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

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 generator of $$Z(\Omega_{2m}^{\epsilon}(q))$$ a square element in $$\Omega_{2m}^{\epsilon}(q)$$

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

The following follows from 'simple groups of Lie type' by Roger W. Carter.

A Lie algebra is a vector space $$L$$ over a field $$K$$ on which a product operation $$[xy]$$ is defined satisfying the following axioms:

(i) $$[xy]$$ is bilinear for $$x, y\in L$$.

(ii) $$[xx]=0$$ for $$x\in L$$.

(iii) $$[[xy]z]+[[yz]x]+[[zx]y]=0$$ for $$x, y, z\in L$$.

for each element $$x$$ of a Lie algebra $$L$$ we define a map $${\rm ad}~x$$ of $$L$$ into itself by $${\rm ad}~x.y=[xy],~~~y\in L.$$

For each $$x,y\in L$$ we define the scalar product $$(x,y)=tr(ad~x.ad~y)$$, which is called the Killing form.

The dimension of the Cartan subalgebras $$H$$ of $$L$$ is called the rank of $$L$$, and will usually be denoted by $$l$$.

Although the roots are defined as elements of the dual space of $$H$$ they can, by considering the Killing form, be regarded as elements of $$H$$ itself.

Each element of the dual space of $$H$$ is expressible in the form $$h\rightarrow (x, h)$$ for a unique element $$x\in H$$. The element $$x$$ is associated with the map $$h\rightarrow r(h)$$ may be identified with the root $$r$$. Thus $$r$$ can be regarded either as an element of $$H$$ or an element of its dual space; the relation between these two being given by $$r(h)=(r, h),~~~h\in H.$$

We now define the Dynkin diagram of the Lie algebra $$L$$. This is a graph with $$l$$ nodes, one associated with each fundamental root $$p_i$$, such that the $$i$$th node is joined to the $$j$$th node by a bond of strength $$n_{ij}$$.

Let $$L$$ be a Lie algebra over a field of characteristic $$0$$ and $$\delta$$ be a derivation of $$L$$ which is nilpotent, i.e. satisfies $$\delta^n=0$$ for some $$n$$. Then $${\rm exp}~\delta=1+\delta+\frac{\delta^2}{2!}+...+\frac{\delta^{n-1}}{(n-1)!}$$ is an sutomorphism of $$L$$.

We write $$x_r(\zeta)={\rm exp}(\zeta ad~e_r)$$ for $$\zeta\in \mathbb{C}$$.

We shall write $$h_r$$ for $$\bar{h}_r$$, $$e_r$$ for $$\bar{e}_r$$, $$x_r(t)$$ for $$\bar{x}_r(t)$$, and $$A_r(t)$$ for $$\bar{A}_r(t)$$. This omission of the bars will not lead to confusion or inconsistency since the objects originaly called $$h_r$$, $$e_r$$, $$x_r(t)$$, $$A_r(t)$$ are special cases of $$\bar{h}_r$$, $$\bar{e}_r$$, $$\bar{x}_r(t)$$, $$\bar{A}_r(t)$$ when $$K=\mathbb{C}$$.

The Chevalley group of type $$L$$ over the field $$K$$, denoted by $$L(K)$$, is defined to be the group of automorphisms of the Lie algebra $$L_K$$ generated by the $$x_r(t)$$ for all $$r\in \Phi$$, $$t\in K$$.

We now consider the special case in which the base field $$K$$ is the finite field $$GF(q)$$ with $$q$$ elements, where $$q$$ is an arbitrary prime power. $$G$$ is then a group of non-singular linear transformations of a space over a finite field, so is a finite group. The Chevalley group of type $$L$$ over $$GF(q)$$ will be denoted by $$L(q)$$.

Theorem 12.1.1 Let $$L$$ be a simple Lie algebra with $$L\neq A_1$$ and let $$K$$ be a field. For each root $$r$$ of $$L$$ and each element $$t$$ of $$K$$ introduce a symbol $$\bar{x}_r(t)$$. Let $$\bar{G}$$ be the abstract group generated by the elements $$\bar{x}_r(t)$$ subject to relations $$\bar{x}_r(t_1)\bar{x}_r(t_1)=\bar{x}_r(t_1+t_2),$$ $$[\bar{x}_s(u),\bar{x}_r(t)]=\prod_{i,j>0}\bar{x}_{ir+js}(C_{ijrs}(-t)^iu^j),$$ $$\bar{h}_r(t_1)\bar{h}_r(t_2)=\bar{h}_r(t_1t_2),~~~t_1t_2\neq 0,$$ and $$\bar{n}_r(t)=\bar{x}_r(t)\bar{x}_{-r}(-t^{-1})\bar{x}_{r}{(t)}.$$ Let $$\bar{Z}$$ be the centre of $$\bar{G}$$. Then $$\bar{G}/{\bar{Z}}$$ is isomorphic to the Chevalley group $$G=L(K)$$.

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

A bilinear form on $$V$$ is a function $$B: V\times V\rightarrow F$$ that becomes a linear fouctional in either variable if the other variable is fixed.

Suppose that $$B$$ is a bilinear form on $$V$$. Say that $$B$$ is symmetric if $$B(v,w)=B(w,v)$$ for all $$v,w\in V$$.

Suppose that $$B_1,B_2$$ are bilinear forms on spaces $$V_1$$, $$V_2$$, respectively. An isometry relative to $$B_1$$ and $$B_2$$ is an $$F$$-isomorphism $$\sigma: V_1\rightarrow V_2$$ satisfying $$B_2(\sigma v,\sigma w)=B_1(v,w)$$ for all $$v, w\in V_1$$.

If $$B$$ is a symmetric form on $$V$$ define $$Q: V\rightarrow F$$ via $$Q(v)=B(v,v)$$, and call $$Q$$ the associated quadratic form.

Through this chapter we assume that $$V$$ is a quadratic space of dimension $$n\geq 2$$ over a filed $$F$$, with $${\rm char}~F\neq 2$$. The isometries of $$V$$ are called orthogonal (linear) transformations; they comprise the orthogonal group $$O(V)$$. Thus $$O(V)=\{\tau\in {\rm GL}(V): B(\tau u,\tau v)=B(u,v),~{\rm all} u,v\in V\}\leq {\rm GL}(V).$$ Equivalently $$\tau\in O(V)$$ if and only if $$Q(\tau v)=Q(v)$$, all $$v\in V.$$

If $${\rm det}~\tau=1$$ call $$\tau$$ a rotation, or a proper orthogonal transformation. If $${\rm det}~\tau=-1$$ call $$\tau$$ a reversion, or (more commonly) an improper orthogonal transformation.

The rotations in $$O(V)$$ form a subgroup which will be denoted $$SO(V)$$, called the special orthogonal group-it is often denoted by $$O^+(V)$$ or $$O_+(V)$$ in the leterature.

Let $$u\in V$$ be any nonzero anisotropic vector (i.e. $$Q(u)\neq 0$$) and define a linear transformation $$\sigma_u$$ via $$\sigma_u(v)=v-2\frac{B(v,u)}{Q(u)}\cdot u$$ for all $$v\in V$$. Then $$B(\sigma_uv,\sigma_uw)=B(v,w),$$ so $$\sigma_u\in O(V)$$. We call $$\sigma_u$$ the (orthogonal) reflection along $$u$$ or through the hyperplane $$u^{\perp}$$. Often in literature $$\sigma_u$$ is called a symmetry, e.g. by Arin in .

If $$B$$ is a symmetric form on $$V$$, with quadratic form $$Q$$, then $$v\in V$$, $$v\neq 0$$, is called isotropic if $$Q(v)=0$$, anisotropic if $$Q(v)\neq 0$$. For convenience, $$v=0$$ is taken to be anisotropic, even though $$Q(0)=0$$. If there exists an isotropic vector then $$B$$ and $$V$$ (and $$Q$$) are called isotropic; otherwise they are anisotropic.

The dimension $$m$$ of a maximal totally isotropic subspace of a quadratic space $$V$$ is called the Witt index of $$V$$; it is denoted varioulsy as $$m(V)$$, or $$m(B)$$, or $$m(Q)$$, or sometimes in the literature by $${\rm ind}(V)$$, etc.

Suppose that $$A$$ is an (associative) $$F$$-algebra with $$1$$. Say that $$A$$ is compatible with the quadratic $$Q$$ on $$V$$ if there is a linear transformation $$\varphi: V\rightarrow A$$ such that $$\varphi(v)^2=Q(v).1_A$$ for all $$v\in V$$.

Proposition 8.3. Suppose $$A$$ is compatible with $$Q$$ via $$\varphi$$ and $$0\neq v\in V$$. Then $$\varphi v$$ is unit in $$A$$ if and only if $$v$$ is anisotropic in $$V$$

A Clifford algebra for a quadratic space $$V$$ is an associative algebra $$C$$ (with 1) that is compatible with the quadratic form $$Q$$ of $$V$$, via a map $$\epsilon: V\rightarrow C$$, and such that give any algebra $$A$$ compatible with $$V$$ (i.e. with $$Q$$) via $$\varphi: V\rightarrow A$$, then there exists a unique algebra homomorphism $$\hat{\varphi}:C\rightarrow A$$ so that the diagram $$\epsilon: V\rightarrow C, \varphi: V\rightarrow A, \hat{\varphi}: C\rightarrow A$$ is commutative.

The decomposition $$C=C_0\oplus C_1$$ of the Clifford algebra determines a $$\mathbb{Z}_2$$-grading of $$C$$. Note that $$F\subseteq C_0$$ and $$V\subseteq C_1$$. If $$r, s\in \{0, 1\}$$ the $$C_rC_s=C_{r+s}$$, the sum $$r+s$$ being taken $${\rm mod}~2$$. Thus $$C_0$$ is a subalgebra of $$C$$ (the even subalgebra); the elements of $$C_0$$ are called even homogeneous of degree 0, and the elements of $$C_1$$ are called odd homogeneous of degree 1.

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

By definition there is then a unique isomorphism $$\alpha: C\rightarrow C^{op}$$ such that $$\alpha|V=1_V$$, which means that $$\alpha:C\rightarrow C$$ is an anti-automorphism.

Proposition 9.1. Suppose that $$u_1, u_2, ..., u_k$$ are nonzero anisotropic vectors in $$V$$ such that $$\sigma_{u_1}\sigma_{u_2}...\sigma_{u_k}=1_V$$. Then $$\prod_{i=1}^k Q(u_i)\in {F^{\times}}^2$$.

Proposition 9.1 is the basis for an important definition. Take $$\sigma\in O(V)$$. Then $$\sigma$$ is a product of reflections in many ways, say e.g. that $$\sigma=\sigma_{u_1}...\sigma_{u_r}$$ and $$\sigma=\sigma_{v_1}...\sigma_{v_s}$$. Then $$\sigma_{u_1}...\sigma_{u_r}\sigma_{v_s}\sigma_{v_1}=1$$ and consequently $$Q(u_1)...Q(u_r)Q(v_s)...Q(v_1)\in {F^{\times}}^2$$ by Proposition 9.1. It follows that $$Q(u_1)...Q(u_r)$$ and $$Q(v_1)...Q(v_s)$$ lie in the same coset of $${F^{\times}}^2$$ in $$F^{\ast}$$, and hence determin a well-defined map $$\theta: O(V)\rightarrow F^{\ast}/{{F^\times}^2},$$ viz. $$\theta(\sigma)=Q(u_1)...Q(u_r){F^{\times}}^2$$

The map $$\theta$$ is evidently a homomorphism; it is called the spinor norm on $$O(V)$$.

The kernel of the restriction to $$SO(V)$$ of the spinor norm $$\theta$$, denoted by $$\mho=\mho(V)$$ is called the reduced orthogonal group of $$V$$ (relative, of course, to the given quadratic form).

Theorem 9.7. Suppose that $$V$$ is isotropic. Then $$\mho(V)=\Omega(V),~{\rm and}~SO(V)/\Omega\cong F^{\ast}/{{F^\times}^2}.$$

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 $$SO(V)\leq \Gamma_0$$.

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 $${\rm Ker}(\chi|{\Gamma_0})=F^\ast$$.

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 from 'Basic algebra I' by Nathan Jacobson.

Let $$(e_1, e_2, ..., e_n)$$ be a base for $$V$$ and let $$\eta\in O(V,Q)$$. Then we have $$B(\eta e_i, \eta e_j)=B(e_i,e_j)$$ for all $$i, j=1,2, ..., n.$$ conversely if these conditions hold for a linear transformation $$\eta$$, then for any $$x=\sum a_ie_i$$ we have $$Q(\eta x)=Q(x)$$. Thus a linear transformation $$\eta$$ of $$V$$ into $$V$$ is orthogonal if and only if $$(26)~~~B(\eta e_i,\eta e_j)=B(e_i,e_j)~1\leq i, j\leq n.$$ Now let $$b=(B(e_i, e_j))$$ the matrix of $$B$$ relative to the base $$(e_1,e_2,...,e_n)$$ and let $$\eta e_i=\sum h_{ij}e_j$$. Then the conditions (26) are that $$B(\sum_k h_{ik}e_k, \sum_l h_{jl}e_l)=B(e_i,e_j)$$ for all $$i$$ and $$j$$. In matrix form these conditions are $$(27)~~~h\cdot b\cdot {}^th=b~~~{\rm for}~h=(h_{ij}).$$

An orthogonal transformation is called proper or a rotation if $${\rm det}~h=1$$; otherwise, the transformation is improper. If we choose an orthogonal base for $$V$$, then the matrix $$b$$ of $$B$$ is diagonal. Then it is clear that any diagonal matrix with diagonal entries $$1$$ or $$-1$$ satisfies (27) and hence determines an orthogonal transformation. Moreover, this is proper or improper according as the number of -1's is even or odd. It is clear that the set $$O^+{V,Q}$$ of rotations is a normal subgroup of index two in $$O(V,Q)$$.

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.

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

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

Definition 4.5. Let $$V$$ be a vector space voer a field $$F$$, $$Q$$ a quadratic form on $$V$$. Let $$T(V)=F\oplus V\oplus(V\otimes V)\oplus (V\otimes V\otimes V)\oplus ...$$ be the tensor algebra defined by $$V$$ (p. 140) and let $$K_Q$$ be the ideal in $$T(V)$$ generated by all of the elements of the form $$(45)~~~x\otimes x-Q(x)1,~~~x\in V$$ Then we definet the Clifford algebra of the quadratic form $$Q$$ to be the algebra $$C(V,Q)=T(V)/{K_Q}.$$

In the remaider of this section, we shall give a brief indiction of some applications of Clifford algebras to the strudy of orthogonal groups. For this purpose, we need to introduce the even (or second) Clifford algebra $$C^+(V,Q)$$ defined to be the subalgebra of $$C(V,Q)$$ geneated by all of the products $$uv$$, $$u,v\in V$$ (that is, by $$V^2$$).

Definition 4.6. The Clifford group $$\Gamma(V,Q)$$ is the subgroup of invertible elements $$u\in C(V,Q)$$ such that $$uxu^{-1}\in V$$ for all $$x\in V$$. Clearly this is a subgroup of the multiplicative group of invertible elements of $$C(V,Q)$$. The even Clifford group is $$\Gamma^+(V,Q)=\Gamma(V,Q)\cap C^+(V,Q)$$.

If $$x\in V$$ and $$u\in \Gamma=\Gamma(V,Q)$$, then $$uxu^{-1}\in V$$ and $$(uxu^{-1})^2=ux^2u^{-1}=Q(x)1$$. Hence the linear transformation $$x\mapsto uxu^{-1}$$ of $$V$$ is in the orthogonal group $$O(V,Q)$$. The map $$\chi$$, where $$\chi(u)$$ is $$x\mapsto uxu^{-1},x\in V$$, is a homomorphism of $$\Gamma(V,Q)$$ into $$O(V,Q)$$ called the vector representation of the Clifford group.

Definition 4.7. The kernel of the homomorphism $$N$$ of $$\Gamma^+$$ into $$F^*$$ is called the spin group $${\rm Spin}(V,Q)$$. Its image $$O'(V,Q)$$ under the vector representation $$\chi$$ is called the reduced orthogonal group. If $$f$$ is any rotation and $$\chi(u)=f$$, then $$N(u){F^{*}}^2$$ is called the spinorial norm of $$f$$.

The spinorial norm map is also a homomorphism (of $$O^+(V,Q)$$ into $$F^*/{{F^*}^2}$$). The spinorial norm of a rotation can be defined directly, without the intervention of the Clifford algebra. If $$f$$ is a given rotation, then we can write $$f=S_{v_1}...S_{v_2}$$. and we can simply define the spinoial norm of f to be the coset $$\prod_1^{2r}Q(v_i){F^*}^2$$. Since $$\chi(v_1...v_{2r})=f$$ and $$N(v_1...v_{24})=\prod_1^{2r}Q(v_i)$$, this is the same element of $$F^*/{{F^*}^2}$$, which we have called the spinorial norm.

Now the reduced orthogonal group can also be defined as the kernel of the spinorial norm map.

The reduced orthogonal group contains the commutator subgroup $$\Omega$$ of $$O$$.

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 $$O^+(V,Q)/{O'(V,Q)}\cong F^{\ast}/{F^{\ast 2}}.$$

By '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.

The following follows from 'twisted wreath products and Sylow 2-subgroups of Classical simple groups' by Warren J. Wong.

If $$O_n$$ is an orthogonal group in dimension $$n$$ over $$GF(q)$$, we shall take the quadratic form on the underlying space to be fixed, so that we can define the spinor norm on the whole group $$O_n$$, not just on the proper orthogonal group $$O_n^{+}$$. If $$K_n$$ is the kernel of the spinor norm, then $$K_n$$ has index 2 in $$O_n$$, and the intersection $$K_n\cap O_n^+$$, which has index $$4$$ in $$O_n$$, is the commuttor subgroup $$\Omega_n$$ of $$O_n$$.

First we consider the Sylow 2-subgroup of $$K_{2n}$$. For $$n=1$$, the group $$O_2$$ is dihedral, so that its Sylow $$2$$-subgroup $$R$$ is dihedral of order $$2^{t+1}$$; $$R=\langle u, v\rangle$$, where $$(23)~u^{2^t}=w^2=1, u^w=u^{-1}.$$ If $$q\equiv 1(\rm mod)~4$$, we may take $$u=\begin{pmatrix}\epsilon^{-1}&0\\0&\epsilon\end{pmatrix}, w=\begin{pmatrix}0&1\\1&0\end{pmatrix},$$ where $$\epsilon$$ is a primitive $$2^t$$-th root of 1 in $$GF(q)$$ If $$q\equiv -1(\rm mod)~4$$, we may take $$u=\begin{pmatrix}a&b\\-b&a\end{pmatrix}, w=\begin{pmatrix}-1&0\\0&1\end{pmatrix},$$ where $$a+b\sqrt{-1}$$ is a primitive $$2^t$$-th root of $$1$$ in $$GF(q^2)$$. In either case, $$w$$ has spinor norm $$1$$, being a symmetry with respect to the hyperplane (line) orthogonal to a vector on which the value of the quadratic form is a square in $$GF(q)$$. Thus the Sylow $$2$$-subgroup $$W$$ of $$K_2$$ contained in $$R$$ is $$\langle u,w\rangle$$, where $$v=u^2$$. This is a dihedral group of order $$2^t$$: $$(24)~v^{2^t-1}=w^2=1,~v^w=v^{-1}.$$ Put $$e=uw$$. Then $$e$$ has order $$2$$, and $$R=W\langle e\rangle$$. where $$(25)~v^e=v^{-1},~w^e=vw.$$ For a general value of $$n$$, let the direct sum matrix of the above matrix $$e$$ with the identity matrix of degree $$2n-2$$ be again denoted as $$e$$: $$(26)~\begin{pmatrix}e&0\\0&I\end{pmatrix},$$ Then $$e$$ is an element of order $$2$$ in $$O_{2n}$$ which is not in $$K_{2n}$$. If $$W$$ is a Sylow $$2$$-subgroup of $$K_{2n}$$ normalized by $$e$$, and $$E=\langle e\rangle$$, then te semi-direct product $$WE$$ is a Sylow 2-subgroup of $$O_{2n}$$, and we call the natural representation $$E\rightarrow {\rm Aut}~W$$ as $$s_2$$-system of $$K_{2n}$$ in $$O_{2n}$$.

We now conside the Sylow 2-subgroups of $$\Omega_n$$ and of $$P\Omega_{2n}=\Omega_{2n}/Z$$, where the centre $$Z$$ of $$\Omega_{2n}$$ is a group of order 2 generated by $$z=-I$$. For $$n=1$$, take the Sylow $$2$$-subgorup $$R=\langle u, w\rangle$$ of $$O_2$$ as in (23). The corresponding Sylow 2-subgroup $$T$$ of $$\Omega_2$$ is the commutator subgroup $$\langle v\rangle$$ of $$R$$, $$v=u^2$$, and $$z$$ is the unique element of order $$2$$ in $$T$$. In this case, $$R$$ is not a split extension of $$T$$. However, for $$n>1$$, take $$e$$ as in (26) and take $$(27)~f=\begin{pmatrix}I_2&0&0\\0&w&0\\0&0&I_{2n-4}\end{pmatrix}$$ where $$I_2$$ and $$I_{2n-4}$$ are identity matrices of degrees $$2$$ and $$2n-4$$. Then $$f$$ lies in $$K_{2n}$$ but not in $$\Omega_{2n}$$, and $$e$$ and $$f$$ generate a non-cyclic group $$F$$ of order $$4$$ which meets $$\Omega_{2n}$$ in the identity. If now $$T$$ is a Sylow 2-subgroup of $$\Omega_{2n}$$ normalized by $$F$$, then the semi-direct product $$TF$$ is a Sylow 2-subgroup of $$O_{2n}$$, and we cal the natural representation $$F\rightarrow {\rm Aut}~T$$ an $$s_2$$-system of $$\Omega_{2n}$$ in $$O_{2n}$$.

When $$n=2$$, we can take $$T$$ to be the group generated by $$d=\begin{pmatrix}u&0\\0&u^{-1}\end{pmatrix}, g=\begin{pmatrix}u&0\\0&u\end{pmatrix}, h=\begin{pmatrix}0&I\\I&0\end{pmatrix}, k=\begin{pmatrix}0&w\\w&0\end{pmatrix},$$ where $$u$$ and $$w$$ are as before and $$I$$ is the identity matrix of degree 2. We have the relations $$(28)~ d^{2^{t-1}}=g^{2t-1}=z, z^2=h^2=k^2=1,$$ $$d^h=d^{-1}, g^k=g^{-1}, [d,g]=[d,k]=[h,g]=[h,k]=1,$$ so that $$T$$ is a central product of two dihedral groups of order $$2^{t+1}$$, and $$z$$ is the generator of the centre of $$T$$. The action of $$F$$ and $$T$$ is given by $$(29)~d^e=g^{-1}, g^e=d^{-1}, h^e=gk, k^e=dh,$$ $$d^f=g, g^f=d, h^f=k, k^f=h.$$

Theorem 10. Let $$2^{t+1}$$ be the great power of $$2$$ dividing $$q^2-1$$, $$F=\langle e,f\rangle$$ a non-cyclic group of order $$4$$, $$T=\langle d, g, h, k\rangle$$ the group of order $$2^{2t+1}$$ given by (28), $$z$$ the generator of the centre of $$T$$, and $$\rho : F\rightarrow {\rm Aut}~T$$ the representation given by (29). Let $$C=\langle a, b, c\rangle$$ be an elementary Abelian group of order $$8$$, $$J=\{1,2\}$$ and $$\varphi=(F\rightarrow {\rm Aut}~C, CF\rightarrow F\sim S_J)$$ the $$r$$-system given by $$a^e=a^f=a, b^e=b^f=b, c^e=ac, c^f=bc,$$ $$a\rightarrow \begin{pmatrix}e&0\\0&e\end{pmatrix}, b\rightarrow \begin{pmatrix}f&0\\0&f\end{pmatrix}, c\rightarrow \begin{pmatrix}0&1\\1&0\end{pmatrix}, e\rightarrow \begin{pmatrix}e&0\\0&1\end{pmatrix}, f\rightarrow \begin{pmatrix}f&0\\0&1\end{pmatrix}.$$ If $$n=2^m, m\geq 1$$ and $$M_{m-1}$$ is the twisted wreath product of $$m-1$$ copies of $$\varphi$$, then $$\rho{\ast}M_{m-1}$$ is equivalent with an $$s_2$$-system of $$\Omega_{2n}$$ in $$O_{2n}$$, where $$O_{2n}=O_{2n}(1,q)$$, and $$z\ast 1$$ correpsponds to the generator of the centre of $$\Omega_{2n}$$. In particular, the Sylow 2-subgroup of $$\Omega_{2n}$$ is isomorphic with a twisted wreath product $$U$$ of $$T$$ with $$m-1$$ copies of $$C$$, and the Sylow 2-subgroup of $$P\Omega_{2n}$$ is isomorphic with $$U/K$$ where $$K$$ is genertated by $$z\ast 1\ast...\ast 1$$.

The following follows from 'the geometry of the classical groups' by Donald E. Taylor.

If $$u$$ is singular and $$v\in \langle u\rangle^\perp$$, we put $$\rho_{u,v}(x):=x+\beta(x,v)u-\beta(x,u)v-Q(v)\beta(x,u)u~~~(11.17)$$ and note that $$\rho$$ is a well-defined element of $$SO(V)$$ for fields of all characteristics. Moreover, if $$v$$ is singular, we obtain the transformation of case II whereas if $$v$$ is non-singular, we obtain those of case III of the previous section. Following Tamagawa (1958) and Higman (1978) we call $$\rho_{u,v}$$ a siegel transformation.

11.46 Theorem. If $${\rm dim}~V\geq 3$$, then Witt index of $$V$$ is at least $$1$$, and $$\Omega(V)\neq \Omega^+(4,2)$$, then $$\Omega(V)$$ is generated by the Siegel transformations of $$V$$

The following follows from 'the subgroup structure of finie classical groups' by Peter B. Kleidman and Martin W. Liebeck

$$P\Omega_{2m}^+(q)\cong D_m(q)$$ for $$m\geq 3$$.

$$P\Omega_{2m}^-(q)\cong {}^2D_m(q)$$ for $$m\geq 2$$.

The following follows from 'atlas of finite groups' by John Horton Conway, R. T. CURTIS, S. P. NORTON, R. A. PARKER, R. A. WILSON and with computational assistance from J. G. THACKRAY.

For a $$\mathbb{C}G$$-module $$V$$ with irreducible character $$\chi$$ we have the Frobenius-Schur indicator $$\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^2),$$ and $$\nu(\chi)$$ takes one of the values {+1, -1, 0}, as $$\chi$$ is afforded by a real representation or is real-valued but not afforded by a real representation or is not real-valued, respectively.

The following are conseqences by number theory and representation theory of finite groups.

It is well known, and easy to derive from the orthogonality relations for group characters and properties of the Frobenius-Schur indicator $$\nu$$ that $$z$$ is a square in $$G$$ if and only if $$\sum_{ \chi \in {\rm Irr}(G)} \nu(\chi) \chi(z) > 0.$$

When $$m$$ is an even integer and $$q$$ a power of a odd prime, then $$q^m\equiv 1~{\rm mod~4}$$.

The following websites may be useful to my quesion:

What do Sylow 2-subgroups of finite simple groups look like?

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

https://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory?rq=1

https://mathoverflow.net/questions/273929/kernel-of-a-double-cover-of-group-as-stem-extension

https://mathoverflow.net/questions/355150/is-every-generator-of-z-rm-spin-n-epsilonq-a-square-element-in-rm

https://mathoverflow.net/questions/356879/is-1-neq-a-in-z2-e-7q-cong-z-2-a-square-element-in-2-e-7q

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

• Can you please define your notation? – Michael Albanese Mar 3 at 1:05
• The notation is standard. But "any" is ambiguous. Does it mean "some" or "every"? Avoid using the word "any" in formal mathematics, particularly if you are not a native English speaker. – Derek Holt Mar 3 at 8:28
• Could I suggest that people don't vote to close the question just because they don't understand it? – Derek Holt Mar 3 at 8:29
• Perhaps you should try asking this on mathoverflow. You might find somebody with enough technical knowledge of the spin groups to answer the question. I would expect the answer to all of your questions to be yes. – Derek Holt Mar 17 at 8:43
• @Chickenmancer OK！ – Yi Wang May 23 at 2:03