TITLE: Centralizer of involutions in simple groups.
QUESTION [5 upvotes]: I need some information about centralizer of involutions in finite simple groups of lie type. Actually I want to know if $G$ is a simple group of lie type over a finite field,

How many conjugacy classes of involution does it have?
If there are more than one, what is their sizes?

I would be grateful, if someone answers these question or introduce a good reference.

REPLY [2 votes]: Here is an answer based on Suzuki's Group Theory Chapter 6.§5, which I recommend.
Definition: A projective involution of $G$ is an element $t \in G \setminus Z(G)$ such that $t^2 \in Z(G)$. The projective centralizer of $t$ in $G$ is $C_G^*(t) = \{ g \in G : [g,t] \in Z(G) \}$. The projective conjugacy class of $t$ in $G$ is $\{ t^g z:g \in G, z \in Z(G) \}$.
It is not hard to verify that the projective involutions $t$ of $H$ are exactly the pre-images of involutions of $H/Z(H)$, the pre-images of their centralizers are exactly the projective centralizers, and the pre-images of their conjugacy classes are exactly the projective conjugacy classes.
We begin by finding the involutions and centralizers for the groups $\operatorname{PGL}_n(F)$. In GLS these are known as $\operatorname{Inndiag}(K)$ where $K$ is a version of the type $A_{n-1}(p^f)$ where $p^f=|F|$. In Carter and most other books, these are known as the adjoint type groups. However, in GLS, the word “adjoint” is used to mean $O^{p'}(H)=\operatorname{PSL}_n(p^f)$ where $H$ is of adjoint type in the usual sense, $H=\operatorname{PGL}_n(p^f)$.
Proposition (PGL): If $F$ is a field of characteristic not $2$, then $\operatorname{GL}_n(F)$ has exactly the following projective conjugacy classes of projective involutions with projective centralizers:
$$\begin{array}{ll}
\text{Name} & \text{Condition} & \text{Representative} & \text{Centralizer} \\ \hline
t_r & 1 \leq r < \tfrac n2 & \begin{bmatrix} I_r & 0_{r,n-r} \\ 0_{n-r,r} & -I_{n-r} \end{bmatrix} & \operatorname{GL}_r(F) \times \operatorname{GL}_{n-r}(F) \\
t_k & n=2k \text{ even} & \begin{bmatrix} I_k & 0_k \\ 0_k & -I_k \end{bmatrix} & \operatorname{GL}_k(F) \wr \operatorname{Sym}(2) \\
t_k' & {n=2k \text{ even}} \atop {\lambda \notin (F^\times)^2}& \begin{bmatrix} 0_k & I_k\\ \lambda I_k & 0_k \end{bmatrix} & \operatorname{\Gamma L}(k, F[\sqrt{\lambda}] / F )
\end{array}$$
The class $t_k$ and $t_k'$ only exist if $n$ is even. If $n$ is even, then $t_k'$ is at most one class (take any particular $\lambda$; if none exist because every element of $F$ is a square, then the class $t_k'$ does not exist). Here $\operatorname{\Gamma L}(k, F_2/F_1)$ for a Galois extension of fields $F_2/F_1$ is the group $\operatorname{Gal}(F_2/F_1) \ltimes \operatorname{GL}(k,F_2)$ where the Galois group acts on a matrix entry-wise.
GLS's table mentions that $\operatorname{Aut}(G) \geq H$ has additional involutions: $\gamma_1$ (the inverse transpose) a new conjugacy class if $n > 2$, and $\gamma_2$ if $n > 2$ is even (composition of $\gamma_1$ and any $t_i$, I think).
Proof (sketch): The center consists of scalar matrices. If $t$ is a projective involution, then $t^2 = \lambda I_n$ for some $\lambda \in F^\times$. If $\lambda=\rho^2$ is a square, and $t$ is diagonalizable with diagonal entries $\pm\rho$. Since $t$ is not a scalar matrix, some are positive and some are negative; we assume most are negative. Diagonalizing, and multiplying by the scalar matrix $\tfrac1\rho$ we get an element in the same projective conjugacy class of the form $t_r$ for some $1 \leq r \leq \tfrac n2$. We can explicitly compute the projective centralizer of $t_r$ to be all block $2\times 2$ matrices $g$ with $g t_r = ( \mu I_n) \cdot t_r g$. We easily get that $g_{11}=\mu  g_{11}$, $g_{12} = -\mu g_{12}$, $g_{21}=-\mu g_{21}$, and $g_{22}=\mu g_{22}$. If $g_{11} \neq 0$, then $\mu=1$, $g_{12}=0$, $g_{21}=0$, $g = \begin{bmatrix} z_{11} & 0 \\ 0 & z_{22} \end{bmatrix} \in \operatorname{GL}_r(F) \times \operatorname{GL}_{n-r}(F)$ (and conversely any such matrix commutes). If $\mu \neq 1$, then $g_{11}=0$, $g_{22}=0$, and for $g$ to be invertible $\mu =-1$, but also $g_{12}$ and $g_{21}$ must be square ($g$ would have rank $2\min(r,n-r)$, so $r=\frac n2$). This case is exactly $t_k$, and these extra matrices are all products of $\operatorname{GL}_k(F) \times \operatorname{GL}_k(F)$ with the permutation matrix $\begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix}$, hence they all live in $\operatorname{GL}_k(F) \wr \operatorname{Sym}(2)$ (and conversely every such matrix is in the projective centralizer). This concludes the case where $\lambda$ is a square. If $\lambda$ is not a square, then multiplying by a scalar matrix $\mu I_n$ we can replace $\lambda$ with any element of $\lambda (F^\times)^2$. Also such a matrix has minimal polynomial the irreducible $x^2-\lambda$, and so is conjugate to $t_k'$. Hence $t$ is in the same projective conjugacy class as $t_k'$ (no matter the choice of $\lambda$). The centralizer again follows form a simple calculation, and is the union of the cases $\mu=1$, which gives $\begin{bmatrix} A & B \\ \lambda B & A \end{bmatrix}$ which is equivalent to $A+\sqrt{\lambda} B \in \operatorname{GL}_k(F[\sqrt{\lambda}])$, and the case $\mu=-1$, which gives a matrix of the former type multiplied by $\sigma = \begin{bmatrix} I_k & 0_k \\ 0_k & - I_k \end{bmatrix}$, which acts on the matrices by conjugation to take $A+\sqrt{\lambda}B$ to $A-\sqrt{\lambda}B$, that is, $\sigma$ is the non-identity element of $\operatorname{Gal}(F[\sqrt{\lambda}]/F)$. $\square$
Proposition (PSL): If $F$ is a finite field of size $q$, $q$ odd, then $\operatorname{SL}_n(F)$ has exactly the following projective conjugacy classes of projective involutions with projective centralizers: $t_r$ (including $t_k$ if applicable) if $r$ is even or $(q-1)_2 > n_2$, $t_k'$ if $(q-1)_2 < \max( n_2,2)$.
Proof (sketch): The elements of the projective conjugacy class of $t_r$ (including $t_k$ if applicable) have determinant $(-1)^{n-r} \mu^n$ for $\mu \in F^\times$. Hence we just need to know if $\mu^n - (-1)^{n-r}$ has a root in $F$. In $F^\times$, $-1$ is the unique element of order 2, so if $(q-1)_2 > n_2$, then $-1$ is an $n$th power, and we can find $\mu$ no matter what $r$ is. Otherwise, $n$ must be even, so if $r$ is even we can take $\mu=1$. Otherwise, $(-1)^{n-r}=-1$ is not an $n$th power, no such $\mu$ exists, and the projective conjugacy class of $t_r$ does not intersect $\operatorname{SL}_n(F)$. The matrices in the projective conjugacy class of $t_k'$ have determinant $(-\lambda)^k / \mu^n$ so we just need to know if $(\mu^k)^2 - (-\lambda)^k$ has a root. If $(q-1)_2 = 1$, then $-1$ is not a square, so $-\lambda=\mu^2$ is a square (a product of two non-residues), and that $\mu$ is a root of the polynomial. If $n_2 > (q-1)_2$, then $k_2 \geq (q-1)_2$, so setting $k=2^i j$, $(-\lambda)^{2^i}$ has odd order, so it has a $2^{i+1}$st root, $\mu$. Then $\mu^{2k} = (\mu^{2^{i+1}})^j = ((-\lambda)^{2^i})^j = (-\lambda)^k$ as desired. If $2\leq n_2 \leq (q-1)_2$, then $-\lambda$ is not a square, so if $k=2^i j$, then $(-\lambda)^{2^i}$ has even order, so does not have a $2^{i+1}$st root (the order of which would be divisible by $2^{i+1}$ contrary to $2^i$ being the highest power of two dividing $q-1$). Hence there is no such $\mu$, and the projective conjugacy class of $t_k'$ does not intersect $\operatorname{SL}_n(F)$. $\square$
The structure of the centralizers is basically easy: intersect with determinant 1 (which produces messy tops) and mod out by the center of $\operatorname{SL}_n(F)$ (which produces messy bottoms). The result is proposition 6.5.2 on page 424.
Suzuki works through the cases of the other classical groups, and the results are similar: in each case the centralizers of the involutions have a generalized Fitting subgroup that is (modulo a central-in-the-generalized-Fitting subgroup of odd order) a semi-simple group of Lie type in the same characteristic, with a fairly obvious relation to the diagram of the original group.
Suzuki also handles the characteristic 2 involutions where the results are very different. Here the centralizers have a generalized Fitting subgroup that is a normal 2-subgroup (of nilpotency class 1 or 2) that is self-centralizing. The quotient then has a generalized Fitting subgroup of the expected type. Basically you get things similar to parabolic subgroups.
I didn't find a detailed discussion like this for the exceptional groups. I believe Aschbacher's Classical Involution Theorem is one way to do this, but I have not finished absorbing it.