Why does 9 qubit Shor code have distance 3?

Question

Distance of a Stabilizer code($S$) is defined by minimum weight of an element in $N(S)-S$ .

Why 9 qubit Shor code has distance 3?

Is there a way to directly see $N(S)-S$ from $S$?

Answer 1

Distance 3 means that an error operator acting on 3 qubits maps a valid code-state to another valid code-state. For the Shor code, an example would be a simultaneous phase flip on the 1st, 4th and 7th qubits ($Z_1Z_4Z_7$).

You can easily check that $Z_1Z_4Z_7∉S$, but that it commute with $S$, hence is part of $N(S)$.

Answer 2

Another way to see this is by using $$\frac{N(S)}{S} \cong P_k $$ Here $k=1$ and we actually know that $\bar{X}=ZZZZZZZZZ$ and $\bar{Z}=XXXXXXXXX$ also we know $S$ and thus we know $N(S)$

Hence we can directly find distance by "minimum weight element in $N(S)-S$"