I'm trying to understand a paragraph from Keith Conrad's expository paper on the dihedral group. The paragraph is:
Pick two adjacent vertices of a regular $n$-gon, and call them $A$ and $B$ as in the figure below. An element $g$ of $D_n$ is a rigid motion taking the $n$-gon back to itself, and it must carry vertices to vertices (how are vertices unlike other points in terms of distance relationships with all points in the polygon?) and $g$ must preserve adjacency of vertices, so $g(A)$ and $g(B)$ are adjacent vertices of the polygon.
The point I've bolded is what is causing me confusion. I can't figure out how vertices differ from other points. Is the idea that a symmetry has to send vertices to vertices, and adjacent vertices have to be mapped to adjacent vertices? I'm struggling to answer this in a way that ties back to the definition of a rigid motion.
Can someone help me to understand this?
[We now have a definitive answer in the form of @kcd's comment below, so there is no need to read this answer. It was also more elaborate than it needed to be, because I misinterpreted the term 'polygon' to mean what is more accurately termed a 'solid polygon'. This usage does have some currency, at least according to Wikipedia: "In contexts where one is concerned only with simple and solid polygons, a polygon may refer only to a simple polygon or to a solid polygon."]
Although I don't know what Conrad has in mind, one possible definition is this: a vertex of a set $P$ - in a Euclidean plane $E,$ say - is a point $v \in P$ that does not lie in any "open line segment" with endpoints in $P.$
This property can be defined in terms of the distance function $d \colon E \times E \to E,$ $(x, y) \mapsto \|x - y\|_2$ - itself usually defined in terms of the Euclidean norm function $\|\cdot\|_2$ - as follows.
If $x$ and $y$ are points of $E,$ the open line segment $]x, y[$ - this is a temporary, ad hoc technical term, with an equally temporary notation - is the set of all points $z \in E \smallsetminus\{x, y\}$ such that $d(x, z) + d(z, y) = d(x, y).$
Thus, a vertex of $P$ is defined as a point $v \in P$ such that there do not exist distinct points $x, y \in P \smallsetminus \{v\}$ such that $d(x, v) + d(v, y) = d(x, y).$
By definition (see the footnote on page 1 of Conrad's paper), a rigid motion is a function $T \colon E \to E$ - it is necessarily a bijection, whose inverse is also a rigid motion - such that $d(Tx, Ty) = d(x, y)$ for all $x, y \in E.$
Lemma. If $T$ is a rigid motion of $E,$ and $P$ is a subset of $E,$ then for all $v \in E,$ the point $v$ is a vertex of the set $P$ if and only if $Tv$ is a vertex of the image set $TP = \{Tx : x \in P\}.$
Proof. Because $T$ is a bijection, it is enough to prove the "if" part. Equivalently: if $v$ is not a vertex of $P,$ then $Tv$ is not a vertex of $TP.$ Suppose, then, that $v$ is not a vertex of $P.$ By definition, there exist distinct points $x, y \in P,$ neither equal to $v,$ such that $d(x, v) + d(v, y) = d(x, y).$ Because $T$ is a bijection, the points $Tv, Tx, Ty$ of $TP$ are pairwise distinct. Also, because $T$ is a rigid motion, $$ d(Tx, Tv) + d(Tv, Ty) = d(x, v) + d(v, y) = d(x, y) = d(Tx, Ty). $$ Therefore, $Tv$ is not a vertex of $TP.$ $\square$
It follows that if $T$ is a rigid motion of $E$ that maps a set $P$ to itself, i.e., $TP = P,$ then a point $v$ is a vertex of $P$ if and only if $Tv$ is a vertex of $P.$ In other words, $T$ restricts to a permutation of the set of vertices of $P.$
In the case where $n$ is an integer $\geqslant 3,$ and $P$ is a regular $n$-gon, it is clear that the set of vertices of $P,$ when defined in this way, is what we would expect. (But it might be a little tedious to write out a strict proof.) If $s$ is the common length of all the sides of $P,$ then two vertices $v, w$ of $P$ are said to be adjacent if and only if $d(v, w) = s.$ It is clear that if $T$ maps $P$ to itself, then $v, w$ are adjacent if and only if $Tv, Tw$ are adjacent.