## Subgroups generated by powers of r are normal in Dih(2n)

Let $D_{2n} = \langle r,s \ |\ r^n = s^2 = 1, rs = sr^{-1} \rangle$ be the usual presentation of the dihedral group of order $2n$ and let $k$ be a positive integer dividing $n$.

1. Prove that $\langle r^k \rangle$ is normal in $D_{2n}$.
2. Prove that $D_{2n}/\langle r^k \rangle \cong D_{2k}$.

1. By a previous exercise it suffices to note that $rr^kr^{-1} = r^k \in \langle r^k \rangle$ and $sr^ks^{-1} = r^{-k} = (r^k)^{-1} \in \langle r^k \rangle$.
2. Define a mapping $\overline{\varphi} : \{ r,s \} \rightarrow D_{2n}/\langle r^k \rangle$ by $r \mapsto r \langle r^k \rangle$ and $s \mapsto s \langle s \langle r^k \rangle$. Note the following.
1. Note that $1 \leq k \leq n$. Now $(r \langle r^k \rangle)^a = r^a \langle r^k \rangle = \langle r^k \rangle$ if and only if $r^a \in \langle r^k \rangle$. The least integer $a$ such that $(r \langle r^k \rangle)^a = \langle r^k \rangle$ is thus $k$.
2. $\overline{\varphi}(s)^2 = s^2 \langle r^k \rangle = \langle r^k \rangle = \overline{1}$, so that $|\overline{\varphi}(s)| = 2$.
3. $\overline{\varphi}(r) \overline{\varphi}(s) = r \langle r^k \rangle \cdot s \langle r^k \rangle = (rs) \langle r^k \rangle = sr^{-1} \langle r^k \rangle$ $= s \langle r^k \rangle \cdot (r \langle r^k \rangle)^{-1} = \overline{\varphi}(s) \overline{\varphi}(r)^{-1}$.
4. Suppose by way of contradiction that $\overline{\varphi}(s) \in \langle \overline{\varphi}(r) \rangle$; then we have $s = r^m$ for some integer $m$, a contradiction.

By a lemma to a previous exercise, $\overline{\varphi}$ induces an injective group homomorphism $\varphi : D_{2k} \rightarrow D_{2n}/\langle r^k \rangle$. Moreover, $\varphi$ is surjective because $(s^ar^b) \langle r^k \rangle = \varphi(s^ar^b)$.