I am working on classifying the stabilisers of quadratic binomials in $GL(\mathbb{C}^{n})$ but struggling to identify the groups which are appearing.
One example is the binomial $$x_{1}x_{2}-x_{1}x_{3}.$$ I have found the stabiliser to be the set group generated by matrices of the from $$\begin{bmatrix} 0 & a & -a\\ b & c & d\\ \frac{ab-1}{a} & c & d \end{bmatrix},\quad \begin{bmatrix} i & 0 & 0\\ j & k & l\\ j & \frac{ik-1}{i} & \frac{il+1}{i} \end{bmatrix},$$ where $a,b,c,d,i,j,k,l\in\mathbb{C}\setminus\{0\}.$ But I cannot pin down any more information about the group.
My usual method of approach is to try and find relations between the matrices to express the group as the quotient of a finitely presented group via. Von Dyckes' Theorem. An example of this is for the binomial $$x_{1}^{2}-x_{1}x_{2}$$ which gave matrices of the form $$\begin{bmatrix} a & 0\\ \frac{a^{2}-1}{a} & \frac{1}{a} \\ \end{bmatrix},\quad \begin{bmatrix} b & -b \\ \frac{b^{2}-1}{b} & -b\\ \end{bmatrix},$$ for $a,b\in\mathbb{C}\setminus\{0\}$. If we let $$A=\begin{bmatrix} a & 0\\ \frac{a^{2}-1}{a} & \frac{1}{a} \\ \end{bmatrix}$$ and $$B=\begin{bmatrix} b & -b \\ \frac{b^{2}-1}{b} & -b\\ \end{bmatrix}$$ for some $a,b\in\mathbb{C}\setminus\{0\}$ then we can see $B^{2}=I_{2}$ and $BAB=A^{-1}$ so for fixed $a,b$ the group is isomorphic to a quotient of the infinte dihedral group $$Dih_{\infty}=\langle x,y\vert y^{2}=1, yxy=x^{-1}\rangle,$$ by Von Dyckes Theorem. Hence the entire group is isomorphic to a quotient of $Dih_{\infty}$ direct product 2 copies of $\mathbb{C}^{\times}$ (the multiplicative group of complex numbers), where the two copies of $\mathbb{C}$ correspond to the choice of $a,b$, i.e $$G\cong Dih_{\infty}/N\times(\mathbb{C}^{\times})^{2},$$ for some $N\triangleleft Dih_{\infty}.$
I have also thought of finding a normal subgroup $N\triangleleft G$ and trying to find its complement in $G$ to express $G$ as a semi-direct product but this has not led anywhere either. In terms of my original example of the binomial $x_{1}x_{2}-x_{2}x_{3}$ I have not found any relations nor a normal subgroup which would allow me to construct a (semi-)direct product.
My question is; Is there a better method than just messing around with matrices until I find some relations and recognise a group presentation? Is there a better way to recognise the underlying group structure?
As a final comment the other binomials I am trying to work with are $$x_{1}^{2}-x_{2}^{2},\; x_{1}^{2}-x_{2}x_{3},\; x_{1}x_{2}-x_{3}x_{4}.$$
I appreciate all responses.
The stabilizers of quadratic homogeneous forms are very easily understood once you pass to symmetric matrices: Every quadratic form $f\in\Bbb C[x_1,\ldots,x_n]$ can also be seen as a symmetric matrix $A\in\Bbb C^{n\times n}$ where the relation is given by the condition $f(x)=x^TAx$, here I write $x=(x_1,\ldots,x_n)^T$ for the vector of variables.
Now the action of $S\in\operatorname{GL}_n(\Bbb C)$ on $f$ is given (up to possibly inversion) by $f\mapsto f\circ S$, and this is the same as considering the action of $S$ on the symmetric matrix $A$ via $A\mapsto S^TAS$. By means of a Takagi factorization, you can see that $A$ is the identity, up to a change of basis. Therefore, the stabilizer of $f$ is always conjugate to the group $$ \operatorname O_n(\Bbb C) = \left\{~ S\in\operatorname{GL}_n(\Bbb C) ~\middle\vert~ S^TS=I ~\right\}. $$ If the matrix $A$ has lower rank, you can make a similar argument and easily deduce the structure of the stabilizer up to conjugation.