# Creating the Cayley table for $\mathbb{Z}_2 \times S_3$

Create the Cayley table for $\mathbb{Z}_2 \times S_3$

I know that the $\mathbb{Z}_2$ is:

\begin{array}{c|cc} + & 0 & 1 \\\hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array}

And that the Cayley table of $S_3$ is

\begin{array}{c|cccccc} \cdot & e & (12) & (13) & (23) & (123) & (132) \\\hline e & e & (12) & (13) & (23) & (123) & (132) \\ (12) & (12) & e & (132) & (123) & (23) & (12) \\ (13) & (13) & (123) & e & (132) & (12) & (23) \\ (23) & (23) & (132) & (123) & e & (13) & (12) \\ (123) & (123) & (13) & (132) & (12) & (132) & e \\ (132) & (132) & (23) & (12) & (13) & e & (123) \\ \end{array}

Though I do not know how to multiply a Cayley table by a Cayley table. Any help will be appreciated.

• How would you multiply two elements in $\Bbb{Z}_2 \times S_3$? This should give a good idea of how to construct the Cayley table. – Qudit Mar 18 '15 at 0:56
• I guess I would multiply an integer by a set permutation. For example, something like $0 \times (123)$ and $0 \times (23)$ to get the "sides" of the Cayley table. – ineedanewnames Mar 18 '15 at 1:12
• No, multiplication in a direct product is element-wise. – Qudit Mar 18 '15 at 1:17
• It's going to be a "pretty big table" with $144$ entries. – David Wheeler Mar 18 '15 at 9:01

The operation which takes the Cayley tables of $G_1$ and $G_2$ and produces the Cayley table of $G_1\times G_2$ is sometimes called the Kronecker product, or tensor product, of matrices. The elements of $G_1\times G_2$ are ordered pairs $(g_1,g_2)$, and the group operation is coordinatewise.
For your example, you're working with a permutation representation of $S_3$, so it might be convenient to also use a representation of ${\Bbb Z}_2$ as the two permutations $(4,5)$ and $\textrm{id}=(4)(5)$. The point is that the set $\{4,5\}$ is disjoint from $\{1,2,3\}$, so the action of $S_3$ on $\{1,2,3\}$ and ${\Bbb Z}_2$ on $\{4,5\}$ together generate a (faithful) action of $S_3\times{\Bbb Z}_2$ on $\{1,2,3,4,5\}$.