Why is a field with characteristic 2 required for the Cartan-Dieudonne theorem? Is there a simple counter-example?
The Cartan-Dieudonne theorem is stated as in Wikipedia :
Let $(V, q)$ be an $n$-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to $2$. Then, every element of the orthogonal group $O(V, q)$ is a composition of at most $n$ reflections.
Reflection with respect to the hyperplane perpendicular to a non-null vector $x$ is defined with geometric product $\rho_x: y\mapsto -xyx^{-1}$. If the characteristic of the field is not 2, the reflectino becomes $\rho_x: y \mapsto y-2 \frac{y\cdot x}{x\cdot x}x$, where the dot product is the symmetric bilinear form associated to $q$. Explicitly, $x \cdot y = \frac{1}{2}(q(x+y)-q(x)-q(y))$. If the characteristic of the field is 2, the reflection becomes $\rho_x: y \mapsto y-\frac{q(x+y)-q(x)-q(y)}{q(x)}x$.
For instance, let $V = \mathrm{span}_{F_2}(e_1, e_2)$, and $q(0)=q(e_1)=q(e_2)=0$, $q(e_1+e_2)=1$. Then $\rho_{e_1+e_2}:0\mapsto0;e_1\mapsto e_2;e_2\mapsto e_1;e_1+e_2\mapsto e_1+e_2$. It is also the only linear isometry on $(V,q)$ other than the identity. It does not contradict the Cartan-Dieudonne theorem.