# Is $1\neq a\in Z(2.E_7(q))\cong Z_2$ a square element in $2.E_7(q)$?

When $$q$$ is a power of some odd prime, is $$1\neq a\in Z(2.E_7(q))\cong Z_2$$ a square element in $$2.E_7(q)$$?

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 $$n_{ij}$$.

The Dynkin diagram of simple Lie algebra $$E_7$$ is as follows:

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 $$L(q)$$.

$$|E_7(q)|=q^{63}(q^18-1)(q^{14}-1)(q^{12}-1)(q^{10}-1)(q^8-1)(q^6-1)(q^2-1)/{(2,q-1)}.$$

The points over a finite field with $$q$$ elements of the (split) algebraic group $$E_7$$, whether of the adjoint (centerless) or simply connected form (its algebraic universal cover), give a fnite Chevalley group. This closely connected to the group written $$E_7(q)$$, however there is ambiguity in this notation, which can stand for several thing:

1. The finite group consisting of the points over $$F_q$$ of the simply connected form of $$E_7$$ (for clarity, this can be written $$E_{7, sc}(q)$$ and is known as the "universal" Chevalley group of type $$E_7$$ over $$F_q$$)

2. (rarely) the finite group consisting of the points over $$F_q$$ of the adjoint form of $$E_7$$ (for clarity, this can be written $$E_{7, ad}(q)$$, and is known as the "adjoint" Chevalley group of type $$E_7$$ over $$F_q$$), or

3. the finite group which is the image of the natural map from the former to the latter: this is what will be denoted by $$E_7(q)$$ in the following, as is most commmon in texts dealing with finite groups.

$$E_7(q)$$ is simple for any $$q$$ and $$E_{7,sc}$$ is its Schur cover, and we often write $$E_{7,sc}(q)$$ as $$2.E_7(q)$$ when $$q$$ is odd.

$$\bar{K}$$ denotes a semisimple algebraic group, with maximal torus $$\bar{T}$$ and root system $$\sum$$.

If $$\sum=E_7$$, then genertors of $$Z(\bar{K})$$ are $$h=h_{\alpha_4}(-1)h_{\alpha_5}(-1)h_{\alpha_7}(-1)$$.

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 $$h_2=h_{\alpha_{2m-1}}(-1)h_{\alpha_{2m}}(-1)$$

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.

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 $$G=L(K)$$.

Let $$S$$ be a Sylow 2-subgroup of $$E_7^u(q)$$ (universal Chevalley group). Then $$Z(S)=\langle h_e(-1), h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1), h_{s_2}(-1)h_{s_3}(-1)\rangle.$$ Since $$h_e(-1)=h_{s_2}(-1)h_{s_5}(-1)h_{s_7}(-1),$$ we conclude that $$Z(S)=\langle h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1), h_{s_2}(-1)h_{s_3}(-1)\rangle\cong C_2\times C_2.$$ Recall that the center of $$Z(E_7^u(q))$$ is $$Z_0=Z(E_7^u(q))=\langle h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1)\rangle$$. It follows that $$\bar{S}=S/{Z_0}$$ is a Sylow $$2$$-subgroup of $$E_7(q)$$.

The center of $$D_6^u(q)$$ is $$\langle h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1), h_{s_2}(-1)h_{s_3}(-1)\rangle$$.

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

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 $$2\cdot \Omega_n^\epsilon (q)$$. 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 $$4.P\Omega_n^\epsilon(q)$$ if $$m$$ is odd, and the structure $$2^2.P\Omega_n^{\epsilon}(q)~$$(necessarily with $$\epsilon +$$) if $$m$$ is even.

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

Analysis:

$$\pi: 2.E_7(q)\rightarrow E_7(q)$$

The following websites may be useful to my quesion:

https://math.stackexchange.com/questions/785603/what-do-sylow-2-subgroups-of-finite-simple-groups-look-like

Kernel of a double cover of group as stem extension

Square roots of elements in a finite group and representation theory

• The expression you have for $z=h_e(-1)$ is valid in the "diagram" $D_6^u(q)$ subgroup of $E_7^u(q)$. Hence, if you can show that $z$ is a square iin $D_6^u(q)$ it will a fortiori be a square in $E_7^u(q)$. This should be possible, perhaps by using the fact that $\Omega_{12}^+(q)$ contains the direct product of three copies of $\Omega_4^+(q)$. May 3, 2020 at 17:28
• @Richard Lyons, Thank you very much! I think if we should check $h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1)$? May 4, 2020 at 9:57
• 27th version of this question! May 6, 2020 at 12:32

The answer is always, yes. Note that there are three classes of involutions in the simply connected version of the algebraic group $$E_7$$: the central involution $$a$$, an involution $$t$$ with centralizer of type $$A_1D_6$$, and the product $$at$$. If $$a$$ were not a square, then in the simple group $$E_7(q)$$, we would only see involutions with centralizer type $$A_1D_6$$. However, in the adjoint group we find centralizers of type $$E_6T_1$$ and $$A_7$$. This website: http://www.math.rwth-aachen.de/~Frank.Luebeck/chev/23elts.html lists the classes of involutions in the simply connected and adjoint groups, or consult the 3rd volume of Gorenstein-Lyons-Solomon.
Here is a general remark about whether a central involution $$z$$ in a finite group $$G$$ is a square : 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 $$\sum_{ \chi \in {\rm Irr}(G)} \nu(\chi) \chi(z) > 0.$$ Since $$\nu$$ vanishes on irreducible characters which are not real-valued, the sum may be restricted to the real-valued complex irreducible characters of $$G$$. Note that the set $$S$$ of real-valued irreducible characters which make a positive contribution to the sum contains those $$\chi$$ which have $$z$$ in their kernel and $$\nu(\chi) = 1,$$ (contribution $$\chi(1)$$) and those $$\chi$$ which do not contain $$z$$ in their kernel and $$\nu(\chi) = -1$$ (contribution also $$\chi(1)$$). Any real-valued irreducible character $$\chi$$ of $$G$$ which lies outside $$S$$ makes a contribution $$- \chi(1)$$ to the sum. Hence $$z$$ is a square in $$G$$ if and only if $$\sum_{ \chi \in S} \chi(1) > \sum_{ \chi \in {\rm Irr}_{\mathbb{R}}(G) \backslash S } \chi(1)$$, where $$Irr_{\mathbb{R}}(G)$$ denoted the set of real-valued complex irreducible characters of $$G$$, and $$S$$ denotes the set of real-valued irreducible characters $$\chi$$ of $$G$$ with $$\nu(\chi) \chi(z) = \chi(1)$$.