Let $R(i):=\{k\in\mathbb N:p_{k}(i|i)>0\}$ then the period is $d(i)=\gcd R(i)$. Denote also $r(i)$ be the minimum of $R(i)$. Why is it then that $R(i)=r(i)+\mathbb N_0 d(i)$ ?
It is clear that for a chain $p_{d(i)}(i|i)>0$ does not necessarily hold, for example if $R(i)=\{4,8,10,\dots\}$ (gcd is $2$ but it is not in the set). Now, what forces $6$ to be also in $R(i)$ ?
Notations: $p_k(i|i)=\Pr(X_k=i|X_0=i)$ and $d(i)$ is the period of the state $i$.
Where did you get the statement? It looks wrong to me. A counter-example follows.
Consider a chain that looks like an "$\infty$"-symbol and consists of two linked cycles of length 1000 and 1001. Let $i$ be the node that links the two cycles. Whenever the chain is at the node $i$, it goes to the left cycle or the right cycle with probability $1/2$, and then has to go around the entire cycle until it reaches $i$ again. It holds: $R(i) = \{1000, 1001, 2000, 2001, 2002, 3000, 3001, 3002, 3003, ...\}$. The period $d(i)$ is then clearly $1$. However, nothing forces $1500$ to be in the set $R(i)$, so $R(i) \neq 1000 + \mathbb{N}_0$.
Determining the right offset $r$ such that $R(i) = \{e_1, \dots, e_n\} \cup (r + d(i) \mathbb{N}_0)$, where $e_1, \dots, e_n$ are finitely many exceptions from the simple rule, is not trivial: it's related to the Coin problem, and is NP-hard.