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I am trying to compute the multiplication table of $(S_1 \wr D_2) \times (S_1 \wr D_3)$.

My effort:

I understand that $S_1$ is the trivial group consisting only the unit element i.e. $e$. $e$ will be the unit element for all groups discussed here. $D_2$ is isomorphic to $C_2 \times C_2 = \langle e, x, y, z | x^2 = y^2 = z^2 = xyz = e \rangle$. The group elements of $C_2 \times C_2$ are $\left\{e, x, y, z\right\}$. I also know that $D_3 = \langle a, b | a^2 = b^2 = (ab)^3 = e \rangle$. If we define the additional shorthands $c := aba$, $d := ab$ and $f := ba$, the group elements are $\left\{e, a, b, c, d, f\right\}$.

$(S_1 \wr D_2) \times (S_1 \wr D_3)\\ \implies ((S_1)^2 \rtimes D_2) \times ((S_1)^3 \rtimes D_3)\\ \implies ((S_1 \times S_1) \rtimes D_2) \times ((S_1 \times S_1 \times S_1) \rtimes D_3)\\ \implies (S_1 \rtimes D_2) \times ((S_1 \times S_1) \rtimes D_3)\\ \implies D_2 \times D_3 $

So, the some of the group elements are $(e,e)$, $(e,a)$, $(e,b)$, $(e,c)$, $(e,d)$, $(e,f)$, $(x,e)$, $(x,a)$, $(x,b)$, $(x,c)$, $(x,d)$, $(x,f)$, $(y,e)$, $(y,a)$, $(y,b)$, $(y,c)$, $(y,d)$, $(y,f)$, $(z,e)$, $(z,a)$, $(z,b)$, $(z,c)$, $(z,d)$, $(z,f)$. So, the order of the group is at least $24$.

If we put them in a multiplication table we can easily determine the rest of the group elements.

My question: Am I doing it right?

PS. I have updated the question after following to advice from Tobias' comment.

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    $\begingroup$ It will help you a lot to notice that whenever you take a product (whether it is direct or semidirect) with the trivial group, you can just remove it. $\endgroup$ – Tobias Kildetoft Jan 31 '16 at 9:37

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