# An automorphism of an affine space preserves parallelism: question on proof

Definitions:

• An automorphism of an affine space is a permutation of the set $$\mathcal{P}$$ of points that preserves lines and planes (if the dimension it at least 3).

This is the proof given in my book (with exponential notation):

Consider an automorphism $$\phi$$. Suppose that $$L$$ and $$L'$$ are parallel lines, than $$L$$ and $$L'$$ are disjoint but contained in a plane $$V$$. $$\underline{\text{This means that } L^{\phi} \text{ and } L'^{\phi} \text{ are also disjoint and contained in the plane } V^{\phi}}$$, implying $$L^{\phi} || L'^{\phi}$$.

Why can claim that $$L^{\phi}$$ and $$L'^{\phi}$$ are disjoint? Also, I don't see why these lines are contained in the plane $$V^{\phi}$$.

The parallel lines $$L$$ and $$L'$$ are contained in a plane $$V$$. That $$L^{\phi}$$ and $$L'^{\phi}$$ are contained in $$V^{\phi}$$ is immediate; all points of $$L$$ and $$L'$$ are points of $$V$$, so because $$\phi$$ is a function, all points of $$L^{\phi}$$ and $$L'^{\phi}$$ are points of $$V^{\phi}$$.
To see that $$L^{\phi}$$ and $$L'^{\phi}$$ are parallel, suppose towards a contradiction that $$L^{\phi}$$ and $$L'^{\phi}$$ are not parallel. Then they meet in a point, say $$p$$. Because $$\phi$$ is a permutation of the set of points, there exists a point $$q$$ such that $$p=q^{\phi}$$. Because $$q^{\phi}$$ is contained in both $$L^{\phi}$$ and $$L'^{\phi}$$ and $$\phi$$ is a permutation, applying $$\phi^{-1}$$ shows that $$q$$ is contained in both $$L$$ and $$L'$$. This contradicts the fact that $$L$$ and $$L'$$ are parallel.
An automorphism is a permutation which is by definition a bijection from $$\mathcal{P}$$ to $$\mathcal{P}$$. Let $$\phi$$ be an automorphism and $$L,L'$$ parallel lines. Suppose $$\phi(L)$$ and $$\phi(L')$$ intersect in $$y$$, then $$\phi^{-1}(y)\in L,L'$$ from which it follows that $$L$$ and $$L'$$ intersect and thus are not parallel.