Let $A$ be a finite alphabet, and let $X \subset A^{\mathbb Z}$ be a subshift of finite type. Then there exist a finite alphabet $B$, a zero-one matrix $M \in \{0,1\}^{B \times B}$, and a topological conjugacy from $X$ onto the topological Markov chain
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\begin{align*}
X_M := \{y \in B^{\mathbb Z} : M_{y_k y_{k+1}} = 1 \text{ for every } k \in \mathbb Z\}.
\end{align*}
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Equivalently, every subshift of finite type is topologically conjugate to a one-step subshift of finite type.