Checking that $\cdot: D_{2n}\times \{1,2,\ldots,n\} \rightarrow\{1,2,\ldots,n\}$ given by $(\alpha , i) \mapsto\sigma_{\alpha}(i)$ is a group action My question is on an example of a group action given in Dummit and Foote's Abstract Algebra (example 4, p.43):

If we fix a labeling of the vertices of a regular $n$-gon, each element $\alpha$ of $D_{2n}$ gives rise to a permutation $\sigma_{\alpha}$ of $\{1,2,\ldots,n\}$ by the way the symmetry $\alpha$ permutes the corresponding vertices. The map of $D_{2n} \times \{1,2,\ldots,n\}$ onto $\{1,2,\ldots,n\}$ defined by $(\alpha,i) \rightarrow \sigma_{\alpha}(i)$ defines a group action of $D_{2n}$ on $\{1,2,\ldots,n\}...$

I'm just trying to verify that the map $(\alpha,i) \rightarrow \sigma_{\alpha}(i)$ is indeed a group action. If we let $1$ denote the identity element of $D_{2n}$, then clearly $\sigma_1$ should correspond to the identity permutation, so $\sigma_1(i) = i$ for all $i \in \{1,2,\ldots,n\}$. This confirms the first property of a group action. Then if we let $*$ denote the binary operation on $D_{2n}$ and $\cdot$ the group action, we just have to verify that for all $\alpha_1, \alpha_2 \in D_{2n}$ and all $i \in \{1,2,\ldots,n\}$,
$$ \alpha_1 \cdot (\alpha_2 \cdot i) = (\alpha_1 * \alpha_2) \cdot i.$$
Now
\begin{align*}
    \alpha_1 \cdot (\alpha_2 \cdot i) &= \sigma_{\alpha_1}(\sigma_{\alpha_2}(i)) && (\text{by defn of } \cdot) \\[4pt]
     &= (\sigma_{\alpha_1} \circ \sigma_{\alpha_2})(i) && (\text{defn of } \circ)
\end{align*}
And
\begin{align*}
   (\alpha_1 * \alpha_2) \cdot i = \sigma_{\alpha_1 * \alpha_2}(i). && (\text{Defn of } \cdot)
\end{align*}
Now am I allowed to assert that $\sigma_{\alpha_1 * \alpha_2} = \sigma_{\alpha_1} \circ \sigma_{\alpha_2}$, and thus $\sigma_{\alpha_1 * \alpha_2}(i) = (\sigma_{\alpha_1} \circ \sigma_{\alpha_2})(i)$?
If I had to justify this, I would say something like the following: Pick an $n$-gon and label its vertices; call this the "original" $n$-gon with the "original labeling" of the vertices. If we rotate or reflect the original $n$-gon, this corresponds to a permutation $\alpha_1$ of the original labeling of the vertices. Then if we apply another rotation/reflection to this newly rotated/reflected $n$-gon, the resulting symmetry (relative to the original $n$-gon) corresponds to the map $\alpha_2 \circ \alpha_1$ (relative to the original labeling of the vertices). Is my reasoning here correct?
Is my argument solid enough, or is there a more rigorous way of expressing this? Is there anything that I'm missing?
(I think my uncertainty stems partly from the fact that Dummit and Foote distinguish between the group $D_{2n}$ and the group of permutations to which the elements of $D_{2n}$ correspond, which I find a tad confusing. Perhaps I am just overthinking things...)
A follow-up question: 
The text goes on to say

Note that this action is faithful: distinct symmetries of a regular $n$-gon induce distinct permutations of the vertices.

Is this supposed to be accepted as "obvious" (I would be ok with that), or is there a rigorous algebraic way of showing this? i.e. if $\alpha_1, \alpha_2 \in D_{2n}$ and $\sigma_{\alpha_1} = \sigma_{\alpha_2}$, how would we show that this implies $\alpha_1 = \alpha_2$? Could we just say that $D_{2n}$ is isomorphic to the set of permutations to which the elements of $D_{2n}$ correspond (hence showing that book's claim)?
(Apologies if these are silly questions. Algebra is not my strong suit...)
 A: Here are some rows trying to make (more) clear the situation described in the question. First of all, we need a handy notation, and clear definition of the action.

Some words on the described action first. It involves $D_n$ in the notation from Dihedral group, wiki page, the dihedral group of all plane symmetries that invariate the set of the vertices of a fixed $n$-gon. And these vertices, labelled by $1,2,\dots,n$. Let us denote by $[n]$ this set of $n$ vertices. Then the given (to-be-)action is a map first:
$$
D_n\times [n]\to [n]\ .
$$
If this is (soon) an action, then it is an action of the group $D_n$ on the set $[n]$. And there is no need to consider permutations of $[n]$ (and/or the group of permutation of $[n]$ and/or the image of $D_n$ inside this group of permutations).
Maybe the example in loc. cit. wants to do something with the permutation $\sigma_\alpha$ associated to some symmetry $\alpha$ of the plane containing the vertices of the $n$-gon. But for the question, we just do not need this notational complication.
The map above is defined by $(\alpha,i)\to\alpha\cdot i:=\alpha(i)$.
Here, $\alpha(i)$ is the result of applying the symmetry (= isometric transformation, which is in particular a function) $\alpha$ on the vertex $i$.
Let us check, that this leads to an action. For the identic map id (the neutral element in $D_n$), and for symmetries $\alpha,\beta$ and some vertex $i$ we have:
$$
\begin{aligned}
\operatorname{id}\cdot i
&=
\operatorname{id}(i)=i\ ,
\\[3mm]
(\alpha\circ\beta)\cdot i
&=(\alpha\circ\beta)(i)\\
&=\alpha(\beta(i))\\
&=\alpha\cdot(\beta(i))\\
&=\alpha\cdot(\beta\cdot i)\ .
\end{aligned}
$$
So we have indeed an action, and the above is all regarding a proof.

The action is faithful for $n>1$. (For $n\ge 3$ this follows from the fact that an affine plane is "determined by a triple of points", i.e. by fixing an origin and two unit vectors from the origin, a first one and a second one. For $n=2$ - just in case we really want to consider a "$2$-gon", the reflection w.r.t. the line through the two points is fixing pointwise the two vertices, so the action is not faithful. The case $n=1$ is also not coming with a faithful action.)

In case we really want to work with permutations, then the group of permutations acting on $[n]$ is the image $\sigma(D_n)$ of $D_n$ inside the permutation group of the set $[n]$ via the map $\alpha\to\sigma_\alpha$, and this action is defined by transport of structure, i.e. $\sigma_\alpha\cdot i :=\alpha\cdot i:=\alpha(i)$. (By transport, it is also an action on the set $[n]$.)
