A problem about groups of matrices I found on the web the following problem and I have no idea how to solve it.
Let $X\subseteq \mathbb C^4$ be the set defined by the equation $$x^2=y^3+z^3+w^3.$$ Let $\tau$ be the involution over $\mathbb C^4$ given by $\tau(x,y,z,w)=(-x,y,z,w).$ Let $G$ be a finite subgroup of $GL(4,\mathbb C)$ such that:
1) $g(X)\subseteq X,$ $\forall g\in G,$
2) $G$ contains the (matrix whose action is the same as the) involution $\tau.$ 
Prove that $\tau$ belongs to the center of $G.$
Prove that the following assertion is false: "there exists a subgroup $H$ of $G$ of index two such that $\tau\notin H.$" 
 A: I proved the proposition as follows.
All the linear spaces of dimension one contained in $X$ are of the form
$$\left( \begin{array}{c} x  \\
y \\
z\\
w
\end{array} \right) =t\cdot \left( \begin{array}{c} a  \\
b \\
c\\
d
\end{array} \right)$$ with $$(ta)^2=(tx)^3+(ty)^3+(tz)^3, \quad \forall t$$ where $t$ is a complex parameter and $(a,b,c,d)\in \mathbb C^4$ is fixed. Since the previous equation is true for every $t,$ $a$ must be $0$ and $b,c,d$ such that $b^3+c^3+d^3=0.$
Since every matirx $g\in G$ takes a linear one-dimensional subspace contained in $X$ into another, if $g=(g_{i,j})_{1\leq i,j\leq 4},$ the first row of $$g\cdot \left( \begin{array}{c} 0  \\
b \\
c\\
d
\end{array} \right)=\left( \begin{array}{c} g_{1,2}b+g_{1,3}c+g_{1,4}d  \\
...\\
...\\
...
\end{array} \right)$$ must be zero. Hence $g_{1,2}b+g_{1,3}c+g_{1,4}d=0$ for every
$b,c,d\in \mathbb C$ with $b^3+c^3+d^3=0.$ This implies that $g_{1,2}=g_{1,3}=g_{1,4}=0.$
Similarly, since the vector $(t,\sqrt[3]{t^2},0,0)\in X$ $\forall t\in \mathbb R$ we have that $\forall g\in G$ and $\forall t\in\mathbb R,$ $$g\cdot \left( \begin{array}{c} t  \\
\sqrt[3]{t^2} \\
0\\
0
\end{array} \right) \in X$$ so that $$(tg_{11})^2=(tg_{2,1}+\sqrt[3]{t^2}g_{2,2})^3+(tg_{3,1}+\sqrt[3]{t^2}g_{3,2})^3+(tg_{4,1}+\sqrt[3]{t^2}g_{4,2})^3\quad \forall t\in \mathbb R.$$ This implies that $g_{2,1}=g_{3,1}=g_{4,1}=0.$
Hence every matrix $g\in G$ has the form $$g=\left(\begin{array}{c|ccc}
\ast &      &      & \\ \hline
     & \ast & \ast & \ast \\
     & \ast & \ast & \ast \\ 
     & \ast & \ast & \ast \\
\end{array}\right)$$ and $\tau $ is in the center of $G.$
For the second part of the question, consider the gruop $G$ of matrices of the form 
$$\left(\begin{array}{c|ccc}
\pm 1 & 0 & 0 & 0 \\
\hline
0  & \huge{A} \\
0 & & &\\
0 & & & 
\end{array}\right)$$ or $$\left(\begin{array}{c|ccc}
\pm i & 0 & 0 & 0 \\
\hline
0  & \huge{-A} \\
0 & & &\\
0 & & & 
\end{array}\right)$$ where $A$ is a $3\times 3$ permutation matrix. It is easily seen that all the subgroups $H$ of index 2 of this group $G$ contain the involution $\tau.$
