Consider $Z_n := (X_n,Y_n)$ where $(X_n)_{n\in \mathbb{N}}$ and $(Y_n)_{n\in \mathbb{N}}$ are irreducible markov chains with periods $\lambda$ and $\mu$. We know that $(Z_n)_{n\in \mathbb{N}}$ is a Markov Chain then. Work on the following problem:
$\mu$ and $\lambda$ have no common divisor $\iff$ $(Z_n)_{n\in \mathbb{N}}$ is irreducible; further $(Z_n)_{n\in \mathbb{N}}$ has period $\lambda \mu$ then.
I do not manage. See below:
We need to show that $\forall i,j,i',j' \exists n \in \mathbb{N}: P(Z_n = (i',j') | Z_0 = (i,j)) > 0$ $\iff$ $\operatorname{gcd}(\mu,\lambda) = 1$.
$\Rightarrow$ direction:
Consider irreducibilty, then the probabilities are always greater than zero. For an $n \in \mathbb{N}$ especially we have $P(Z_n = (i,j) | Z_0 = (i,j)) > 0$. So we have $n|\mu$ and $n|\lambda$.
But how does this impy that $\lambda$ and $\mu$ have no common divisor?
$\Leftarrow$ direction:
I do not know. It seems that if you have no common divisor you will manage to reach every tuple $(i',j')$ starting from any $(i,j)$ since you can reach every $i'$ from $i$ since X is irreducible and the same goes for the $j$'s. But I do not manage to get this formally written down..
Hints? Is this the right way to go?