The <graph-state stabilizer generators> of the path are
$$
S_1^A=X_1Z_2,
\qquad
S_2^A=Z_1X_2Z_3,
\qquad
S_3^A=Z_2X_3,
$$
and those of the triangle are
$$
S_1^B=X_1Z_2Z_3,
\qquad
S_2^B=Z_1X_2Z_3,
\qquad
S_3^B=Z_1Z_2X_3.
$$
Any pair of distinct generators has Pauli factors $X$ and $Z$ in exactly two common positions. Each such position contributes one minus sign on exchange, so the two signs cancel. Thus
$$
\boxed{[S_r^A,S_s^A]=[S_r^B,S_s^B]=0
\quad\text{for all }r,s}.
$$
This is the general commutativity mechanism for graph-state stabilizers associated with an undirected graph.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 324 / 4 / b / ii / Solution

The graph-state stabilizer generator at vertex $v$ is $S_v=X_v\prod_{u\in N(v)}Z_u$. Two such generators commute because adjacent vertices contribute two Pauli anticommutations and nonadjacent vertices contribute none.


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