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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 $\tau\notin H.$"

1 Answers 1

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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 $\tau.$