3
$\begingroup$

The outer automorphism group of the Suzuki simple group ${}^2B_2 (2^{2m+1})$, $m \geq 1$ is cyclic of order $2m+1$ and is generated by a field automorphism $\varphi$ of order $2m+1$. For any almost-simple group $S \leq H \leq {\rm Aut}(S)$ with $S={}^2B_2 (2^{2m+1})$, the group $H/S$ is cyclic. I would like to know

  1. the action of $\varphi$ on the conjugacy classes of the group $S$, and

  2. the set of complex character degrees of the group $H$.

Thanks for your help.

$\endgroup$
  • 1
    $\begingroup$ For insight into this kind of computation, it's probably useful to look at the small cases Sz(8) and especially Sz(32) included in the Atlas of Finite Groups. For example, $S=$Sz(32) involves $m=2$, while $\varphi$ has order 5; here the Atlas seems to answer both of your questions, if you unpack the notational conventions. But in general what you are asking for probably gets much longer computationally even though the character degrees for the Suzuki groups are worked out in Suzuki's original papers. (Those papers may also be worth consulting.) $\endgroup$ – Jim Humphreys Oct 11 '15 at 19:34
3
$\begingroup$

The action of $\varphi$ on the conjugacy classes of the Suzuki groups ${\rm Sz}(2^{2m+1}) = {}^2B_2 (2^{2m+1})$ is described by Lemma 3.3 in:

Further, denoting by $\omega(G)$ the number of orbits on a group $G$ under the action of its automorphism group, Theorem 3.4 in that paper says that we always have $$ \omega({}^2B_2 (2^{2m+1})) \ = \ \omega({\rm PSL}(2,2^{2m+1})) + 2. $$ A graphical illustration of the correspondence between the orbits on ${}^2B_2 (2^{2m+1})$ and the ones on ${\rm PSL}(2,2^{2m+1})$ is shown in Figure 1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.