Markov Chain Periodicity GCD

Question

I am not understanding the concept of period. Why is {b,c,d,e,f} transient with period 2? If I start in b I can go around c, d, f, d, b. That entire cycle is 6. Alternatively I can go b, c, d, b - 4 steps. Does the period just mean the gcd(4,6) which is 2?

On the other hand: return times for state j can occur in multiples of three (first k, then l, then j) or in multiples of two (first l then j). Since two and three are relatively prime, their great common divisor is one. Hence, d(j) = d(k) = d(l) = 1.

Following same argument for {h,i} shouldn't it be that you can either go in one or two steps. So gcd(1,2)=1. Why is this recurrent with period 2?

Answer 1

How would you get from $h$ to itself in one step? You have to go to $i$ and then after that you're forced to go straight back to $h$.

As for the first thing, yes, the period is $2$ because of the existence of the $4$ path and the $6$ path together.

Answer 2

Because once you enter the subclass $\{ h, i \}$, you cannot "get out" of it, and the chain will alternate between $h$ and $i$, i.e., $h \to i \to h \to i$ and so on. Thus, the period will be exactly $2$.