Find the order of an element
$\left(\bigl(\begin{smallmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 6 & 1
\end{smallmatrix}\bigr),b\right)$
in the direct product G x H where G is the symmetric group $S_6$ and H is the Klein-4-group.
What I did:
A corollary from Lagrange's Theorem tells us that the order of an element divides an order of a group. I will denote it as $ |g|$ divides $|G|$.
The order of $ |H|=4,|G|=6!=720$.
Now I have no idea of the possible next step.
Best Answer
Hint:
Decompose the permutation in $S_6$ as a product of disjoint cycles. The order of a cycle is its length, and the order of a product of disjoint cycles is the l.c.m. of the orders of its cycle components. Finally, the order of $(\sigma, b)$ is the l.c.m. of the orders of $\sigma$ and $b$.