Question on transition probability matrices

Question

Question: $P$ is the transition matrix of a finite state space Markov chain. Which of the following statements are necessarily true?

$1.$ If $P$ is irreducible, then $P^2$ is irreducible.

$2.$ If $P$ is not irreducible then $P^2$ is not irreducible.

$3.$ If $P$ is irreducible then $\frac{I+P}{2} $ is irreducible.

$4.$ If $P$ is irreducible and aperiodic, then $P^3$ is irreducible.

Mt Attempt: For $(1)$, I found a counterexample, take $P= \left[ \begin{matrix} 0&1 \\ 1&0 \end{matrix} \right]$. Here, $P$ is irreducible but $P^2=I$, which is reducible. I tried some examples for $(2),(3)$ and $(4)$ but they could not find any counterexample for these, I guess that last three options are correct, however, I was unable to prove them.

Kindly help me to prove the last three options. Is there any theorem regarding these results? Can these results be generalized for $P^n$? Thank you so much for your effort(s)!

Answer 1

It's not actually a proof, but maybe it helps.

2. If $P$ is not irreducible, there are some states $i$ and $j$, such that you can't get from $i$ to $j$. If $P^2$ is not irreducible, there are some states $i$ and $j$, such that you can't get from $i$ to $j$ with an even number of steps.
   Since you aren't able to get from $i$ to $j$ with any number of steps, especially you can't get from $i$ to $j$ with an even number of steps.

3. If $P$ is irreducible, then you can get from any state to any other. Further observe that in the matrix $\frac{I+P}2$ the same entries are non-zero as before, since from $p_{ij}>0$ follows $\frac{p_{ij}}{2}>0$. It means we can still get from any state to any other, but the probabilities change. The underlying graph is almost the same, except that we add loops for every node.

4. If $P$ is irreducible, then you can get from any state $i$ to any other state $j$. If $P^3$ is irreducible, then you can get from any state $i$ to any other state $j$, such that the number of steps from $i$ to $j$ is divisible by 3. I am not sure about this, but I guess it is true. Sure you need $P$ to be aperiodic, otherwise $\begin{bmatrix} 0 &1&0\\0&0&1\\1&0&0\end{bmatrix}$ is a counterexample.