Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 2 / a / Solution

Every element of the $n$-qubit Pauli group squares to either $1$ or $-1$. If a stabilizer generator $S_j$ had $S_j^2=-1$, closure of the group would put $-1$ in $S$, contrary to the definition. Thus $S_j^2=1$. Since Pauli operators are unitary,and $S_j^{-1}=S_j$ proves $S_j=S_j^{†}$.

If two stabilizers $g,h$ anticommute and a nonzero vector $|\psi ⟩$ belonged to the codespace, thena contradiction. Hence a nonzero stabilizer code requires $S$ to be Abelian. Its centralizer of a stabilizer group is