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.
Here the ground field is a finite field $F_q$ with $q$ a power of some prime $p$.
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 $C_{(4, q^m+1)}$.
$P\Omega_6^+(q)=PSL_4(q)$.
$P\Omega_6^-(q)=PSU_4(q)$.
Suppose that $2^{t+1}$ is the $2$-part of $q^2-1$. Let $T$ be a central product of two dihedral groups $D_1=\langle d, h: h^{-1}dh=d^{-1}$ and $D_2=\langle g, k: k^{-1}gk=g^{-1}$ 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 $$C=\langle {(e_2, 0; 0, e_2)}, (f_2, 0; 0, f_2), (0, I_2; I_2, 0)\rangle$$ ($e_2$ and $f_2$ are $2\times 2$ diagonal matrices of the forms ${\rm diag}(e,1)$ and ${\rm diag}(f,1)$), is the group $$tw_1(T)=\langle (T, 0; 0, I_2), (I_2, 0; 0, T), {(e_2, 0; 0, e_2)}, (f_2, 0; 0, f_2), (0, I_2; I_2, 0)\rangle.$$ 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 $$U=\langle (tw_n(T), 0; 0, I_{2^n}), V=(I_{2^n}, 0; 0, tw_n(T))\rangle \cong tw_n(T)\times tw_n(T)$$ and $$\langle (e_{2^n}, 0; 0, e^{2^n}), (f_{2^n}, 0; 0, f^{2^n}), (0, I_{2^n}; I_{2^n}, 0)\rangle \cong C$$ 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 $P\Omega^\epsilon(2^n,q)$.
Analysis:
- $\pi: \Omega_6^-(3)\rightarrow P\Omega_6^-(3)$.
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.
- 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