposable as direct product of two non-trivial groups. – Mahmood Al Jun 19 '14 at 21:34

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.)

• Dear James: I am not an algebraist, so I had never seen this representation of dihedral group of order $8$. Could you just briefly explain it with respect to the usual representations of a rotation $r$ and a symmetry action $s$. – Mahmood Al Jun 19 '14 at 21:32
• @Darkmoon. This just sets up the dihedral group in a convenient form, generated by three involutions with one central. As a permutation group the dihedral group $D_{4}$ of order $8$ is generated by the permutations $(1,3)$ and $(1,2,3,4)$. Here, we have $a_{n}\mapsto (2,4)$, $b_{n}\mapsto (1,2)(3,4)$ and $c_{n}\mapsto (1,3)(2,4)$, for each $n$. The centre of $D_{4}$ is $\langle (1,3)(2,4)$. Think of the vertices of a square labelled $1,2,3,4$, in clockwise order, to visualise these. – James Jun 20 '14 at 3:13
• I guess in $\uparrow 12$, you mean $g_{i_1}\neq g_{j_1}$ and similarly $g_{i_k}\neq g_{j_m}$. – Mahmood Al Jul 7 '14 at 16:35
• @Darkmoon Yes, thank you for noticing the slip. I've fixed it now. – James Jul 8 '14 at 1:55