In Daniel Gottesman's thesis "Stabilizer Codes and Quantum Error Correction", he makes the following claim:

The set of elements in $G$ that commute with all of $S$ is defined as the centralizer $C(S)$ of $S$ in $G$. Because of the properties of $S$ and $G$, the centralizer is actually equal to the normalizer.

My understanding of the centralizer $C(S)$ of $S$ is that elements in $C(S)$ commute with every element in $S$ i.e. $u \in C(S) \rightarrow us=su \text{, } \forall s \in S$.

Whereas, for the normalizer $N(S)$ of $S$ is that the elements in $N(S)$ commute with the subgroup $S$, meaning that $v \in N(S) \rightarrow vS=Sv$. This means that $vs=s'v$, where $s,s' \in S$, but $s$ not necessarily equal to $s'$, meaning elements in the normaliser do not necessarily commute with every element of $S$.

I am wondering what are the properties of $S$ and $G$ that Gottesman is referring to, that make $C(S) = N(S)$?

Is it to do with the stabilizing property of $S$?


1 Answer 1


The property of $S$ is: $$ s\in S\implies -s\notin S\tag1. $$ The property of $G$ is: $$ \forall u,v\in G\quad uv=vu\,\lor\,uv=-vu\tag2. $$

Claim. If a group $G$ of linear operators$^1$ satisfies $(2)$ and a set $S\subset G$ satisfies $(1)$, then $C_G(S)=N_G(S)$.

Proof. It's clear that $C_G(S)\subset N_G(S)$, so assume $v\in N_G(S)$. Then $vs=s'v$ for some $s,s'\in S$. By $(2)$, either $s'=s$ or $s'=-s$. However, if $s'=-s$, then by $(1)$ we have that $s'\notin S$. The contradiction means that $s'=s$, but then $v\in C_G(S)$. $\square$

$^1$ I'm assuming the elements of $G$ are linear operators so that negation is well defined. One can extend the claim to more general groups at the expense of clarity.


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