In the algebraic group $G = \operatorname{PGL}_4(\mathbb{C})$, let $E$ denote the subgroup of elements of order dividing 2 in the diagonal maximal torus; it is generated by the images of the three matrices $$ \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix},\quad \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & I_2 \end{bmatrix},\quad \begin{bmatrix} 1 & 0 \\ 0 & I_3 \end{bmatrix}. $$ Direct computation shows that $N_{G}(E)/C_{G}(E) \cong S_4\,$, which is the Weyl group of $G$. Analogous result can be obtained for $n = 3, 5, 6, 7, 8, 9, 10$ in $\operatorname{PGL}_{n}(\mathbb{C})$ by direct calculations. Is this observation true for any $n$? Is there any reference relevant?
1 Answer
$\newcommand{\ZZ}{{\mathcal Z}_G} \newcommand{\NN}{{\mathcal N}_G} \newcommand{\zz}{{\mathfrak z}_G} \newcommand{\Lie}{{\rm Lie\,}} \renewcommand{\tt}{{\mathfrak t}} \renewcommand{\gg}{{\mathfrak g}} \newcommand{\X}{{\sf X}} \newcommand{\Z}{{\Bbb Z}}$ Yes, this is true for any $n\ge 3$.
Let $G$ be a semisimple group of adjoint type over an algebraically closed field $k$ of characteristic 0. Let $T\subset G$ be a maximal torus. Write $$E=T^{(2)}:= \{t\in T(k)\ \ t^2=1\}.$$ We wish to compute the centralizer $\ZZ(E)$ and the normalizer $\NN(E)$.
Observe that $\NN(E)\supseteq \ZZ(E)\supseteq T$. We compute $\zz(E):=\Lie \ZZ(E)$.
Lemma 1. $\zz(E)=\tt:=\Lie T$.
Proof. Since $\ZZ(E)\supseteq T$, we have $\zz(E)\supseteq\tt$. Write the root decomposition $$ \Lie G=\tt\oplus\bigoplus_{\beta\in R}\gg_\beta\,,$$ where $R=R(G,T)\subset \X^*(T)$ is the root system. Since $\zz(E)\supseteq\tt$, we have $$ \zz(E)=\tt\oplus\bigoplus_{\beta\in M}\gg_\beta$$ for some subset $M\subseteq R$. Here $$M=\{\beta\in R\ \ \beta(t)=1\ \forall t\in E\}.$$ Let $S\subset R$ be a system of simple roots (a basis of $R$). Since $G$ is of adjoint type, the set $S\subset R\subset \X^*(T)$ is a basis of the character group $\X^*(T)$ of $T$. It follows that for any simple root $\alpha\in S$, there exists $t\in E=T^{(2)}$ such that $\alpha(t)=1$. Let $W=W(G,T)=\NN(T)/T$ be the Weyl group. The group $W$ acts on $E$ and on $R$, and $W\cdot S=R$. Therefore, for any root $\beta\in R$, there exists $t\in E$ such that $\beta(t)=1$, and therefore $M=\varnothing$ and $\zz(E)=\tt$, as required.
We compute $\NN(E)$. Since ${\rm char}(k)=0$, it follows from Lemma 1 that the identity component $\ZZ(E)^0$ of $\ZZ(E)$ is $T$. We see that $\NN(E)$ normalizes $\ZZ(E)$, and hence it normalizes $\ZZ(E)^0=T$. It follows that $\NN(E)\subseteq \NN(T)$. On the other hand, $\NN(T)$ normalizes $T$, and therefore, it normalizes $E=T^{(2)}$, whence $\NN(T)\subseteq \NN(E)$. Thus $\NN(E)=\NN(T)$.
We wish to compute $\ZZ(E)$. Consider the Weyl group $W=\NN(T)/T$. Set $$W_E=\ZZ(E)/T\subseteq \NN(T)/T=W.$$ Then $W_E$ is a finite group, the kernel of the natural homomorphism $W\to{\rm Aut\,} E$, and $\ZZ(E)$ is the preimage of $W_E$ in $\NN(T)$.
Lemma 2. Let $W'=S_n$ (the permutation group on $n$ symbols) naturally acting on the set $$E'=\ker\, \Sigma\colon (\Z/2\Z)^n\to \Z/2\Z,$$ where $\Sigma(x_1,x_2,\dots,x_n)=x_1+x_2+\cdots+x_n\,.$ If $n\ge 3$, then this action is effective (has trivial kernel).
Proof. Left to OP and the reader.
Observe that Lemma 2 is false for $n=2$, when ${\rm Aut\,} E'=\{1\}$ whereas $S_2\neq \{1\}$.
Theorem. Let $G={\rm PGL}_n$ with $n\ge 3$. Then $\NN(E)/\ZZ(E)=\NN(T)/T=W$.
Proof. We have seen that $\NN(E)=\NN(T)$. By Lemma 2, if $n\ge 3$, then $W_E=\{1\}$, whence $\ZZ(E)=T$, and the theorem follows.
Edit. Observe that the analogues of Lemma 2 and the theorem do not hold for the adjoint group ${\rm SO}(2n+1)$ of type ${\sf B}_n$ for $n\ge 1$. Indeed, then $$W\simeq(\Z/2\Z)^n\rtimes S_n\,,$$ and the normal subgroup $(\Z/2\Z)^n\subseteq W$ acts on $E=T^{(2)}\cong (\pm1)^n$ trivially. Therefore, $\ZZ(E)\neq T$.

$\begingroup$ Thank you for the answer. I got confused by Lemma 2. What is the map $\Sigma$? $\endgroup$ Jul 31, 2022 at 1:02

$\begingroup$ I have added the definition of the map $\Sigma$. $\endgroup$ Jul 31, 2022 at 5:51

$\begingroup$ Is the proof that $C_{G}(E)^{\circ} = T$ a must? Thank you, $\endgroup$ Jul 31, 2022 at 6:47

$\begingroup$ Yes. The argument goes as follows: ${\mathcal N}_G(E)$ normalizes ${\mathcal Z}_G(E)$, and therefore, it normalizes ${\mathcal Z}_G(E)^0=T$. We conclude that ${\mathcal N}_G(E)\subseteq {\mathcal N}_G(T)$. $\endgroup$ Jul 31, 2022 at 7:44

1$\begingroup$ @LSpice: In positive characteristic, I cannot write two lines without mistakes. So I suggest that you write, say, in comments, your proof that ${\mathcal Z}_G(E)^0=T$ also in positive characteristic. $\endgroup$ Aug 1, 2022 at 18:03