Lemma 1.6 -- the period is constant on an irreducible chain

For any finite type $V$ (with decidable equality; $V$ may be empty, in which case the statement is vacuous since it takes two elements of $V$ as arguments), any matrix $P$ of real numbers indexed by $V \times V$ that is stochastic — meaning $P_{x,y} \ge 0$ for all $x, y \in V$ and every row sums to one, $\sum_{y} P_{x,y} = 1$ for all $x$ — and that is irreducible in the following sense: for every pair $x, y \in V$ there exists a natural number $t \ge 0$ with $(P^t)_{x,y} > 0$ (note $t = 0$ is allowed, so the diagonal case $x = y$ holds automatically because $P^0$ is the identity matrix), the theorem asserts that for all $x, y \in V$ the period of $P$ at $x$ equals the period at $y$: $\mathrm{per}_P(x) = \mathrm{per}_P(y)$. Here the period at a state $x$ is defined via the return set $R_P(x) = \{\, t \in \mathbb{N} \mid t \ge 1 \text{ and } (P^t)_{x,x} > 0 \,\}$, as the supremum, taken in the natural numbers, of the set of common divisors of $R_P(x)$:

When $R_P(x)$ is nonempty this supremum is the greatest common divisor of the return times (the set of common divisors is finite, and its largest element is the gcd, since $0 \in R_P(x)$ is excluded by the condition $t \ge 1$). When $R_P(x)$ is empty (the state never returns to itself with positive probability at any time $t \ge 1$), every natural number is vacuously a common divisor, the set of common divisors is unbounded, and the supremum operator on $\mathbb{N}$ assigns the junk value $0$, so in that case $\mathrm{per}_P(x) = 0$. The conclusion is exactly the equality of these two natural numbers for the two given states $x$ and $y$, under the stochasticity and irreducibility hypotheses stated above.