Count elements of order 6 in $D_3 \times S_4 \times \mathbb{Z}_4$ I need to count elements of order 6 in $D_3 \times S_4 \times \mathbb{Z}_4$.
I understand that the order 6 of the elements implies that the $\mathrm{lcm}(|d_i|, |s_j|, |z_k|)=6$.
That means next possible combinations of $|d_i|, |s_j|, |z_k|$:


*

*1, 1, 6

*1, 2, 6

*1, 3, 6

*1, 6, 6

*2, 1, 6

*2, 2, 6

*...  


However, I don't understand how to find the amount of elements of required order (1, 2, 3, 6) in each of $D_3, S_4, \mathbb{Z}_4$.
Also it is unclear to me how to exclude combinations like 1, 1, 1: do I have to process each of them or there is some formula to exclude them from final count?
 A: A finite group $G$ possesses an order list $o(G)=(a_1,a_2,\ldots)$, where $a_k$ denotes the number of elements  $x\in G$ having order $k$. It is easy to see that
$$o(D_3)=({\bf1},{\bf3},{\bf2}),\qquad o(S_4)=(1,9,8,6),\qquad o({\mathbb Z}_4)=(1,1,0,2)\ .\tag{1}$$
In order to establish $(1)$ we look at the individual elements of the three involved groups. $D_3$ is the symmetry group of an equilateral triangle. It consists of the identity (order $1$), three reflections across symmetry axes (order $2$), and two $120^\circ$-rotations (order $3$). An element $\pi\in S_4$ is a permutation of the set $\{1,2,3,4\}$. The order of such a permutation is determined by its cycle structure. E.g. this $\pi$ has order $3$ iff it keeps one number fixed ($4$ choices) and computes the other three numbers cyclically ($2$ choices). Therefore there are $8$ elements in $S_4$ having order $3$. Etcetera.
Now a product of three elements $x_1\in D_3$, $\>x_2\in S_4$, $\>x_3\in {\mathbb Z}_4$ has order $6$ iff the orders of the $x_i$ in their mother groups have LCM $6$. Selecting the admissible order triples accordingly we obtain the following number of admissible choices:
$${\bf1}\cdot(8\cdot1)+{\bf3}\cdot\bigl(8\cdot(1+1)\bigr)+{\bf2}\cdot\bigl(1\cdot 1+9\cdot(1+1)+8\cdot1\bigr)=110\ .$$
