Let $P$ be a finite projective plane, $\Gamma$ a collineation group and suppose $\Gamma$ is transitive on the points of $P.$ If $P$ contains a non-trivial elation, show that $P$ is a Moufang plane.
I solved it, but I'm suspicious about the last step. Here's my solution:
Let $\alpha$ be a $(A,l)$ elation, then $|\Gamma_{(l,l)}|>1$ so $l$ is a translation line by the previous problem. Now we use the following result extensively:
If $\alpha$ is a collineation and $l$ a translation line, then $\alpha(l)$ is a translation line.
Given any $B \not\in l,$ pick $\beta$ such that $\beta(A) = B,$ then $\beta(l)$ passes through $B$ so there's a translation line through every point of $P.$ If there's 3 non-concurrent translation lines, then every line is a translation line by a previous result, so $P$ is a Moufang plane. Otherwise, every translation line concurs at $X \in l$ and every line through $X$ is a translation line.
At this point I wasted so much time trying to generate new lines from the existing ones and couldn't ever derive a contradiction. After 1 hour I had the idea to take a line not containing $X$ and turn it into one that passes through $X.$ Pick $Y, Z$ such that $X, Y, Z$ are not collinear and $\beta$ such that $\beta(Y) = X,$ then $\beta(YZ) = X\beta(Z)$ is a translation line, so $\beta^{-1}(\beta(YZ)) = YZ$ is as well.
It seems suspicious that no matter what you do, it's impossible to create any more lines once you have them concur at a point, and the only way to continue is to take a new line, make it concur, and undo the automorphism. Is there another proof or is this essentially the only way?
There is indeed a better way, I discovered it while trying to chase down a Moufang plane for another problem and came back now to settle this. Let $XYZ$ be a triangle and pick $\beta \in \Gamma$ sending $X \to Y.$ Then $\beta(XY), \beta(XZ)$ are 2 distinct lines through $Y,$ so at least one of them isn't $XY$; call it $l.$ Then $XY, XZ, l$ are 3 non-concurrent lines.