Proving that the Spinor-kernel of SO3(K) is simple EDIT: This whole proof fails, as not everything in $SO_3(K)$ can be written as $\sigma_1\sigma_2$. Instead, we have $SO_3(K)\cong SL_2(K)$, which is not simple if $|K|=3$.
I tried to prove something, but I ran into the following problems:
(a) Can someone help me out with (*)? (EDIT: Solved)
(b) Is it a necessary condition that a quadratic form on $K^{*2}$ takes all values of $K^*$ on $K^2$? (see (**))
Thanks in advance.
EDIT:
As a (c), I realized I messed up the last part. Can someone help me there as well? It is probably closely related to (a).
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Let $K$ be a field with $Char(K)\neq 2$. On $V=K^3$, we take take the quadratic form given by the matrix
$$\begin{pmatrix}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}$$
With other words, we take the quadratiq form $Q$ that has a zero locus, and where $-\det{Q}$ is a square. Now consider the following group:
$$G=\{M\subseteq O(V): \det M=1, sn(M)=1\}$$
Here is $sn: O(K)\to K^*/K^{*2}$ the spinor norm. I try to prove that this group is simple. To start, we use the following result that is also used to prove the simplicity of Special Linear groups.

Let $G$ be a group acing on $S$ with trivial kernel, and suppose the action is double transitive; that is, it is transitive on $S^2\setminus\Delta$. Furthermore, assume the following: 
a) There exists an $x\in S$ and a normal divisor $H\triangleleft G_x$ such that $H$ is Abelian, and $G$ is generated by all conjugates of $H$.
b) $[G,G]=G$
Then $G$ is simple.
(less specifically, if (a) is true, then all nontrivial normal subgroups of $G$ contain $[G,G]$).

We let $G$ act on $Z(Q)\subset \mathbb{P}^2(K)$, the zero locus of $Q$. This yields a double transitive action with trivial kernel.
To prove (a), let $P=\langle e_1\rangle$. Let $H\subseteq G_{P}$ be group of elements that keep $e_1$ itself. This is normal, and in fact abelian. To see the latter, it consists of matrices in the form:
$$\begin{pmatrix}
1 & -\frac{c^2}{2} & -c\\
0 & 1 & 0\\
0 & c & 1
\end{pmatrix} (c\in K)$$
Abelian-ness can then be decuced from the image of the right bottom matrix in $GL_2(K)$.
EDIT (used to be (*)): It remains to prove conjugates of $H$ generate $G$. Let $\rho\in G$ be any operator. It can be proven that anything in $O(K)$ with determinant $1$ is a product of two reflections:
$$\rho=\sigma_{a}\sigma_{b}\quad a,b\in V\setminus Z(Q)$$
First assume $Q(a),Q(b)\in K^{*2}$. Then one can construct a $F\in O(K)$ such that $F(a)=e_3$. This means that there exists a $z_1\in Z(Q)$ such that $z_1\perp a$ and similarly a $z_2\in Z(Q)$ such that $z_2\perp b$. Now let $c\in \langle a,z_1\rangle\cap\langle b, z_2\rangle\setminus\{0\}$.
It follows easily that $Q(c)\in K^{*2}$. As $c\perp z_1$, the map $\sigma_a\sigma_c$ preserves $z_1$. Likewise, $\sigma_c\sigma_b$ preserves $z_2$. Since $\rho=(\sigma_a\sigma_c)(\sigma_c\sigma_b)$, it follows that $\rho$ is is in the group generated by conjugates of $H$.
For the general case, we know that $Q(a)Q(b)\in K^{*2}$. As such, we may assume $Q(a)=Q(b)$, and call it $\lambda$. Let $z_1\in Z(Q)$ and let $z_2=\sigma_b(z_1)$, and $z_3=\sigma_a(z_2)$. Assume that we have chosen $z_1$ so that $z_1\neq z_2$ and $z_2\neq z_3$. Then it follows that $Q(z_1-z_2),Q(z_2-z_3)\in \lambda K^{*2}$. So letting $b'=z_1-\lambda z_2$ and $a'=z_2-\lambda^{-1}z_3$, it follows that $Q(a'),Q(b')\in K^{*2}$. On top of that, $\sigma_{a'}\sigma_{b'}(z_1)=z_3$, so $(\sigma_{a'}\sigma_{b'})^{-1}\rho$ preserves $z_1$. As $\sigma_{a'}\sigma_{b'}$ is of the form discussed above, it follows that $\rho$ is in the group generated by conjugates of $H$.
To prove (b), Let $\rho\in G$ be any operator. It can be proven that anything in $O(K)$ with determinant $1$ is a product of two reflections:
$$\rho=\sigma_{a}\sigma_{b}\quad a,b\in V\setminus Z(Q)$$
As $sn(\rho)=1$, it follows that $Q(a)Q(b)\in K^{*2}$. As a result, we may assume $Q(a)=Q(b)$.
We need an extra hypothesis: Any quadratic form in dimension $2$ takes any value(****).**
Under this hypothesis, it is fairly easy to find a $\tau\in G$ such that $\tau(a)=b$ (don't forget to ensure it has the right determinant and norm). We have already shown (or should I say, assumed) that $G$ is generated by conjugates of $H$. As such, we can write $\tau=\tau_n...\tau_1$ where each $\tau_i$ preserves an element of $Z(Q)$. Letting $a_i=\tau_i...\tau_1(a)$, we can write:
$$\rho=\sigma_a\sigma_b=\prod_{i=0}^{n-1}\sigma_{a_i}\sigma_{a_{i+1}}=\prod_{i=0}^{n-1}\sigma_{a_i}\sigma_{\tau_{i+1}(a_{i})}$$
As a result, we can reduce to the case $\tau\in H$. We saw by above that we actually have $H\cong K$. As a result, all elements of $H$ are in fact squares. So write $\tau=\mu^2$, we find that:
$$\rho=\sigma_a\sigma_{\tau(a)}=\sigma_a\sigma_{\mu(a)}\sigma_{\mu(a)}\sigma_{\mu^2(a)}=\sigma_a\sigma_{\mu(a)}\mu \sigma_a\sigma_{\mu(a)}\mu^{-1}=[\sigma_a\sigma_{\mu(a)},\mu]$$
This proves that $\rho\in [G,G]$ and hence $[G,G]=G$.
EDIT: This last trick doesn't work. Sigh. We need to make sure $a\perp \mu(a)$.
 A: Ok, I figured it out myself. It works for any field, so the answer to (b) is no.
To prove that $[G,G]=G$, let $\rho=\sigma_a\sigma_b\in G$. One can actually prove that if $\sigma_a\sigma_b\in H$, that we must have that $Q(a),Q(b)\in K^{*2}$. So $G$ is in fact generated by $\sigma_a\sigma_b$ where $Q(a), Q(b)\in K^{2*}$. So we can assume $\rho$ is of that form as well. In fact, we can assume $Q(a)=Q(b)=1$.
As before, there exist a $z_1, z_2\in Z(Q)$ such that $z_1\perp a$ and $z_2\perp b$. Again, let $c\in \langle a,z_1\rangle \cap \langle b,z_2\rangle$. As we can scale $z_1,z_2$ freely, and change the signs of $a,b$ freely, we can assume $c=a+z_1=b+z_2$.
Next step is to construct a $\tau_1\in G$ such that $\tau(a)=c$ and $\tau(p_1)=p_1$ for a certain $p$ with $Q(p)=1$. As $a$ and $z_1$ essentially take the roles of $e_3$ and $e_1$ respectively, there is a $\tau_1'\in G$ such that $\tau_1'(z_1)=z_1$ and $\tau_1'(a)=\frac{1}{4}z_1+a$. By composing $\tau_1'$ with (a conjugate of) the matrix... $$\begin{pmatrix} 4&0&0\\0&\frac{1}{4}&0\\0&0&1\end{pmatrix}$$ ...we get a $\tau_1\in G$ such that $\tau_1(z_1)=4z_1$ and $\tau(a)=z_1+a=c$. By looking at the matrix form of $\tau_1$ on $\langle z_1,a\rangle$, it is clear that $\tau_1$ keeps a point $p_1$ invariant. As $p_1\notin \langle z_1\rangle$, it also follows $Q(p_1)\in K^{*2}$. This concludes the construction of $\tau_1$.
By choosing $\tau_2,p_2$ similarly for $b$, we find that:
$$\rho=\sigma_a\sigma_b=\sigma_a\sigma_{p_1}\sigma_{p_1}\sigma_c\sigma_c\sigma_{p_2}\sigma_{p_2}\sigma_b=(\sigma_a\sigma_{p_1})\tau_1(\sigma_{p_1}\sigma_{a})\tau_1^{-1}\tau_2(\sigma_b\sigma_{p_2})\tau_2^{-1}(\sigma_{p_2}\sigma_{b})=[\sigma_a\sigma_{p_1},\tau_1][\tau_2,\sigma_b\sigma_{p_2}]$$
So $\rho\in [G,G]$
