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:


*

*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 $F_q$)

*(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

*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
 A: 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 $\nu(\chi) \chi(z) = \chi(1)$.
However, I am not sure whether enough information about the character table and Frobenius-Schur indicators is available for the groups you are considering. 
A: 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.
Notice that this yields explicit elements of the simple group that power to the centre in the central extension.
