For an exercise (self-study) I'm trying to prove the following statement:
Let $K$ be a division ring and $E=K\mathbf{P}^2$ the projective plane over $K$. Let $u$ be a collineation of $E$ onto itself. Suppose that, for every $x\in E$, the points $x,u(x),u(u(x))$ lie on one line. Then, for every $x\in E$, there exists a line $L$ of $E$ such that $x\in L$ and $u(L)=L$.
This has been pretty easy except for this special case: $x$ is the only fixed point of $u$. Do we know something about collineations with exactly one fixed point? One has to show that, for some (probably every) $y\in E-\{x\}$, $y,u(y),x$ lie on one line. I have tried to prove this using projective coordinates, but didn't succeed. Am I missing something obvious here?
I'm glad for any help.
If for two different $y_1,y_2\in E$ the lines $y_1,u(y_1)$ and $y_2,u(y_2)$ are different lines as well, then the point where they intersect has to be a fixed point. This is because the intersection of two fixed lines is always a fixed point. Since every projective plane contains at least four points, no three of them collinear, you know that such points $y_1,y_2\in E-\{x\}$ will exist. And since in your case you know that there is only one fixed point, you can conclude that for every $y\in E-\{x\}$ the line $y,u(y)$ has to coincide with $x$, since otherwise you'd get another fixed point from its intersection with another fixed line. As you can see, this argument holds even for non-Desarguesian planes, since it does not require coordinates.