Claim: Let P be an irreducible transition matrix and let P has period $d>1$. Then, the state space S splits into d sets $A_{1}, \cdot \cdot \cdot A_{d}$
I am unclear why this is true. Since P is irreducible transition matrix of a MC, every state i, j in a state space S communicates; the state space S has only one communicating class.
Since P has period $d>1$, $d$ divides every $n \in \mathbb{Z_{0}^{+}}$ for the transition probability $p^{n}\left(i,i\right)>0$.
How does this leads to the splitting of the state space S into $d$ disjoint sets?
Any help is appreciated.
Pick an arbitrary state $s$ to put in class $A_0$. (I'm going to number starting at $0$ rather than $1$ because it's slightly nicer.) For every other state $t$, find a positive integer $k$ such that $p^k(s,t) > 0$. Then put $t$ in class $A_{k \bmod d}$.
We should check that this is well-defined (up to the choice of class for the initial state $s$). Suppose we have $k_1, k_2$ such that $p^{k_i}(s,t) > 0$, and in the other direction $\ell$ such that $p^\ell(t,s) > 0$. Then $$p^{k_i+\ell}(s,s) \ge p^{k_i}(s,t) p^{\ell}(t,s) > 0,$$ so by periodicity, we know that $k_1 + \ell \equiv k_2 + \ell \equiv 0 \pmod{d}$. In particular, $k_1 \equiv k_2 \pmod d$, which was what we wanted.
We can give a similar proof that if you have states $x \in A_i$ and $y \in A_j$ such that $p^k(x,y) > 0$, then $k \equiv j-i \pmod d$. In particular, from a state in $A_i$ we can only transition to a state in $A_{(i+1) \bmod d}$.