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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 $V^{\phi}$.

Thanks in advance.

MyWorld
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2 Answers2

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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.

Floris Claassens
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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.

Servaes
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