2
$\begingroup$

According to Table 4.5.1 of [1], there should be 10 classes of involutions of type "p" and "e" in $\operatorname{Aut}(K)$ where $K=K_a=P\Omega^{+}(8,3)$. And Table 4.5.1 also gives the centralizer structure of each involution class in $\operatorname{Inndiag}(K)$. I understand that here $\operatorname{Inndiag}(K)=\operatorname{PCSO}^{+}(8,3)$ where $\operatorname{CSO}^{+}(8,3)$ is the conformal special orthogonal group.

My reading of Table 4.5.1 for the centralizer structure: $$\begin{array}{cc} \text{class} & \text{order} \\ \Omega^{+}(6,3).2.2.2 & 48522240 \\ \Omega^{-}(6,3).4.2^2.2 & 209018880 \\ \Omega^{+}(4,3)^2/2.2.(2^2\times 2).2^2 & 2654208 \\ \Omega^{-}(4,3)^2.2.(2^2\times 2).2^2 & 8294400 \\ \text{$P\Omega^{+}(4,9).2.2.2^2$ (two copies)} & 2073600 \\ \text{$\operatorname{SL}(4,3)/2.2.2.2$ (two copies)} & 48522240 \\ \text{$\operatorname{SU}(4,3)/2.4.4.2$ (two copies)} & 209018880 \end{array}$$

Note that the second to last component is the group "Outdiag", to use the notation of [1].

I did some computation using Magma in $\operatorname{PCSO}^{+}(8,3)$. There are 10 classes of involution indeed. But the centralizer orders don't match completely. Which part of my understanding is not correct?

[1] Gorenstein, Daniel, Richard Lyons, and Ronald Solomon. Classification of the Finite Simple Groups, Number 3. 1st ed. Vol. 40. Providence: American Mathematical Society, 1997. Print.

Edit: Computation results: involution centralizers and their orders

[ C4.2A(3,3).D4, C4.2A(3,3).D4, C4.2A(3,3).D4, C2*PSL(4,3).C2^2, C2*PSL(4,3).C2^2, C2*PSL(4,3).C2^2, C2.A6^2.C2^3, C2.A6^2.C2^3, C2.A6^2.C2^3, C2.C2^8.C3^4.C2.C2^4 ]

[ 104509440, 104509440, 104509440, 48522240, 48522240, 48522240, 2073600, 2073600, 2073600, 1327104 ]
$\endgroup$
8
  • $\begingroup$ It seems hard to answer the question without knowing your results. Could you summarise them? $\endgroup$
    – LSpice
    Commented Jun 9 at 13:58
  • $\begingroup$ Yes, yes. Do you mean the computation results? thank you $\endgroup$
    – scsnm
    Commented Jun 9 at 22:22
  • $\begingroup$ Re, yes, if you have specific expected results and actual results, then it would probably help to post them. $\endgroup$
    – LSpice
    Commented Jun 9 at 22:45
  • 1
    $\begingroup$ Many thxs for the editing $\endgroup$
    – scsnm
    Commented Jun 9 at 23:10
  • 1
    $\begingroup$ I suspect that you have the wrong group. If G is the simple group then the multiplier of $G$ is $2\times 2$ while the outer automorphism group is $S_4$. The outer automorphisms act on the $2\times 2$ through the quotient $S_3$. So any particular double cover of $G$ has fewer automorphisms than $G$. And even the double covers that admit involutary automorphisms don't necessarily fit together to make a group of the form $2.G.2$. So my suspicion is that the group is twice as big as you think, but without a lot of work trying to reconcile your notation with GLS, it's hard to comment further. $\endgroup$ Commented Jun 10 at 11:08

0

You must log in to answer this question.