Is the period of an irreducible Markov Chain the same for all states? I'm a bit confused about this theorem about irreducible Markov Chains:

If a Markov chain is irreducible, then all its states have the
same period $d(i) := g.c.d.\{n > 0|P^n(i, i) > 0\}$.[source]

Let's suppose we have the following matrix which represents an irreducible Markov Chain:
\begin{equation}
P = \begin{bmatrix}
0 & 0 & 0.8 & 0.2\\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 
\end{bmatrix}
\end{equation}
if we start from state 1, we can reach it again after two transitions:
\begin{equation}
1\rightarrow 3 \rightarrow 1
\end{equation}
but if we start from state 2, we can only reach it again after 4 transitions:
\begin{equation}
2\rightarrow 3 \rightarrow 1 \rightarrow 4 \rightarrow 2
\end{equation}
So what is the period of the system? Is it 2 or 4? We can argue that it is 4 because 4 is a multiple of two and it is acceptable for both cases. But using the prementioned theorem $d(1) = 2$ and all other states should follow. What is it that I am getting wrong here?

The period is $2$ as given in the formula.  A Markov chain is periodic if the chain can return to the same state only at multiples of some integer $>1$.  See https://www.randomservices.org/random/markov/Periodicity.html
