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 $H \cap \langle r \rangle$.


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 $H\cap \langle r \rangle$.


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