Claim:$i\leftrightarrow j$ implies $d\left ( i \right )=d\left ( j \right )$
Suppose $i\leftrightarrow j$:
This implies that j is accessible from i and i is accessible from j. Then, $\exists n_{1}$ such that $p_{ij}^{n_{1}}>0$ and $\exists n_{2}$ such that $p_{ji}^{n_{2}}>0$
By the chapman kolmogorov equation:
$p_{ii}^{n_{1}+n_{2}} \geq p_{ij}^{n_{1}}p_{ji}^{n_{2}}>0$
since $p_{ii}^{n_{1}+n_{2}}>0$, by definition of periodicity of any state i:
there exists a gcd, say $d_{i}$, that divides $n_{1}+n_{2}$.
Suppose $p_{jj}^{n}>0$.
Then, by the chapman kolmogorov equation:
$p_{ii}^{n_{1}+n_{2}+n} \geq p_{ij}^{n_{1}}p_{jj}^{n}p_{ji}^{n_{2}} >0$
Again, by periodicity, $d_{i}$ is the greatest common divisor of $n_{1}+n+n_{2}$.
How can I convince myself that $d_{i}$ also divides n?
Counterexample, 2 is the gcd for $8=2+1+5$ but 2 does not divide 1