Inclusion of the normalizer in the centralizer

Question

From wiki, a centralizer $C(S)$ and a normalizer $N(S)$ of stabilizer group $S$ are defined as (1), (2):
$$C(S) = \{g \in G \,|\, gs = sg \, \text{for all}\, s \in S\} \tag{1}$$ $$N(S) = \{g \in G \,|\, gSg^{\dagger} = S\} \tag{2}$$ where G is Pauli group and S is subgroup of G.

Here, I'm trying to prove (3):
$$N(S) \subseteq C(S) \tag{3}$$ Suppose $g \in N(S)$, then is it valid to state (4)? My concern is that if it's correct to put the same s on both sides in (4).
$$gsg^{\dagger} = s \, \text{for all}\, s \in S \tag{4}$$

If it's valid, then just multiplying $g$ both sides of (4) from the right gives:
$$gs = sg$$ Hence, $g \in N(S)$ implies $g \in C(S)$, and that proves $N(S) \subseteq C(S)$.

If (4) isn't valid, how do you prove (3)? I'd be grateful for your insight.

Answer 1

We always have that $C(S)\subseteq N(S)$. The reverse inclusion $N(S)\subseteq C(S)$ is false if there are no restrictions on $G$. For a counter example, let $$S = \langle X\otimes X, Z\otimes Z\rangle.$$

The operator $H\otimes H \in N(S)$ and $H\otimes H\not\in C(S)$ since

$$(H\otimes H)(X\otimes X)(H^\dagger\otimes H^\dagger) = Z\otimes Z \\ (H\otimes H)(Z\otimes Z)(H^\dagger\otimes H^\dagger) = X\otimes X. $$

Hence $N(S)\not\subseteq C(S)$.

If you restrict $G$ to be the Pauli group, then it is true that $N(S)\subseteq C(S)$. This is because given $g\in G$ and $s\in S$, we either have $gs = sg$ or $gs = -sg$.

If they anticommute for at least one $s$ i.e. $gs = -sg$, then $g\not\in N(S)$ since $gsg^\dagger = -s$ and $-s\not\in S$. Hence $g\in N(S)$ only if $gs = sg$ for all $s\in S$. For any such commuting $g$, we see that $g\in C(S)$. Hence $N(S)\subseteq C(S)$ if $G$ is the Pauli group.