= Solution
For a state $i$, let $N_i=\{n\geq1:P^n_{ii}>0\}$ and $d_i=\gcd N_i$. If distinct states $i,j$ communicate, choose positive-probability paths $i\to j$ and $j\to i$ of lengths $r,s$. Then $r+s$ is a return time to each. For every $n\in N_i$, concatenate $j\to i$, the $i$-return loop, and $i\to j$ to see that $s+n+r\in N_j$. Since $d_j$ divides this and $s+r$, it divides $n$. Thus $d_j\mid d_i$, and interchanging the states gives $d_i\mid d_j$. This proves \b[<period is constant on a communicating class>].
Read the transition matrix from the original PDF; the converted TeX loses the location of the second row's final entry. The <communicating classes> are
$$
\boxed{\{1,3,4\},\quad\{2,7\},\quad\{6\},\quad\{5\}}.
$$
Within $\{1,3,4\}$ the only internal cycle is $1\to3\to4\to1$. Return lengths are multiples of three, and length three has positive probability, so its period is three. This class is open because it has exits to other classes. The class $\{2,7\}$ is the closed deterministic two-cycle, of period two. The absorbing class $\{6\}$ is closed and has period one.
State 5 has outgoing transitions but no incoming transition at all, so it is an open singleton class and $N_5=\varnothing$. For this <return-free state in a Markov chain>, the convention $\gcd\varnothing=0$ gives \b[period zero]; if period is defined only when positive return times exist, it is undefined instead. In particular, this singleton is not aperiodic: aperiodicity would require period one.
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