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If I have the quantum error group $E$ which contains elements of the form:

{$ \pm w_{1} \otimes \dots \otimes w_{n}, \pm i w_{1} \otimes \dots \otimes w_{n} $}

The centre of $E$:

$Z(E) = <iI> = ${ $\pm I, \pm iI $ }

The quotient group:

$E/Z(E) =$ {$eZ(E): e \in E $}

I am finding it difficult to see how the quotient group is an abelian group. However, I think this is because I can't visualist what cosets of the form $eE(Z)$ might look like? Would someone be able to give an example of one of these cosets?

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The elements of the n-fold Pauli group are either commute or anti-commute. In the quotient over the center they will all commute, since $-1$ wouldn't matter. Note that a general element of the Pauli group has the form $$ i^k w_{1} \otimes \dots \otimes w_{n}, $$ where $k=0,1,2,3$ and $w_j$ are Pauli matrices. Four elements for each $k$ together form a coset, and as a coset representative you can just take $w_{1} \otimes \dots \otimes w_{n}$.

Same reasoning works for subgroups of the Pauli group.

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  • $\begingroup$ Thank you for your response, I understand most of it now, except for this part: " In the quotient over the center they will all commute, since $−1$ wouldn't matter. " Can you explain why $-1$ doesn't matter here? $\endgroup$
    – am567
    Oct 13, 2023 at 15:30
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    $\begingroup$ $-1$ is in the center $\{\pm 1,\pm i\}$ which we factor out. More precisely, it's coset representatives which commute up to factors in the center. $\endgroup$
    – Danylo Y
    Oct 13, 2023 at 15:38

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