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 $\Omega_{2m}^{\epsilon}(q)$？

C. $SO(V)\leq \Gamma^0$, ${\rm Spin}(V)\leq \Gamma^0$, then what are $SO(V)\cap {\rm Spin}(V)$ and $\Omega(V)\cap {\rm Spin}(V)$?


I. The following notations and consequces are followed from page 78 of 'Classical groups and geometric algebra' by Larry C. Grove.

1. Let us sketch an alternative approach to $\theta|_{SO(V)}$. Define the Clifford group $\Gamma=\Gamma(V)$ to be the normalizer in the group $U(C)$ of units in $C$ of the subspace $V$ of $C$, i.e.
$$\Gamma=\{x\in U(C): xvx^{-1}\in V, {\rm all}~\in V\}.$$
Define the even Clifford group to be $\Gamma_0=\Gamma\cap C_0$. If $x\in V$ is nonzero and anisotropic, and $v\in V$, then $xvx^{-1}=-\sigma_xv\in V$ by Proposition 8.3, so $O(V)\leq \Gamma$. It follows easily that $SO(V)\leq \Gamma_0$.

2. If $x\in \Gamma$ and $v\in V$ define $\chi_x(v)=xvx^{-1}\in V$. Then 
$$Q(\chi_xv)=Q(xvx^{-1})=(xvx^{-1})^2=xv^2x^{-1}=xQ(v)x^{-1}=Q(v),$$
so $\chi(x)\in O(V)$. In fact $\chi$ is a homomorphism from $\Gamma$ into $O(V)$, called the vector representation of $\Gamma$. It can be shown that $\chi$ maps $\Gamma_0$ onto $SO(V)$, with ${\rm Ker}(\chi|{\Gamma_0})=F^\ast$.

3. If $\alpha$ is the unique anti-automorphism of $C$ with $\alpha_V=1_V$ discussed in Proposition 8.15, then we may define a homomorphism $N: \Gamma_0\rightarrow F^{\ast}$ via $N(x)=\alpha(x)x$. The kernel of $N$ is called the spin group ${\rm Spin}(V)$. Then $\chi({\rm Spin}(V))=\mho(V)$, the reduced orthogonal group. If $\sigma\in SO(V)$, then there exists $x\in \Gamma_0$ with $\chi(x)=\sigma$, and then $\theta(\sigma)=N(x)F^{\times 2}$ gives the spinor norm of $\sigma$.

The following consequence follows form page 241 of 'Basic algebra II' by Nathan Jacobson. 

4. Theorem 4.16. Let $Q$ be of positive Witt index. Then the reduced orthogonal group $O'(V,Q)$ coincides with the commutator subgroup $\Omega$ of $O(V,Q)$ and $$O^+(V,Q)/{O'(V,Q)}\cong F^{\ast}/{F^{\ast 2}}.$$

The following consequence follows from page 392 of 'Basic algebra I' by Nathan Jacobson. 

5. In the case of a finite field the Witt index is always positive if $n\geq 3$.

The following notation follows form page 359 of 'Basic algebra I' by Nathan Jacobson. 

6. If $B$ is a non-degenerate symmetric bilinear form, then $B$ is called isotropic or a null form if there exists a vector $u\neq 0$ such that $B(u,u)=0$. Such a vector is called isotropic.

The following is Proposition 8.15 (page 73) of 'Classical groups and geometric algebra' by Larry C. Grove. 

7. Proposition 8.15. If $V$ is a quadratic space there is a unique anti-automorphism $\alpha: C(V)\rightarrow C(V)$ with $\alpha_V=1_V$.

II. By page 80 of 'the finite simple groups' of Robert A. Wilson, we have the following the two results:

1. 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 $2.{\Omega_n^\epsilon}(q)$.

2. 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 $\epsilon=+$) is $m$ is even.

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

The following problem and its solution may be useful to my question.

https://mathoverflow.net/questions/140568/double-covers-of-the-orthogonal-groups#