Suppose a Markov chain has transition matrix $P$, is irreducible and positive recurrent with period $D$ (greater than $1$). Consider a new Markov Chain with transition matrix $P^D$. Is this also irreducible?
I think there will be exactly $D$ communicating classes so it is not irreducible. We can partition our original Markov Chain into $D$ sets, $T_i$ for $i = 0, 1, 2, ..., D - 1$ such that any state in $T_i$ must go to $T_{i+1}$ and $T_D = T_0$. Then for our new Markov Chain, these will be our classes (not sure how to prove this). But at the very least, $T_0$ and $T_1$ are not in the same class because things in $T_0$ can only go to things in $T_0$ in a multiple of $D$ steps. Is this correct? How come we are not using the positive recurrent criterion?
Also, I've forgotten the construction needed for this cyclic decomposition, so it would be nice if someone could link a proof.
Positive recurrent is reserved for another subpart of the problem, which was probably related to the periodic extension. The statement that you said is true even without recurrent property. Cyclic decomposition is probably not required for this question. We may fix a state $a$ and decompose the state space by the number of steps $nd+i$ from $a$, where $n\in\mathbb{N}$. Try to show this is a disjoint decomposition and closed (in the notion of communicating class). Then, it will be enough to show the question's statement is true.