# Centralizer of involutions in simple groups.

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,

1. How many conjugacy classes of involution does it have?

2. If there are more than one, what is their sizes?

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

• Carter's book (finite groups of Lie type) might have something about this, though I don't have it handy here to check. I know that at least it has information about the number of conjugacy classes with elements whose order is not divisible by the characteristic of the field, which is a related but possibly also easier question than your. – Tobias Kildetoft Feb 5 '14 at 8:51
• If you are interested, I've now got a very good handle on centralizers of involutions in classical groups. For the "similarity" groups (PGL, PCSp, PGU, PGO), I have nice exact formulas for everything. For the smaller simple groups, there are lots of details that depend on specific properties of $q$. I suspect I could program the conditions, but I don't think it would look very pretty written out. – Jack Schmidt May 23 '14 at 22:26

## 2 Answers

There is a table in GLS vol 3, page 172, but I haven't too much luck getting the information out of it. In the low rank cases that I've tried, its been hard to match things up nicely. Also I don't think the table is enough to compute the exact size.

## Example of using the GLS table

For example: $\operatorname{Aut}(\operatorname{PSL}(3,5)) = A_2(5)$ has involutions $t_1$ and $\gamma_1$ (every other row in the table does not apply). The Out-class of $t_1$ is determined by the 2-adic valuation of $q-\epsilon$ where $\epsilon=1$ indicates PSL instead of PSU. The 2-adic valuation of $q-\epsilon=5-1=4$ is 2 which is larger than the 2-adic valuation of $n=3$ (which is 0). Hence the Out-class of $t_1$ is "1", inner, as in, $t_1$ is in PSL itself, not just Aut(PSL). The Out-class of $\gamma_1$ is determined by the residue of $m$ and $q$ mod 4: now $m\not\equiv -1 \mod 4$ does not apply, but $q\equiv \epsilon \mod 4$ does ($5\equiv 1 \mod 4$), so we are in the first case "g", and $\gamma_1$ is the graph automorphism $x \mapsto x^{-T}$. In particular, it is not inner, so we probably don't care about it.

Now the centralizer of $t_1$ has $5'$-residual $A^\epsilon_{m-1}(q) = A_1(5)$ -- the version is "u" for universal, which I believe means $\operatorname{SL}(2,5)$. It may also have extra factors of $2$ associated with it. In this example, it has one such factor, the centralizer is $\operatorname{GL}(2,5)$. (In unrelated examples I've done, these extra factors can be predicted, but are complicated; if one chooses the "right" group instead of the simple one, then everything is fine.) I believe no extra complications arise in this particular example. At any rate, $\operatorname{PSL}(3,5)$ contains exactly one class of involutions, and the centralizer of an involution is isomorphic to $\operatorname{GL}(2,5)$.

## Machine calculations

Here are some explicit results for low rank groups over small fields of odd characteristic: Some of the groups have a single class of involution, but typically the number of classes of involutions is about half the Lie rank for the large rank groups. I list each class on its own line. Not all classes exist in all fields. The table in the book also includes classes in the automorphism group. I've been able to accurately count how many classes are in the simple group from the table, but the centralizer structures have not matched up for me.

• @ jack Schmidt: At the first, I really thank you for your interesting answer. I saw the table (in reference), however I didn't read all the book, but I couldn't calculate myself! For example in $PSL(3,5)=A_2(5)$, according the table centralizer of involution is $A_1(5)$, that its size is 60. But this is wrong! Would you please explain for me how I should use the table. Thanks a lot again. – Adeleh Feb 6 '14 at 5:39
• @ jack Schmidt: Thanks for your explanation. – Adeleh Feb 7 '14 at 10:09

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.

• For classical groups, the $t_r$ have centralizers that are (conformal) classical groups of the same type, $t_k'$ is basically the same, and there is one other type $t_k^o$ (totally isotropic) whose centralizer is always GL (instead of whatever classical group is in play). Things are even clearer over algebraically closed fields where the $t_k'$ case does not occur. – Jack Schmidt May 30 '14 at 1:35
• Thanks a lot. I need a little time to understand it. – Adeleh Jun 12 '14 at 6:07