Dihedral group generated by $\langle r,s\rangle$ for all $n$ Under wikipedia for Dihedral groups it claims the following:
The $2n$ elements in $D_n$ can be written as $\{e,r,r^2,r^3,\ldots,r^{n-1},s,rs,r^2s,\ldots,r^{n-1}s\}$.
I know why this is true and it intuitively make sense, but where is wikipedia coming up with this set with? From what I know I think it's from $D_n = \langle r,s\rangle$ that is, the dihedral group is generated by a rotation $r$ and a reflection $s$.
Question: How can I prove that the $D_n$ can be written as $\{e,r,r^2,r^3,\ldots,r^{n-1},s,rs,r^2s,\ldots,r^{n-1}s\}$?
So I tried to use induction on $n$, and proved the base case but got stuck on the induction step I had no clue how to use the induction hypothesis.
Also can someone confirm with me whether $\langle r,s\rangle=\{r^is^j:i,j \in \mathbb{Z}\}$ correct?
 A: In general, if you have $\langle r,s\rangle$ in a group, you get expressions of the form
$$r^{a_1}s^{b_1}r^{a_2}s^{b_2}\cdots r^{a_m}s^{b_m}$$
where $a_i\in\mathbb{Z}$, and all are nonzero except perhaps for $a_1$ and/or $b_m$. This because the subgroup generated by a set is the collection of all finite products of elements of the set and their inverses.
Now, in the special case of the dihedral group $D_n$, you have some other relations among $r$ and $s$ that allow you to “rewrite” a lot of these expressions. For example, the three most common defining relations among $r$ and $s$ in the dihedral group $D_n$ are:


*

*$r^n = 1$.

*$s^2=1$.

*$sr = r^{n-1}s$.


So, suppose you have an expression of the form 
$$r^{a_1}s^{b_1}r^{a_2}s^{b_2}\cdots r^{a_m}s^{b_m}.$$
The first thing to notice is that by adding multiples of $n$ to the $a_i$ you don’t change the value of the expression, since $r^n=1$. So, you may assume that each $a_i$ lies between $0$ and $n-1$, inclusively. Similarly, adding multiples of $2$ to the $b_j$ doesn’t change anything, so you may assume that any $b_j$ lies between $0$ and $1$, inclusively. This uses the relations 1 and 2 above.
As to the latter one, that tells you that you can “shuffle” the $s$’s all the way to the right, by replacing any instance of $sr$ with $r^{n-1}s$. You can show inductively that this means that $sr^k = r^{n-k}s$ for any integer $k$.
So, suppose you have an expression in $D_{12}$ given by
$$s^5r^{27}s^4r^2sr^{-3}s^2rsr^2$$
Then we can start by replacing the exponents: the $s^5$ can be replaced with $s^1$ (because $5\equiv 1\pmod{2}$; the $r^{-3}$ can be replaced by $r^{9}$, since $9\equiv -3\pmod{12}$; and the $r^{27}$ by $r^3$; etc. Then we simplify if some of the $r$s or $s$ “cancel”; and finally we can rewrite by shuffling $s$’s to the right, and using the fact that $s^2=1$, to rewrite this expression:
$$\begin{align*}
s^5r^{27}s^4r^2sr^{-3}s^2rsr^2 &= sr^31r^2sr^91rsr^2\\
&= sr^3r^2sr^9rsr^2\\
&= sr^5 sr^{10}sr^2 = (sr^5)sr^{10}sr^2\\
&= (r^{12-5}s)sr^{10}sr^2 = r^7s^2r^{10}sr^2\\
&= r^7r^{10}sr^2 = r^{17}sr^2 = r^5sr^2\\
&= r^5(sr^2) = r^5r^{12-2}s = r^5r^{10}s\\
&= r^{15}s = r^3s.
\end{align*}$$
You should establish, then, that in $D_n$, any expression as above can be re-written using the relations 1, 2, and 3, so that in the end you end up with a bunch of $r$s followed by a bunch of $s$s, and then by using 1 and 2, into an expression of the form
$$r^is^j,\qquad 0\leq i\leq n-1,\quad 0\leq j\leq 1.$$
This is called a normal form for the elements of $D_n$: every element, no matter how it is generated using $r$ and $s$, is equal to one and only one product of this form.
There are many ways of trying to establish this. You can do induction, but not on $n$: fix $n$, and work only on $D_n$, but with $n$ arbitrary. For example, you could do induction on $m$, where the product using powers of $r$ and $s$ has length $2m$.
A: First, note that $\langle r\rangle$ is a normal subgroup of $D_n$, and so we get a map
$$\pi: D_n \to D_n/\langle r\rangle.$$
By considering products of subgroups, we can write $D_n$ as
$$D_n = \pi^{-1}(D_n/\langle r\rangle) = \pi^{-1}(\pi(\langle s\rangle)) = \langle s\rangle\langle r\rangle = \langle r\rangle\langle s\rangle$$
Thus each element in $D_n$ takes the form $r^k$ or $r^ks$.
