Skip to main content
43 of 59
added 122 characters in body
Yi Wang
  • 271
  • 1
  • 7

Is every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ a square element in ${\rm Spin}_n^{\epsilon}(q)$?

A. I wonder if every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ is a square element in ${\rm Spin}_n^{\epsilon}(q)$?

B. When $Z(\Omega_{2m}^{\epsilon}(q))\cong C_2$, is the unique element of order two of $Z(\Omega_{2m}^{\epsilon}(q))$ a square element in $\Omega_{2m}^{\epsilon}(q)$

Notes.

  1. Here the ground field is a finite field $F_q$ with $q$ a power of some prime $p$.

  2. We always set $n=2m\geq 6$ and $q^m\equiv \epsilon~({\rm mod}~4)$.

Some quoted results:

It is easy to find elements of the spin group which square to $-1$, and hence the spin group is a proper double cover of the orthogonal group. We write ${\rm Spin}_n^\epsilon(q)$ for this group of shape $2.\Omega_n^\epsilon(q)$.

If $n$ is odd, or if $n=2m$ and $q^m\equiv -\epsilon~{\rm mod}~4$, then $\Omega_n^\epsilon(q)$ is already simple and the spin group has the structure $2.\Omega_n^\epsilon(q)$.

If $n=2m$ and $q^m\equiv \epsilon~({\rm mod}~4)$, then $\Omega_n^\epsilon(q)$ has a centre of order 2, and the spin group has the structure $4.{\rm P\Omega}_n^\epsilon (q)$ if $m$ is odd, and the structure $2^2.{\rm P\Omega}_n^\epsilon (q)$ (necessarily with $\epsilon=+$) if $m$ is even.

If $m\geq 3$, then $P\Omega^{\epsilon}(2m,q)$ is a finite simple group.

$P\Omega_{2m}^+(q)=D_m(q)$ for $m\geq 3$ and its Schur multiplier is $C_{(4, q^m-1)}$ if $m$ is odd and $C_{(2, q^m-1)}\times C_{(2,q^m-1)}$ if $m$ is even.

$P\Omega_{2m}^{-}(q)={}^2D_m(q)$ for $m\geq 2$ and its Schur mulitiplier is $C_{(4, q^m+1)}$.

$P\Omega_6^+(q)=PSL_4(q)$.

$P\Omega_6^-(q)=PSU_4(q)$.

Analysis:

  1. $\pi: \Omega_6^-(3)\rightarrow P\Omega_6^-(3)$.

If $x^2=-1$, then $f(x)$ is an element of order 2, however $o(x)=4$, a contradiction by Richard Lyons's notes below.

  1. Let $S$ be a Sylow $2$-subgroup of ${\rm GO}_{2m}^{+}(q)$, we have $S\cong D_4\times (D_4\wr C_2)$, when $m=3$, $q=5$.

Therefore $Z(S)\cong C_2\times C_2$. Hence there is an element $a\in S$ such that $a^2=-1$ by the structure of $D_4$, Now however is $a\in \Omega_6^+(5)$?

The following websites may be useful to my question.

Double covers of the orthogonal groups

http://brauer.maths.qmul.ac.uk/Atlas/v3/

About the number of their conjugacy classes in some classes of finite simple groups

https://math.stackexchange.com/questions/3567274/is-every-generator-of-z-rm-spin-n-epsilonq-a-square-element-in-finite

Yi Wang
  • 271
  • 1
  • 7