Problem Statement
Let $n \geq 3$. Prove that $$R_n = \langle r \rangle = \{I,r,r^{2},...,r^{n-1}\} \subset D_n$$ Is a normal subgroup.
Now I know there is a fairly well known fact that any subgroups of index $2$ are normal and in this case we have $\lvert D_n \rvert = 2n$ and $\lvert R_n \rvert = n$ so from Lagrange's Theorem we have $$\frac{\lvert D_n \rvert}{\lvert R_n \rvert} = [D_n : R_n] = 2$$ However this problem is from a section which precedes even the definition of the index of a subset, let alone Lagranges Theorem or the other referenced fact. So I suspect they want us to solve this problem using some other means. I've written up a draft of a proof that I would like to be critiqued. I think it makes sense however I often miss small details and tend not to write the most streamlined proofs as I'm still a relatively nascent proof-writer.
Relations between elements given for $D_n$
Below are $3$ relations between the elements of $D_n$ given to us during the section. Let $r^k = r_{\frac{2\pi k}{n}}$ be a rotation by $\frac{2\pi k}{n}$ counterclockwise from the $x$-axis and let $r^{k}j = j_{\frac{\pi k}{n}}$ be a reflection through the line that makes an angle $\frac{\pi k}{n}$ with the $x$-axis (counterclockwise). Then we have:
- $r^k r^l = r^{k+l}$
- $r^k j = jr^{-k}$
- $r^k j r^l j = r^{k-l}$
Proof
The subgroup $R_n$ is normal iff $\forall a \in D_n$ we have $$aR_n a^{-1} \subset R_n$$ Thus we must show that $\forall r \in R_n$ and $\forall a \in D_n$ $$ar^l a^{-1} \in R_n,\ \text{where}\ 0\leq l \leq n-1$$ Since $R_n$ is the cyclic subgroup generated by $r$ we can say $r^l$ is an arbirtrary element of $R_n$. Now all elements of $D_n$ are generated by one of two elements:
- $n$ "rotations" are generated by $\langle r \rangle = \{r^k \mid 0\leq k \leq n-1 \} = R_n$
- $n$ "reflections" are generated by $\langle rj \rangle = \{r^k j \mid 0\leq k \leq n-1 \}$
So first let $a \in \langle r \rangle$. For $k = 0$, $r^0 = I$ the identity element for the group $D_n$ then we have: $$Ir^lI^{-1} = I(r^lI) = Ir^l = r^l \in R_n$$ For $1 \leq k \leq n-1$ we have $$r^kr^lr^{-k} = r^k(r^lr^{-k}) = r^kr^{l-k} = r^{k+l-k} = r^{k-k+l} = r^l \in R_n$$ By the first relation and the fact that addition in the integers is commutative and associative. This makes sense since $R_n$ is a subgroup and hence closed under products.
Now let $a \in \langle rj \rangle$. Then we have $$(r^kj)r^l(r^kj)^{-1} = (r^kj)r^l(jr^{-k}) = (r^kjr^lj)r^{-k} = (r^{k-l})r^k = r^{k-l-k} = r^{-l} \in R_n$$ By relations $2$, $3$ and $1$ and the fact that the operation in $D_n$ is associative as well as closure of inverses in $R_n$ since $R_n$ is a subgroup and hence a group. Thus $aR_na^{-1} \subset R_n$ and consequently $R_n \vartriangleleft D_n$ $\square$