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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 in $Z(\Omega_{2m}^{\epsilon}(q))$ 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:

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

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.

$P\Omega_{2m}^+(q)=D_m(q)$ 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 $C_{(4, q^m+1)}$.

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

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

Let $P$ be a $2$-group of cyclic center $\langle a \rangle$ and let $\omega_n(P)=P\wr C_2\wr C_2\cdot\cdot\cdot \wr C_2$ be the wreath product of $P$ and $n$ copies of $C_2$, where $n\geq 2$.

Suppose that $2^{t+1}$ is the $2$-part of $q^2-1$. Let $T$ be a central product of two dihedral groups

$$D_1=\langle d, h: h^{-1}dh=d^{-1}\rangle$$

and

$$D_2=\langle g, k: k^{-1}gk=g^{-1}\rangle$$

of order $2^{t+1}(d^{2^{t-1}}=g^{2^{t-1}})$ and let $e, f\in {\rm Aut}T$ be chosen such that

$$o(e)=o(f)=2, [e,f]=1,$$

$$d^e=g^{-1}, g^e=d^{-1}, h^e=gk, k^e=dh, d^f=g, g^f=d, h^f=k, k^f=h.$$

The twisted wreath product $tw_1(C)$ of $T$ and $C$, where

$$C=\langle \left( \begin{array}{cc} e_2 & 0 \\ 0 & e_2 \\ \end{array} \right), \left( \begin{array}{cc} f_2 & 0 \\ 0 & f_2 \\ \end{array} \right), \left( \begin{array}{cc} 0 & I_2 \\ I_2 & 0 \\ \end{array} \right) \rangle$$

($e_2$ and $f_2$ are $2\times 2$ diagonal matrices of the forms ${\rm diag}(e,1)$ and ${\rm diag}(f,1)$), is the group

$$tw_1(T)=\langle \pmatrix{T & 0\\ 0 & I_2}, \pmatrix{I_2 & 0\\ 0 & T}, \pmatrix{e_2 & 0\\ 0 & e_2}, \pmatrix{f_2 & 0\\ 0 & f_2}, \pmatrix{0 & I_2\\ I_2 & 0}\rangle.$$

Note that $C$ is elementary Abelian of order 8. In general, $tw_{n+1}$ (the twisted wreath product of $T$ and $n+1$ copies of $C$) is generated by

$$\langle U=\pmatrix{tw_n(T) & 0\\ 0 & I_{2^n}}, V=\pmatrix{I_{2^n} & 0\\ 0 & tw_n(T)}\rangle\cong tw_n(T)\times tw_n(T)$$

and

$$\langle \pmatrix{e_{2^n} & 0\\ 0 & e_{2^n}}, \pmatrix{f^{2^n} & 0\\ 0 & f^{2^n}}, \pmatrix{0 & I_{2^n}\\ I_{2^n} & 0}\rangle\cong C$$

where $e_{2^n}$ and $f_{2^n}$ are $2^n\times 2^n$ diagonal matrices of the form

$${\rm diag}(e, 1, ..., 1)$$

and

$${\rm diag}(f, 1, ..., 1).$$

Let $z$ be the generator of the center of $T$ and let $$E=\prod_{a\in tw_{n+1}(T)}\langle z^a\rangle.$$ Then $E$ is elementary Abelian of order $2^{n+1}$. Suppose that $E=\prod\langle z_i\rangle$ (direct product). Then $z_0=\prod z_i$ generates the center of $tw_{n+1}(T)$.

$\omega_{n-2}(T)$ is a Sylow $2$-subgroup of $\Omega^{\epsilon}(2^n,q)$. Further, $\omega_{n-2}(T)/Z$, where $Z=\langle z_0\rangle$, is a Sylow 2-subgroup of $P\Omega^\epsilon(2^n,q)$.

Let $S$ be a Sylow $2$-subgroup of $P\Omega^\epsilon(2m,q)$, where $m\geq 4$, $q$ is a power of an odd prime and $q^m\equiv \epsilon (\rm mod~4)$.

Let $F_q$ be the field of $q$ elements. Let $\Phi_1$ be the determinant mapping and $\Phi_2$ be the spinorial norm mapping $\Phi_2: O^\epsilon\rightarrow F_q^\times/{F_q^\times}^2\cong C_2$. It is clear that

$${\rm ker \Phi_1}\cap {\rm ker \Phi_2}=P\Omega^\epsilon(2m,q)$$

Let $2m=2^{r_1}+2^{r_2}+...+2^{r_t}$. Then $T=W_{r_1}\times W_{r_2}\times ...\times W_{r_k}$ is a Sylow 2-subgroup of $O^\epsilon(2m,q)$.

Denote by $Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

$T=S'W_{r_i}$ for all $i$.

$Lie(r)$ is the set of finite groups possessing a $\sigma$-setup $(\bar{K},\sigma)$ over $\bar{F}_r$ such that $\bar{K}$ is simple. Furthermore, $$Lie=\bigcup_r Lie(r),~~~{\rm the~union~over~all~primes}~r$$

If $\sum=D_{2m}$, then $Z(\bar{K}_u)$ is $\bar{F}^{(2)}\times \bar{F}^{(2)}$.

If $\sum=D_{2m}$, then the generators of $Z({\bar{K}})$ are $h_1=h_{\alpha_1}(-1)h_{\alpha_3}(-1)...h_{\alpha_{2m-1}}(-1)$ and $h_2=h_{\alpha_{2m-1}}(-1)h_{\alpha_{2m}}(-1)$.

Analysis:

Let

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

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.

The following websites may be useful to my question.

Double covers of the orthogonal groups

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

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    $\begingroup$ What is the ground field? Since you're specifying a quadratic form, I guess not $\mathbb C$. Is it $\mathbb R$? $\endgroup$
    – LSpice
    Mar 18, 2020 at 3:22
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    $\begingroup$ The answer to B is "not necessarily." The smallest counterexample I know is $\Omega_6^-(3)$, which has a center $Z$ of order $2$. In the group $P\Omega_6^-(3)=\Omega_6^-(3)/Z$, there is a single conjugacy class of involutions. The inverse image of any of these involutions is a pair of involutions in $\Omega_6^-(3)$. You can find these facts in the {\it Atlas of Finite Groups} by Conway et al. Note that $P\Omega_6^-(3)\cong PSU_4(3)$. $\endgroup$ Mar 19, 2020 at 22:29
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    $\begingroup$ 39 versions of this question! $\endgroup$ May 7, 2020 at 1:53
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    $\begingroup$ I’m voting to close this question because 57 insignificant edit is an incorrect way to use MO. $\endgroup$
    – abx
    May 14, 2020 at 3:50
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    $\begingroup$ @DerekHolt: The question is reasonable, the OP's use of the MO resources is not. Worse, even after abx's remarks about the edits, the OP persisted in his behaviour. There were 57 edits noted by abx, now there are 59. Plus, there is something else wrong with the question: with all these edits, it has become a moving target. Not voting to close, but flagging to the attention of a moderator. $\endgroup$
    – Alex M.
    May 14, 2020 at 7:55

2 Answers 2

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Since nobody has answered Question A, I computed a few examples with small dimensions (up to $14$), and small finite fields (up to order 9, depending on the dimension) in Magma.

The results were consistent and can be summed up as follows. All generators of $Z({\rm Spin}^\epsilon_n(q))$ of order $2$ are squares of elements in ${\rm Spin}^\epsilon_n(q)$. But generators of order $4$ are not. That is also consistent with the negative answer to Question B.

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  • $\begingroup$ Thanks a lot! Is what you mean when $Z({\rm Spin}_n^\epsilon(q))\cong C_4$, then, for question A, all geneartors of order 4 are not squares, and all generators of order 2 are yes, and for question $B$, the answer is also negative; when $Z({\rm Spin}_n^\epsilon(q))\cong C_2\times C_2$, then for question A, all generators of order 2 are squares, and for question B, the answer is also yes? $\endgroup$
    – Yi Wang
    Jun 3, 2020 at 3:58
  • $\begingroup$ Yes, you seem to have just repeated what I said in my answer. $\endgroup$
    – Derek Holt
    Jun 3, 2020 at 7:29
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With the help of both Professors Robert Guralnick and Frank Lübeck, I get following two answers extracted from their reply to my email.

  1. The first one due to Professor Robert Guralnick.

Suppose $x^2=-1$. Then eigenvalues of $x$ are $\pm i$ (with the same multiplicity). Then we can decompose the space into an orthogonal sum of $x$-invariant two dimensional subspaces with $x^2=-1$ on each.

If $m$ is even, then $x$ has spinor norm 1 in any case. This determines $\epsilon=+$.

If $q\equiv 1~{\rm mod}~4$ and $m$ is odd, these are each of $+$ type. In ${\rm SO}(2,q)^+$, the torus has order $q-1$, so we see -1 is a square. Now compute the spinor norm of $x$, if $q\equiv 1~{\rm mod}~8$, then $x$ is a square in ${\rm SO}(2)$ and so has spinor norm 1.

If $q\equiv 3~{\rm mod}~4$ and $m$ is odd, these are each of $-$ type. Then the torus on ${\rm SO}^{-}_2$ has order $q+1$. If $q\equiv -1~{\rm mod}~8$, then again we have $x$ with spinor norm 1.

  1. The second one due to Professor Frank Lübeck.

Problem. Compute $$MP(H,I)=\{h\in H||h|~{\rm is~4}, h^{|h|/2}\in I\},$$ where $I$ is consisting of an involution in the center of $H$, when $H=\Omega_{2m}^\epsilon(q)$ or $H={\rm Spin}_{2m}^\epsilon(q)$ for $q^m\equiv \epsilon~{\rm mod}~4$.

Solution. For $H={\rm Spin}_{2m}^\epsilon(q)$ the set $MP(H,I)$ is not empty for any $I$ (it is obvious for $n$ odd because the center is cyclic of order 4).

For $H=\Omega_{2n}^+(q)$, there is only one $I$ and the set $MP(H,I)$ is not empty iff $n$ is even or $n$ is odd and $q\equiv 1~{\rm mod}~8$.

For $H=\Omega_{2n}^-(q)$, there is only one $I$ and the set $MP(H,I)$ is not empty iff $n$ is odd and $q\equiv -1~{\rm mod}~8$.

For a proof we can use that an element of order 4 ($q$ odd) is contained in a maximal torus $T$ (which is isomorphic to a direct product of copies of the multiplicative group of the field $k$ (an algebraic closure of $F_q$).

The element in $I$ is contained in $T$ and there are always elements of order 4 squaring to this involution, this is an element of ${\rm Spin}_{2m}^\epsilon(k)$ or ${\rm SO}_{2m}^\epsilon(k)$, respectively. The question is now if such an element of order 4 is conjugate to an element in the finite group ${\rm Spin}_{2m}^\epsilon(q)$ or $\Omega_{2m}^\epsilon(q)$.

The Spin case should not be so difficult, but Omega is more difficult. One approach would be to consider $\Omega_{2m}^\epsilon(q)$ as the image of ${\rm Spin}_{2m}^\epsilon(q)$ under the projection map ${\rm Spin}_{2m}^\epsilon(k)\rightarrow {\rm SO}_{2m}(k)$. The case $\Omega_{2m}$ with even $m$ then follows from the Spin case. But for odd $m$ one has to argue with elements of order 8 in Spin.

For computing with elements in maximal tori one could refer to section 2 of the following paper https://arxiv.org/abs/1211.3692

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