The following question is from a paper I am reading called "Quantum Error Correction Via Codes Over GF(4)" It says:

Let $E$ be the quantum error group.

Let $S' \leqslant E$ which specifies undetectable errors.

Let $S \leqslant S'$ consist of errors that have no effect on the encoded state.

Every element of $S'$ commutes with $S$. This implies that $S$ is abelian.

I am a little bit confused by the above statement. If $S \leqslant S'$ and $\forall$ $s \in S$, $s' \in S'$ $$s * s' = s' * s$$ would this not instead imply that $S'$ is the abelian group as $S$ is a subgroup of $S'$ and as such all $s \in S$ imply $s \in S'$ ?

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    $\begingroup$ Your title question and the question in the body are very distinct. So I am editing your question a bit. The answer to the title question is addressed here. $\endgroup$ Oct 13, 2023 at 2:44

1 Answer 1


would this not instead imply that S′ is the abelian group

No. It does not. We have the following definitions.

(1) Every element of $S'$ commutes with every element of $S$.

(2) $S$ is a subgroup of $S'$.

(3) Putting (1) and (2) together implies that every element of $S$ commutes with every element of $S$. This makes $S$ abelian. I think this much you understand.

But nowhere in the definitions is it stated that the elements of $S'$ outside $S$ commute amongst themselves. To reiterate symbolically: the set elements of $S'$ outside $S$ are denoted by $S'\backslash S$. From fact (1) we know that both elements of $S$ and $S'\backslash S$ (which together equal $S'$) commute with elements of $S$. But elements of $S'\backslash S$ are not guaranteed to commute with each other. So $S'$ is not guaranteed to be abelian from these definition.

In fact, as you will learn, $S'$ (more commonly known as the normalizer of $S$ and denoted $\mathcal{N}(S)$), is not abelian. At the logical level elements of $S'\backslash S$ act like logical $X$, $Y$, and $Z$ operators. As you know, these operators don't commute with each other. So $S'$ can't be abelian. More details are found here.

  • $\begingroup$ Thank you, I understand why $S'$ is not abelian now! I am not sure if my interpretation here is right, so (assuming an $[[n,k,d]]$ code) $S'/S$ $=$ $P_{k}$ $=$ $\{ \pm w_{1} \otimes \dots \otimes w_{k}, \pm i w_{1} \otimes \dots \otimes w_{k}\}$ where $$w_{i} \in \{I, X, Y, Z\}$$ and so $E$ would be equal to $P_{n}$? $\endgroup$
    – am567
    Oct 13, 2023 at 10:15
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    $\begingroup$ I fixed my notation for clarity. Set difference should be denoted by backslash. The cosets of $S$ in $S'$ are denoted by $S'/S$. Then, you are correct, except $S'/S$ is isomorphic to $P_k$, not equal. $\endgroup$ Oct 13, 2023 at 17:32

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