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In $G=\operatorname{PGL}_{2n}(\textbf{C})$, WLG, we assume all the toral elementary abelian 2-subgroups in discussion are in $T$, the image in $G$ of the group of diagonal matrices in $\operatorname{GL}_{2n}(\textbf{C})$.

I'm aware that the conjugacy classes of involutions in $G$ have representatives $$a_{j}=\left[ \begin{pmatrix} -I_{j} & 0 \\ 0 & I_{2n-j} \end{pmatrix} \right]$$ where $j=1..n$ and $C_{G}(a_{j})=C_{G}(a_{j})^{\circ}$ for $j=1..n-1$ and $C_{G}(a_{n})=\langle C_{G}(a_{n})^{\circ},b_{n}\rangle$ where $b_{n}$ is an order 2 element. At last, $C_{G}(\langle e_{1},e_{2}\rangle)^{\circ} = C_{G}(e_{1})^{\circ}\cap C_{G}(e_{2})^{\circ}$ for any two involutions.

As a result, for an elementary abelian 2-subgroup $E$, we have $C_{G}(E)=C_{G}(E)^{\circ}$ when none of the generators of $E$ is conjugate to $a_{n}$. If there is one generator of $E$ conjugate to $a_{n}$, after we identify this generator with $a_{n}$, we see if the $b_{n}$ associated with $a_{n}$ centralizes all the other generators, then $C_{G}(E) \neq C_{G}(E)^{\circ}.$

I hope to show for an elementary abelian 2-subgroup $E$, we have $C_{G}(E)=C_{G}(E)^{\circ}$ if and only if there is a choice of generators of $E$ s.t. none of them is conjugate to $a_{n}$.

The right implying the left is clear. For the other direction, I'm thinking to show if there is one generator conjugate to $a_{n}$, we can replace it with an element not conjugate to $a_{n}$, using the connectedness of the centralizer somehow. So far this has led me nowhere.. Any help would be appreciated.

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  • $\begingroup$ I do not follow your claim about elementary Abelian 2-groups. In answer to another question of yours, I showed that $C_G(e_1, e_2)^\circ$ equals $C_G(e_1)^\circ \cap C_G(e_2)^\circ$ if $e_1$ and $e_2$ lie in a common torus. But I do not know that the claim is true if $e_1$ and $e_2$ do not lie in a common torus, and I know that it can fail if you replace two involutions by three, even if they lie in a common torus. So how do you deduce your claim about elementary Abelian 2-groups? $\endgroup$
    – LSpice
    Commented Dec 17, 2022 at 20:01
  • $\begingroup$ Yes, if all your elements come from a common torus, then you can describe centralisers and connected centralisers easily. It's clear in complete generality that $H \mathrel{:=} \bigcap_{i \in I} C_G(e_i)^\circ$ is between $C_G(e_i : i \in I)$ and its identity component, but there is no reason in general that $H$ should be connected. $\endgroup$
    – LSpice
    Commented Dec 17, 2022 at 21:23
  • $\begingroup$ Thank you! So we agree that my question makes sense given that I assumed all the toral groups have been conjugated into $T$? $\endgroup$
    – user488802
    Commented Dec 17, 2022 at 21:27
  • $\begingroup$ Re, no. First, while every individual involution can be conjugated into a fixed torus, they cannot all be conjugated simultaneously into a common torus (not even necessarily if they commute). But maybe you meant to assume this. Even if so, then it need not be true that the intersection of the connected centralisers is connected. $\endgroup$
    – LSpice
    Commented Dec 17, 2022 at 21:30
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    $\begingroup$ I appreciate all your input and time! Merry Christmas in advance! $\endgroup$
    – user488802
    Commented Dec 18, 2022 at 0:29

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