# 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?

• When specifying a group, it is enough to specify generators and relations that the given generators satisfy. In writing $D_n = < r,s >$, you've told me that $D_n$ is generated by two things, but you haven't told me how those generators behave under the law of composition. A more complete description is $D_n = <r,s; r^2 = 1, rsr^{-1} = s^{-1}>$. Given these relations, you can convince yourself that the given set is indeed a correct representation of the dihedral group. – Aniruddh Agarwal Jun 3 at 23:31
• If you'd like to read about it, this is related to free groups, although this concept might be a little too abstract for you at this stage. – Aniruddh Agarwal Jun 3 at 23:31
• I think the main point is that $s^2 = e$ and $r^n = e$ – Mann Jun 3 at 23:33
• @AniruddhAgarwal Yes, sorry I forgot to mention the fact of $r^n = e$ and $s^2 = e$ and $srs = r^{-1}$. But it's the induction step that I'm having trouble with, if anyone can guide me a little for the induction step that'll be greatly appreciated. Thanks. – Skypanties Jun 3 at 23:43

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:

1. $$r^n = 1$$.
2. $$s^2=1$$.
3. $$sr = r^{n-1}s$$.

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 $$2m$$.

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 $$r^ks$$.

• The answer linked to is incorrect. It claims that the group will be the "internal direct product" of the subgroups. So you would be claiming here that $D_n\cong \langle s\rangle \times \langle r\rangle$, which is patently false. The linked-to answer misuses the term "direct product". For $G$ to be the internal direct product of subgroups $H$ and $K$, you must have $G=HK$, and $H\cap K=\{1\}$, and that $hk=kh$ for each $h\in H$ and $k\in K$. The only thing you check in that answer is $G=HK$, and the rest do not follow. – Arturo Magidin Jun 4 at 21:15
• @ArturoMagidin Yup, I commented there and I’ll edit this answer. The result seems right if you say “product of subgroups” and not “internal direct product” (as I mistakenly did). Obviously $D_n$ is not abelian. It seems as though the commutativity problem isn’t an issue for my answer here, however. It’s still true that $D_n = \langle s\rangle\langle r\rangle$ as a product of subgroups. – Santana Afton Jun 4 at 21:22
• The answer is then incomplete. While it does get you that every element of $D_n$ can be written as $s^ir^j$, it does not show that the expression is unique, and it does not give the desired answer, which is to express every element os $r^js^i$, $0\leq j\leq n-1$, $0\leq j\leq 1$, uniquely. – Arturo Magidin Jun 4 at 21:29
• @ArturoMagidin I’ve edited the answer — it should no longer be incomplete. The question simply asks to show that each element of $D_n = \{e,r,r^2,\dots,r^{n-1},s,rs,\dots,r^{n-1}s\}$ — proving that $D_n = \langle r\rangle\langle s\rangle$ achieves this. – Santana Afton Jun 4 at 21:49