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...)
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 $[n]$.)