From the definition, if a state's period is one, we call this state aperiodic. It's pretty counterintuitive.
For example, the naive case
Obviously, it will come back to state a after $1, 2, 3, \cdots$ iterations. Why will we call this state aperiodic? Are there some intuitive considerations?
The definition of aperiodic state can be found:
In this website, the definition is below Fig 11.11
Markov Chains and Mixing Times By David Asher Levin, in page 7.
For @geetha290krm 's comment, I've posted the following part in another post.
I have another related question is why define period of state i as $d_i \equiv perd(i) = \gcd\{n | P^n_{ii} > 0\}$. Since we cannot necessarily go back to state i after $d_i$ iterations.
For example,
If we start from state $1$, we can only go back to $1$ after $\{6, 9, 12, 15, 18, \cdots\}$ iterations. Although $\gcd = 3$, $(P^3)_{11} = 0$. However, we still say the period of state $1$ is $3$.
Some intuitive considerations come from comparison with signals.
A signal $x(t)$ with period T satisfies $x(t+T)=x(t)$. The period is defined as the minimum T to satisfy the equality, not any possible multiple of it. Constant signals, similar to your chain, do not have a defined period.
Would the period of a constant be $0$, as the limit of some minimum, or rather the frequency $0$ and therefore the period... what, positive infinite?