I have just started reading Abstract Algebra by Dummit and Foote and I have been stuck on their explanation of dihedral groups.
On page 23 Dihedral Groups there are 3 paragraphs(one paragraph is at beginning of page 24). There the author tries to explain the rotation via permutation of the set of numbers that are assigned to vertices. After that he explains the permutation of rotating the $n$-gon by $1$ step clockwise and says that the permutation is just $1$.
Then each symmetry $s$ can be described uniquely by the corresponding permutation $\sigma$ of $\{1,2,3,…,n\}$ where if the symmetry $s$ puts vertex $i$ in the place where vertex $j$ was originally, then $\sigma$ is the permutation sending $i$ to $j$. For instance, if $s$ is a rotation of $2\pi/n$ radians clockwise about the center of the $n$-gon, then $\sigma$ is the permutation sending $i$ to $i+1$, $1 \le i \le n-1$, and $\sigma(n) = 1$.
Can anyone check out those 3 paragraphs and tell me how to understand the author's explanation of dihedral groups?
Best Answer
A permutation of a set is not that same set but rearranged. Instead it is the act of rearranging; it is a function from that set to itself. In this example the set is $\{1,\ldots,n\}$ and the permutation is the function $$\sigma:\ \{1,\ldots,n\}\ \longrightarrow\ \{1,\ldots,n\}:\ i\ \longmapsto\ \begin{cases}i+1&\text{ if }i\neq n\\1&\text{ if }i=n\end{cases}.$$ That is, te particular permutation described here maps each number to the next number, and the last number to the first. When applied repeatedly, it sends the numbers around in a circle. After $n$ times, we are back to where we started.
This is precisely what happens to the vertices of a regular $n$-gon when rotating by $\tfrac{2\pi}{n}$ degrees, which is what the included pictures intends to illustrate; you can visualize this permutation by imagining this rotation of a regular $n$-gon.