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I am stuck at aperiodic property of irreducible markov chain. Let us consider an irreducible markov chain. It's stated herein that for an irreducible markov chain, a single aperiodic state implies that the whole chain is aperiodic. So, how can we proceed to prove it?

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By definition, the period $|x|$ of a state $x$ on a Markov chain is the only positive integer which divides the length of any chain starting and ending at $x$. More precisely, if $S:x=s_0,\ldots,s_n=x$ is a path in the Markov chain which starts and ends at $x$, then $|x|$ divides the length $n$ of $S$.

In what follows, we denote by $|S|$ the length of a path in the Markov chain.

Suppose we are in an irreducible Markov chain, and some state $y$ is aperiodic, i.e., $|y|=1$. Let's show that any other state $x$ is also aperiodic. By irreducibility, there exists a path $S$ connecting $x$ to $y$ and a path $T$ connecting $y$ to $x$. Since $|y|=1$, there exist paths $P_1,\ldots,P_r$, of lengths $|P_1|,\ldots,|P_r|$, such that $\operatorname{gcd}(|P_1|,\ldots,|P_r|)=1$.

Consider the contatenated paths $SP_jT$, which connect $x$ to itself. Then $|x|$ divides $|SP_jT|=|S|+|P_j|+|T|$ for all $j$. But also, $ST$ starts and ends at $x$, so $|x|$ divides $|ST|=|S|+|T|$. Putting these two facts together, we see that $|x|$ divides $|P_j|$ for all $j$, and therefore $|x|=1$, so $x$ is aperiodic.

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