In Devitt et al. 2013's introduction to quantum error correction, the authors mention (bottom of page 12) how the stabilizer group for $N$ qubits is abelian.

More specifically, here is the quote:

An $N$-qubit stabilizer state $\lvert\psi\rangle_N$ is then defined by the $N$ generators of an Abelian (all elements commute) subgroup $\mathcal G$ of the $N$-qubit Pauli group, $$\mathcal G=\{K^i\,|\,K^i\lvert\psi\rangle=\lvert\psi\rangle,\,[K^i,K^j]=0,\forall (i,j)\}\subset \mathcal P_N.$$

I am confused by this. Is the stabilizer subgroup $\mathcal G$ defined as an abelian subgroup of elements of $\mathcal P_N$ that stabilizes $\lvert\psi\rangle$, or is it instead the case that the subgroup of elements of $\mathcal P_N$ that stabilize $\lvert\psi\rangle$ is abelian?

If the latter, doesn't this introduce ambiguity in the definition? There could be other elements that stabilize $\lvert\psi\rangle$ but not commute with $\mathcal G$.

If the former, how is this shown? I can see why the action of $K^i K^j$ and $K^j K^i$ is identical on $\lvert\psi\rangle$, but how do you show that $K^i=K^j$?

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    It they are not abelian, they don't stablize a state (as they don't have a common eigenvector). I seem to me missing the problem. – Norbert Schuch Oct 25 at 19:07
  • @NorbertSchuch mh, couldn't you have $AB\lvert\psi\rangle=BA\lvert\psi\rangle$ for the state to be stabilized, but $AB\lvert\phi\rangle\neq BA\lvert\phi\rangle$ for other states? That is, can it not be the case that they act as commuting on the specific state that defines the subgroup, but not on other states? – glS Oct 25 at 19:09
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    Not for Paulis. Otherwise: Sure. – Norbert Schuch Oct 25 at 19:16

It is not necessary to define the group as commuting —$\def\ket#1{\lvert#1\rangle}$ by virtue of every element in the group stabilising the state $\ket{\psi}$, this property follows.

Because we are considering subgroups of the $N$-qubit Pauli group, any two elements either commute or anti-commute. Let $P \in \mathcal P_N$ be an operator which stabilises some vector $\ket{\psi}$, that is such that $P \ket{\psi} = \ket{\psi}$. Suppose that $Q$ anticommutes with $P$. It then follows that $$ Q \ket{\psi} = Q P \ket{\psi} = - P Q \ket{\psi}. $$ Now, if $Q \ket{\psi} = \lambda_Q \ket{\psi}$ for any scalar $\lambda_Q$ at all, we have $$ \lambda_Q \ket{\psi} = - P Q\ket{\psi} = - \lambda_Q \ket{\psi}. $$ But this implies that either $\lambda_Q = 0$ (which is impossible as $Q$ is unitary) or $\ket{\psi} = \mathbf 0$, the zero vector.

It follows that if $\ket{\psi}$ is actually a state (so that in particular it has norm 1), any operator in $\mathcal P_N$ which has $\ket{\psi}$ as a $\pm1$ eigenvector must commute with all operators which stabilise $\ket{\psi}$. Thus the subgroup of $\mathcal P_N$ which stabilises $\ket{\psi}$ is abelian.

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