How to find the order of a permutation? Given that $x= (1 2 3)(4 5 6 7 8)$, what is the order of $x$? (i.e the smallest integer $k$ such that 
 A: When a permutation is written as a product of disjoint cycles, its order is the least common multiple of those cycles' lengths (easy proof by induction).
In your case the order is $\;l.c.m (3,5)=15\;$
A: The order of a cycle is its length.
The order of a product of disjoint cycles is the least common multiple of the cycles' lengths. 
For example, $$\operatorname{order}(1, 2, 3) = 3,\quad \operatorname{order}(4,5, 6, 7, 8) = 5.$$
$$\operatorname{order}(123)(45678) = \operatorname{lcm}(3, 5) = 15.$$
A: We have this general result:

If in a group $G$ we have two elements $x$ and $y$ that commute  i.e. $xy=yx$ and with  coprime orders $m$ and $n$ respectively : $\gcd(m,n)=1$ then $mn$ is the order of $xy$.

Proof: We solve the equation:
$$(xy)^k=e\iff x^ky^k=e\tag1$$
$$(1)\implies x^{kn}=e\implies m|kn\tag2$$
and
$$(1)\implies y^{km}=e\implies n|km\tag3$$
and since $m$ and $n$ are coprime so with $(2)$ and $(3)$ we get
$$n|k\;\text{and}\; m|k\implies mn|k$$
so the order of $xy$ is $mn$.
Can you apply this result to your case?
