This article shows that every subgroup of $$D_n = \langle r, s \rangle$$ is cyclic or dihedral.

Theorem 3.1. Every subgroup of $$D_n = \langle r, s \rangle$$ is cyclic or dihedral. A complete listing of the subgroups is as follows:

(1) $$\langle r^d \rangle$$, where $$d | n$$, with index $$2d$$;

(2) $$\langle r^d, r^{i}s \rangle$$, where $$d | n$$ and $$0 \le i \le d − 1$$, with index $$d$$.

Every subgroup of $$D_n$$ occurs exactly once in this listing.

The proof is as follows:

It is left to the reader to check $$n = 1$$ and $$n = 2$$ separately. We now assume $$n \ge 3$$.

Let $$H$$ be a subgroup of $$D_n$$. The composite homomorphism $$H \hookrightarrow D_n \to D_n/\langle r \rangle$$ to a group of order 2 is either trivial or onto. Its kernel is $$H \cap \langle r \rangle$$.

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Why is the kernel $$H \cap \langle r \rangle$$? I failed to prove this by showing that the kernel cannot contain reflections such as $$r^{i}s$$.

Let $$G$$ be a group and $$N\triangleleft G$$. Consider the natural homomorphism $$\pi:G\to G/N$$ given by $$g\mapsto gN$$.
Claim: $$\ker (\pi)=N$$.
Proof: Let $$n\in N$$. Then $$\pi (n)=nN=N$$. So $$N\subseteq \ker (\pi)$$. Let $$g\in \ker (\pi)$$. Then $$gN=N$$ or $$g\in N$$. So $$\ker (\pi)\subseteq N$$. Hence $$\ker (\pi)=N\;■$$
Let $$H$$ be a subgroup of $$G$$ and consider the restricted homomorphism $$\pi|_{H}:H\to G/N$$ given by $$h\mapsto hN$$. Clearly, $$\ker (\pi|_{H})=H\cap \ker (\pi)=H\cap N$$.
Note that $$\pi|_{H}$$ is same as $$H \overset {\iota}{\hookrightarrow} G \overset {\pi}{\to} G/N$$.
The kernel of the quotient map is $$\langle r \rangle$$ and the kernel of the overall map is the preimage of this under the inclusion map, which is $$H\cap \langle r \rangle$$.