[proofplan] Part (2) (recurrence is a class property) was established in [Recurrence is a Class Property](/theorems/2214). Part (3) (positive recurrence is a class property) is deferred to the convergence chapter, where the necessary tools involving invariant distributions are developed. We prove parts (1) and (4). For part (1), the key idea is that if $i \leftrightarrow j$, we can "sandwich" return times to $j$ through $i$ using the [Chapman--Kolmogorov Equations](/theorems/2207), showing that $d_i$ divides every element of $D_j$ and vice versa, hence $d_i = d_j$. Part (4) follows immediately from parts (1) and (3). [/proofplan] [step:Establish path lengths between communicating states] Since $i \leftrightarrow j$, there exist integers $m, n \geq 1$ such that \begin{align*} p_{i,j}(m) > 0 \quad \text{and} \quad p_{j,i}(n) > 0. \end{align*} Define $\alpha = p_{i,j}(m)\, p_{j,i}(n) > 0$. By the [Chapman--Kolmogorov Equations](/theorems/2207), the round trip $i \to j \to i$ in $m + n$ steps gives: \begin{align*} p_{i,i}(m + n) \geq p_{i,j}(m)\, p_{j,i}(n) = \alpha > 0, \end{align*} so $m + n \in D_i$, where $D_i = \{r \geq 1 : p_{i,i}(r) > 0\}$ is the set of possible return times to $i$. [/step] [step:Show $d_i \mid d_j$ by routing returns to $j$ through $i$] Let $r \in D_j$ be arbitrary, so $p_{j,j}(r) > 0$. By the [Chapman--Kolmogorov Equations](/theorems/2207), routing the path $i \to j$ (in $m$ steps), $j \to j$ (in $r$ steps), and $j \to i$ (in $n$ steps): \begin{align*} p_{i,i}(m + r + n) \geq p_{i,j}(m)\, p_{j,j}(r)\, p_{j,i}(n) = \alpha\, p_{j,j}(r) > 0. \end{align*} Therefore $m + r + n \in D_i$, which means $d_i \mid (m + r + n)$. Since $m + n \in D_i$ (established above), we also have $d_i \mid (m + n)$. Subtracting: $d_i$ divides $(m + r + n) - (m + n) = r$. Since $r \in D_j$ was arbitrary, $d_i$ divides every element of $D_j$. By the definition of $d_j = \gcd(D_j)$, this means $d_i \mid d_j$. [guided] The strategy is to show that the period of $i$ divides every possible return time to $j$. The mechanism is path routing: any return to $j$ in $r$ steps can be "embedded" in a return to $i$ by prepending the path $i \to j$ (taking $m$ steps) and appending $j \to i$ (taking $n$ steps). This creates a return to $i$ in $m + r + n$ steps. Since $d_i = \gcd(D_i)$, we know $d_i$ divides every element of $D_i$, in particular $d_i \mid (m + r + n)$ and $d_i \mid (m + n)$. The key arithmetic step is the subtraction: $d_i \mid (m + r + n)$ and $d_i \mid (m + n)$ imply $d_i \mid r$. Why does this suffice? The definition of $\gcd$ is: $d_j = \gcd(D_j)$ is the largest integer dividing every element of $D_j$. We have shown that $d_i$ divides every element of $D_j$, so $d_i \leq d_j$ (since $d_j$ is the largest such divisor). In other words, $d_i \mid d_j$. [/guided] [/step] [step:Conclude $d_i = d_j$ by symmetry] By the same argument with $i$ and $j$ interchanged (using $p_{j,i}(n) > 0$ and $p_{i,j}(m) > 0$ in the reversed roles), we obtain $d_j \mid d_i$. Since $d_i \mid d_j$ and $d_j \mid d_i$, and both $d_i, d_j \geq 1$, we conclude $d_i = d_j$. This proves part (1). [/step] [step:Note that part (2) was established previously] Part (2) — that recurrence is a class property — was proved in [Recurrence is a Class Property](/theorems/2214). The argument there shows that if $i \leftrightarrow j$ and $\sum_n p_{jj}(n) = \infty$, then $\sum_n p_{ii}(n) = \infty$, and conversely, using the [Chapman--Kolmogorov Equations](/theorems/2207) and the [Recurrence Summability Criterion](/theorems/2211). [/step] [step:Note that part (3) is deferred to the convergence chapter] Part (3) — that positive recurrence is a class property — requires tools from the theory of invariant distributions. The proof uses the fact that an irreducible chain is positive recurrent if and only if it admits an invariant distribution (see [Existence and Uniqueness of Invariant Distribution](/theorems/2223)), and the invariant distribution $\pi_i = 1/\mu_i$ connects the mean return times of communicating states. This result is established in the convergence chapter. [/step] [step:Derive part (4) from parts (1) and (3)] A state is ergodic if and only if it is both aperiodic ($d = 1$) and positive recurrent. By part (1), $d_i = d_j$, so $i$ is aperiodic if and only if $j$ is aperiodic. By part (3), $i$ is positive recurrent if and only if $j$ is positive recurrent. Therefore $i$ is ergodic if and only if $j$ is ergodic. This proves part (4). [/step]