Is $1\neq a\in Z(\Omega_6^{+}(5))\cong C_2$ a square element in $\Omega_6^{+}(5)$?

Question

Is $1\neq a\in Z(\Omega_6^{+}(5))\cong C_2$ a square element in $\Omega_6^{+}(5)$?

In other words, I wonder if there is an element $x\in S$ such that $x^2=a$ and forthermore $x\in \Omega_6^+(5)$, where $S$ is a Sylow 2-subgroup of the orthogonal group ${\rm GO}_6^+(5)$?

A. 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 $\theta(\sigma)=N(x)F^{\ast 2}$ 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 $\alpha_V=1_V$.

B. Firstly, we suppose that $n=2m$ and $q^m\equiv 1~(\rm mod~4)$, then the following are some quoted and inferred results:

1. ${\rm GO}_{2m}^+(q)=\{T^tA_1T=A_1, T\in {\rm GL}(2m,q)\}$, where $A_1=\left( \begin{array}{cc} 0 & I_m \\ I_m & 0 \\ \end{array} \right) $.

2. Let $S$ be a Sylow $2$-subgroup of ${\rm GO}_{2m}^{+}(q)$, then $S$ is isomorphic to a Sylow $2$-subgroup of ${\rm GO}_{2m+1}^{+}(q)$ by page 147 of 'The sylow 2-subgroups of the finite classical groups' of Roger Carter and Paul Fong.

3. Let $S$ be a Sylow $2$-subgroup of ${GO}_{2m+1}^{+}(q)$ and let $$2m=2^{r_1}+2^{r_2}+...+2^{r_t},~~~r_1<r_2<...<r_t.$$ Then $S\cong W_{r_1}\times W_{r_2}\times ... \times W_{r_t}$, where $W_r$ is a Sylow $2$-subgroup of ${\rm GO}_{2^r+1}^{+}(q)$ by page 146 of 'The sylow 2-subgroups of the finite classical groups' of Roger Carter and Paul Fong.

4. Let $W$ be a Sylow 2-subgroup of ${\rm GO}_3^+(q)$, $W_r=W\wr T_{r-1}$. Then $W_r$ is a Sylow $2$-subgroup of ${\rm GO}_{2^r+1}^+(q)$, where $T_{r-1}=C_2\wr C_2\wr ... \wr C_2$ is the wreath product of $C_2$ $r-1$ times by page 146 of 'The sylow 2-subgroups of the finite classical groups' of Roger Carter and Paul Fong.

5. The orthogonal group ${\rm GO}_2^{+}(q)$ is a dihedral group of order $2(q-1)$ by page 146 of 'The sylow 2-subgroups of the finite classical groups' of Roger Carter and Paul Fong.

6. ${\rm P\Omega}_6^+(q)\cong {\rm PSL}_4(q)$ by page 96 of 'the finite simple groups' of Robert A. Wilson.

7. $\Omega_n^+(q)$ has a centre of order 2 by page 80 of 'the finite simple groups' of Robert A. Wilson.

8. When $n=2m$ is even and $\epsilon=+$, then $|{\rm P\Omega}_n^+(q)|=N/d,$ where $$N=q^{m(m-1)}(q^m-1)(q^{2m-2}-1)(q^{2m-4}-1)...(q^2-1)$$ and $d=(4,q^m-1)$. Therefore $|{\rm P\Omega}_6^+(5)|=2^7\cdot 3^2\cdot 5^6\cdot 13\cdot 31$.

Then in particular, for $\Omega_6^{+}(5)$, we have $S\cong W_1\times W_2=(W\wr T_{1-1})\times (W\wr T_{2-1})=W\times (W\wr C_2)=D_4\times (D_4\wr C_2),$ where $W\cong D_4$ is a Sylow 2-subgroup of ${\rm GO}_3^+(5)$ or ${\rm GO}_2^{+}(5)$.

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

Analysis. If $a\in \Omega_6^+(5)$,

In general, for ${\rm GO}_{2m}^+(q)$, we have

$S\cong W_{r_1}\times ...\times W_{r_t}=(W\wr T_{r_1-1})\times ...\times (W\wr T_{r_t-1})=[D_{q-1}\wr (C_2\wr C_2\wr ...\wr C_2)_{r_1-1}]\times ...\times [D_{q-1}\wr (C_2\wr C_2\wr ...\wr C_2)_{r_t-1}],$

where $W=D_{q-1}$ is a Sylow $2$-subgroup of ${\rm GO}_2^+(q)$ or ${\rm GO}_3^+(q)$ and

$$W_{r_i}=W\wr T_{r_i-1}=[D_{q-1}\wr (C_2\wr C_2\wr ...\wr C_2)_{r_i-1}]$$

is a Sylow $2$-subgroup of ${\rm GO}_{2^{r_i}}^{+}(q)$ or ${\rm GO}_{2^{r_i}+1}^{+}(q)$.