Does the fact that the elements of the normalizer group commute with elements of the stabilizer group imply that the normalizer is abelian?

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 $$s \in S'$$ ?

• 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. Commented Oct 13, 2023 at 2:44

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.

• 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 $P_{n}$? Commented Oct 13, 2023 at 10:15
• 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. Commented Oct 13, 2023 at 17:32