I am stuck with exercice 10.44 on page 466 of Nielsen & Chuang's Quantum Computation and Quantum Information.
I call $G_n$ the group of the n-tensor product of Pauli matrices. An element $x \in G_n$ is thus the tensor product of $n$ Pauli matrices with a coefficient $\pm 1$ or $\pm i$ in front to be able to have a group structure.
For example for $n=3$ we could have: $i X_1 Z_2 Y_3 \equiv i X \otimes Z \otimes Y$.
Let $S$ be a subgroup of $G_n$ not containing the matrix $-I$. Show that $N(S)=G(S)$ where $N(S)$ is the normaliser of $S$ in $G_n$ and $Z(S)$ is the centraliser of $S$ in $G_n$.
I am able to prove $Z(S) \subset N(S)$ by doing:
Let $g \in Z(S)$.
Then $\forall s \in S: g.s=s.g$
Thus $g.s.g^{-1} = s \in S$
So $g \in N(S)$
But for the reciprocal I am completly stuck. Above all I don't see the problem with having $-I$ in the subgroup...
How to do it ?
First of all we have to note that elements of $G_n$ all have order $1,2$ or $4$. The elements $X$ of order $4$ have the property that $X^2 = -I $, so that the subgroups under consideration are elementary abelian $2$-groups.
Closely inspecting the conjugacy classes (the orbits under conjugation) of elements $X$ one finds that they all have size $2$, in fact they are of the form $\{X, -X\}$, except if $X$ is a scalar matrix in which case the class size is $1$.
So if $S$ is an elementary abelian $2$-group with generators $X_1, \ldots, X_m$ then $g \in G$ sends $X_i$ to $-X_i$ unless $g$ centralizes $S$ or $X_i$ is a scalar matrix. A subgroup is then sent by $g$ to itself if either $g$ centralizes $S$, or if $g^{-1}X_ig = -X_j$ for some $i,j$ (where $X_i$ is not a scalar matrix), but that would mean that $X_iX_j^{-1} = -I \in S$, so we can conclude that $g$ normalizes $S$ only if $g$ centralizes $S$.